Ordinary Differential Equations: Methods and Applications

Ordinary Differential Equations Ordinary Differential Equations: Methods and Applications W. T. Ang and Y. S. Park U...

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Ordinary Differential Equations

Ordinary Differential Equations: Methods and Applications

W. T. Ang and Y. S. Park

Universal Publishers Boca Raton

Ordinary Differential Equations: Methods and Applications Copyright © 2008 W. T. Ang and Y. S. Park All rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without written permission from the publisher Universal Publishers Boca Raton, Florida • USA 2008 ISBN-10: 1-59942-975-6/ISBN-13: 978-1-59942-975-5 (paper) ISBN-10: 1-59942-974-8/ISBN-13: 978-1-59942-974-8 (ebook) www.universal-publishers.com

Library of Congress Cataloging-in-Publication Data Ang, W. T., 1961Ordinary differential equations : methods and applications / W.T. Ang and Y.S. Park. p. cm. Includes bibliographical references and index. ISBN 978-1-59942-975-5 (pbk. : alk. paper) 1. Differential equations. I. Park, Y. S., 1964- II. Title. QA372.A598 2008 515'.352--dc22 2008026023

To our parents “Everywhere, we learn from those whom we love” Johann Wolfgang von Goethe

Preface This introductory course in ordinary diﬀerential equations is intended for junior undergraduate students in applied mathematics, science and engineering. It focuses on methods of solutions and applications rather than theoretical analyses. Applications drawn mainly from dynamics, population biology and electric circuit theory are used to show how ordinary diﬀerential equations appear in the formulation of problems in science and engineering. The calculus required to comprehend this course is rather elementary, involving diﬀerentiation, integration and power series representation of only real functions of one variable. A basic knowledge of complex numbers and their arithmetic is also assumed, so that elementary complex functions which can be used for working out easily the general solutions of certain ordinary diﬀerential equations can be introduced. The pre-requisites just mentioned aside, the course is mainly self-contained. The course comprises six chapters. Chapter 1 gives the basic concepts of ordinary diﬀerential equations, explaining what an ordinary diﬀerential equation is and what is involved in solving such an equation. It also illustrates how ordinary diﬀerential equations can be derived from physical laws or basic principles for two specific examples of problems. In Chapter 2, methods of solution are given for some first order ordinary diﬀerential equations. The equations studied include those which can be written in separable form, those which are linear, and the nonlinear Bernoulli diﬀerential equation. Mathematical models which describe population growth are given as examples of applications involving first order ordinary diﬀerential equations. v

In Chapter 3, the mathematical theory for constructing general solutions of second order linear ordinary diﬀerential equations is studied. It is applied to obtain general solutions of second order linear ordinary diﬀerential equations with constant coeﬃcients and the Euler-Cauchy equations. Also discussed is the extension of the theory to higher order linear ordinary diﬀerential equations. Chapter 4 shows how linear ordinary diﬀerential equations with constant coeﬃcients arise in the formulation of problems involving electric circuits and spring-mass systems. Specific examples of problems are solved. Chapter 5 introduces the power series method and the Frobenius method for deriving series solutions of rather general homogeneous second order linear ordinary diﬀerential equations. The methods studied can be applied to solve some well known ordinary diﬀerential equations in mathematical physics, such as the Legendre’s equation and the Bessel’s equation, giving rise to particular special functions, but those equations and the associated special functions are not examined in this course. Chapter 6 describes some simple numerical methods for solving first and second order ordinary diﬀerential equations. For a particular example of applications, the second order nonlinear ordinary diﬀerential equation which governs the motion of a swinging pendulum is solved numerically. Exercises are set not only to test the understanding of students but sometimes also to impart additional insights into the materials studied. Suggested solutions to all the exercises are given at the end of the chapters. To promote the use of this course for self-study, the solutions provided are by and large complete with details. W. T. Ang and Y. S. Park, Singapore, 2008

vi

Contents 1 Basic concepts 1.1 What is an ODE? . . . . . . . 1.2 Solving an ODE . . . . . . . . . . 1.2.1 General solution . . . . . . 1.2.2 Particular solution . . . . . 1.2.3 Exact solution . . . . . . . 1.3 Exercise I . . . . . . . . . . . . . . 1.4 Why study ODEs? . . . . . . . . . 1.4.1 ODE for a body in motion 1.4.2 ODE for a pursuit problem 1.5 Exercise II . . . . . . . . . . . . . . 1.6 Solutions to Exercise I . . . . . . . 1.7 Solutions to Exercise II . . . . . .

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10 10 11 12 12 12 13 14 14 18 20 22 24

2 First order ODEs 2.1 Preamble . . . . . . . . . . . . . . . . . . . . . 2.2 First order ODEs in separable form . . . . . . . 2.3 Linear 1st order ODEs . . . . . . . . . . . . . . 2.3.1 Homogeneous linear 1st order ODEs . . 2.3.2 Nonhomogeneous linear 1st order ODEs 2.4 Bernoulli diﬀerential equation . . . . . . . . . . 2.5 Population dynamics . . . . . . . . . . . . . . . 2.5.1 Malthus theory of unlimited growth . . 2.5.2 Verhulst theory of limited growth . . . . 2.6 Exercise III . . . . . . . . . . . . . . . . . . . . 2.7 Solutions to Exercise III . . . . . . . . . . . . .

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27 27 27 33 34 34 36 38 38 39 41 43

vii

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3 Second order linear ODEs 3.1 Preamble . . . . . . . . . . . . . . . . . . . . . . 3.2 General solution of homogeneous 2nd order linear ODE . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Linearly independent functions . . . . . . 3.2.2 Construction of general solution . . . . . 3.3 Homogeneous 2nd order linear ODEs with constant coeﬃcients . . . . . . . . . . . . . . . . . . 3.4 Euler-Cauchy equations . . . . . . . . . . . . . . 3.5 Exercise IV . . . . . . . . . . . . . . . . . . . . . 3.6 Solving nonhomogeneous ODEs . . . . . . . . . . 3.6.1 Finding a particular solution by guesswork 3.6.2 Method of variation of parameters . . . . 3.7 Extension to higher order linear ODEs . . . . . . 3.7.1 General N -th order linear ODEs . . . . . 3.7.2 General solution of a homogeneous ODE . 3.7.3 General solution of a nonhomogeneous linear ODE . . . . . . . . . . . . . . . . . . 3.8 Exercise V . . . . . . . . . . . . . . . . . . . . . . 3.9 Solutions to Exercise IV . . . . . . . . . . . . . . 3.10 Solutions to Exercise V . . . . . . . . . . . . . . 4 Circuits and springs 4.1 Preamble . . . . . . . . . . . . . . . . . . . . . 4.2 Electric circuits . . . . . . . . . . . . . . . . . . 4.2.1 Basic electrical components . . . . . . . 4.2.2 Voltage across an electric component . . 4.2.3 ODEs in electric circuits . . . . . . . . . 4.3 Exercise VI . . . . . . . . . . . . . . . . . . . . 4.4 Spring-mass systems . . . . . . . . . . . . . . . 4.4.1 A simple spring-mass system . . . . . . 4.4.2 A more complicated spring-mass system 4.5 Exercise VII . . . . . . . . . . . . . . . . . . . . 4.6 Solutions to Exercise VI . . . . . . . . . . . . . 4.7 Solutions to Exercise VII . . . . . . . . . . . .

viii

. . . . . . . . . . . .

49 49 50 50 52 55 64 70 71 72 78 81 81 81 82 83 85 89 96 96 96 96 98 99 109 109 109 115 118 119 124

5 Series solutions 5.1 Preamble . . . . . . . . . . . . 5.2 Review of power series . . . . 5.3 Power series method for ODEs 5.4 Exercise VIII . . . . . . . . . . 5.5 Frobenius method . . . . . . . 5.6 Exercise IX . . . . . . . . . . . 5.7 Solutions to Exercise VIII . . . 5.8 Solutions to Exercise IX . . . .

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128 . 128 . 128 . 131 . 146 . 147 . 161 . 162 . 167

6 Numerical methods 176 6.1 Preamble . . . . . . . . . . . . . . . . . . . . . . 176 6.2 Euler’s method for 1st order ODEs . . . . . . . . 176 6.3 Second order ODEs . . . . . . . . . . . . . . . . . 183 6.4 Oscillation of a pendulum . . . . . . . . . . . . . 187 6.4.1 Nonlinear ODE . . . . . . . . . . . . . . . 187 6.4.2 ODE for ‘very small’ oscillation . . . . . . 188 6.4.3 Numerical solution for ‘larger’ oscillation 189 6.5 Numerical prudence . . . . . . . . . . . . . . . . 191 6.6 Exercise X . . . . . . . . . . . . . . . . . . . . . . 191 6.7 Solutions to Exercise X . . . . . . . . . . . . . . 194

ix

Chapter 1

Basic concepts 1.1

What is an ODE?

An equation which contains the derivative(s) of a yet to be determined function y(x) (a function of one variable) is called an ordinary diﬀerential equation (ODE) in y(x). Below are some examples of ODEs in y(x): (1) (2) (3) (4)

dy − 2x = 0 dx dy d2 y − 2y(x) = x5 +3 2 dx dx d3 y d2 y x3 3 + 6x2 2 − 3xy(x) = sin(2x) dx dx 4y d d2 y 2x2 (y(x))10 4 + 3x 2 = xy(x) dx dx

An ODE in y(x) is said to be of order N if dN y/dxN is the highest order derivative of y(x) present in the ODE. In the examples above, (1) is an ODE of order 1 (or 1st order ODE); (2) is of order 2; (3) is of order 3; and (4) is of order 4. It may be sometimes convenient to use the notation y0 (x) =

dy 00 d2 y d3 y dN y , y (x) = 2 , y 000 (x) = 3 , · · · , y (N) (x) = N . dx dx dx dx 10

Thus, we may write the 4th order ODE in (4) above as 2x2 (y(x))10 y 0000 (x) + 3xy 00 (x) = xy(x), or even more simply as 2x2 y 10 y 0000 + 3xy 00 = xy if it is already understood that y is a function of x. Note that y 10 above refers to y raised to the power 10, not to be mistaken as the 10th order derivative y(10) = d10 y/dx10 . More generally, we may regard an ODE in y(x) as an equation of the general form F (x, y, y 0 , y00 , y 000 , · · · , y (N−1) , y (N) ) = 0. Here F denotes a mathematical expression involving x, y, y 0 , y00 , y 000 , · · · , y (N−1) and y (N) .

1.2

Solving an ODE

Given an ODE in y(x), we are interested in finding functions y(x) that satisfy the equation, that is, we are interested in solving the ODE. For the purpose of illustration, let us now consider solving the ODE y 00 − 6x = 0 which may be rewritten as d2 y = 6x. dx2 We may solve the ODE above by directly integrating it twice with respect to x, that is, Z dy = 6xdx = 3x2 + C dx Z ⇒ y = (3x2 + C)dx = x3 + Cx + D. 11

Here C and D are constants which may have any values (arbitrary constants). Thus, we find that y(x) = x3 + Cx + D (where C and D are arbitrary constants) satisfies the ODE y00 − 6x = 0. Not all ODEs may be easily solved by direct integration like in the example above. In subsequent chapters, we will look at various methods for solving ODEs.

1.2.1

General solution

From the example above, it appears that solving an ODE is really “undoing the derivatives” in the ODE. Roughly speaking, if an ODE is of order N, we are required to integrate it N times in order to solve it. Consequently, N arbitrary constants appear in the solution obtained. For our purpose, we regard any function y(x) with N arbitrary constants in it as a general solution of an N -th order ODE, if the function satisfies the ODE. Thus, we may regard y(x) = x3 + Cx + D as a general solution of the ODE y 00 − 6x = 0.

1.2.2

Particular solution

If some or all of the arbitrary constants in a general solution of an ODE assume specific values, we obtain a particular solution of the ODE. Examples of particular solutions of the ODE y 00 − 6x = 0 are y(x) = x3 + x + D, y(x) = x3 + Cx − 2 and y(x) = x3 − 3x.

1.2.3

Exact solution

We regard a solution of an ODE as exact if the solution can be directly expressed in terms of elementary functions1 . Thus, the general solution y(x) = x3 + Cx + D of the ODE 00 y − 6x = 0 is exact. If the values of C and D are given, we 1

For our purpose, a function is regarded as elementary if it can be calculated directly using the function keys of an ordinary scientific hand calculator. Thus, cos(x), sin(x), exp(x) and g(x) = x3 − 1 are elementary.

12

may readily evaluate y(x) = x3 + Cx + D for any value of x by using an ordinary hand calculator. We may not be able to find exact solutions of some ODEs. For example, take the ODE dy sin(x) = dx x whose solution is theoretically given by Z sin(x) dx. y= x It is, however, not possible to express the integral (on the right hand side) in terms of elementary functions. Thus, the ODE does not have an exact solution.

1.3

Exercise I

1. Check by direct substitution whether y(x) = ex + 2e−x is a solution of each of the following ODEs in y(x) or not. (Substitute y(x) = ex + 2e−x into the left hand side of a given ODE and simplify to see if it is possible to obtain the right hand side.) (a) (b) (c) (d)

y 00 − y = 0

y 000 + 3y 00 − 2y0 = 2ex − 4e−x

2y00 + 2y 0 − 3y = e2x + 5e−x y 000 + 3y 00 − y0 − 3y = 0

2. Find out whether each of the following functions is a solution of the ODE p0000 (x) − 5p00 (x) − 36p(x) = 0 or not. (a)

p(x) = 2 sinh(3x)

(b)

p(x) = cosh(2x)

(c)

p(x) = cos(3x) − sin(3x)

(d)

p(x) = 2 sin(2x) + 3 cos(2x)

13

3. If y(x) = α sin(2x) + β cos(2x) is a solution of the ODE y 00 + 4y 0 + 3y = 3 sin(2x), find the constants α and β. (Hint. Substitute y(x) = α sin(2x) + β cos(2x) into the ODE and choose the constants α and β in such a way that the equation is satisfied for all x.) 4. If y (x) = α + βx + γx2 is a solution of the ODE x3 y 000 + x2 y 00 − y = x2 + 2x + 1, find the constants α, β and γ. 5. Each of the following ODEs in y(x), if rewritten in an appropriate form, can be solved by direct integration. Find general solutions of these ODEs. √ 0 (a) xy = x6 + 5x (b)

y 0 ex = 5

(c)

y 00 + 6x2 − 2 = 0

(d)

xy 0 + y = 4x3 + 2x (Hint.

d dy (xy) = x + y) dx dx

6. Verify that y(x) =

s

5 2 cos(x) + 4 sin(x) + 10e2x

is a solution of the ODE y0 + y − y 3 sin(x) = 0. (Hint. Diﬀerentiate the solution with respect to x to obtain y 0 = (sin(x) − 2 cos(x) − 10e2x )y 3 /5. You may do this by squaring both sides of the solution first.)

1.4

Why study ODEs?

Some problems in science and engineering may be formulated in terms of ODEs. Below are two examples of such problems. More examples will be given in later chapters.

1.4.1

ODE for a body in motion

Consider the motion of a body which is dropped at some height above the ground. 14

We assume that the body, which is pulled by gravity towards the earth, moves along a straight line which is perpendicular to the surface of the earth (that is, it moves along a vertical path). In addition to gravity, the body is also acted upon by a force due to air resistance. A sketch of the situation is given in Figure 1.1. Let s(t) be the vertical downward displacement (in meter) of the body from a fixed point P (which may lie on the path of motion)2 . We are interested in finding s(t), that is, the position of the body at time t.

Figure 1.1 All motion obeys Newton’s law which may be stated as follows. If the mass of a moving body is constant, then the total force (in newton) acting on the body is equal to the product of the mass (in kilogram) and the 2

We define s(t) to be the downward displacement of the object from P at time t. This means that s(t) > 0 if the object is below P at time t, and s(t) < 0 if it is above P. It is, of course, possible to choose P in such a way that s(t) > 0 at all time t during which the object is falling.

15

acceleration (in meter per second per second) of the body. “Force” and “acceleration” are vectors. In simple terms, a vector is a quantity which has a magnitude and a direction. Thus, Newton’s law as stated above implies that the force and the acceleration are in the same direction. For the situation depicted in Figure 1.1, Newton’s law is simply: (total downward force acting on body) = (mass of body) × (downward acceleration of body). As mentioned above, we assume that there are only two types of forces acting on the body, namely gravity and air resistance. The gravitational force has a magnitude given by mg, where m is the mass of the body and g is the acceleration due to gravity (that is, g ' 9.81 meter per second per second near planet earth). The gravitational force acts downward. The magnitude of the force due to air resistance is taken to be given by k|s0 (t)|. Here k is a positive coeﬃcient which depends on, among other things, the shape of the body. The derivative s0 (t) is the downward velocity3 of the body. In this case, s0 (t) is always greater than zero at any time t during the motion as the body is always moving towards the ground. The force due to air resistance acts upward against the motion of the body. It is obvious that (total downward force on body)

= mg − k|s0 (t)| = mg − ks0 (t)

(since s0 (t) > 0).

The downward acceleration of the body is given by s00 (t). Newton’s law now becomes mg − ks0 (t) = ms00 (t). “Downward velocity” implies that s0 (t) > 0 if the object is moving downward, and s0 (t) < 0 if it is moving upward. 3

16

The above equation is an ODE in s(t) which may be rewritten as s00 (t) +

k 0 s (t) − g = 0. m

If we can solve the ODE for s(t), we can predict the position of the body at any time t during which it is moving towards the earth. For the case in which k is a non-zero constant, the ODE may be solved using methods of solution in later chapters (see Problem 3 in Exercise V on page 84). For now, let us consider the special case in which the eﬀect of air resistance on the motion is so small that it may be ignored, that is, we take k/m in the ODE to be zero. For such a case, the ODE which describes Newton’s law of motion reduces to s00 (t) − g = 0 or s00 (t) = g. This simpler ODE may be solved by directly integrating it twice with respect to t. Bearing in mind that g is a constant, we obtain Z ds = gdt = gt + C dt Z 1 ⇒ s(t) = (gt + C)dt = gt2 + Ct + D. 2 Here C and D are arbitrary constants. To calculate the position of the body, we need to know the values of the constants C and D. The values of these constants may be determined if the displacement and velocity of the body are known at a certain time. For example, if we are told that s0 (t) = 3 (meter per second) when t = 0 (second) then 3 = g · 0 + C ⇒ C = 3. Furthermore, if s(t) = 2 (meter) at t = 0, we obtain 1 2 = g · 02 + 3 · 0 + D ⇒ D = 2. 2 17

Thus, if there is no air resistance, and if the displacement and velocity at t = 0 are 2 and 3 (in the appropriate units) respectively, then the displacement of the body is given by 1 s(t) = gt2 + 3t + 2 2 at any time t during which the body is falling towards the ground.

1.4.2

ODE for a pursuit problem

With reference to a Cartesian coordinate system denoted by Oxy, consider the following pursuit problem. At time t = 0, a navy ship is at (a, 0), where a > 0, and it spots a small boat at (0, 0). The boat moves with a constant speed u along the positive y axis. The navy ship decides to chase after the boat by moving at a constant speed v and in a direction that is always directed at the boat. We are interested in finding the path of the navy ship.

Figure 1.2 A geometrical sketch of the problem is given in Figure 1.2. Let the path be given by the curve y = p(x), where p(x) is an 18

unknown function yet to be determined. The position of the boat at time t is B (0, ut) (along the y axis). At time t, the navy ship is located at N (x, p(x)). Now the line NB (dotted line in Figure 1.2) is tangential to the path y = p(x) at N because the navy ship is always directed at the boat. Thus, the gradient of N B is equal to the gradient of the tangent to the curve y = p(x) at the point N (x, p(x)), that is, p(x) − ut x ⇒ xp0 (x) = p(x) − ut. p0 (x) =

Since the ship is moving, x and y are functions of t. Let us diﬀerentiate the above equation once with respect to t. Using the chain rule for diﬀerentiation, we obtain xp00 (x)

dx dx dx + p0 (x) = p0 (x) −u dt dt dt

which simplifies to xp00 (x)

dx = −u. dt

Note that dx/dt and dy/dt are the x and y components of the velocity of the navy ship. Since its speed is v, we may write r dx dy ( )2 + ( )2 = v. dt dt From y = p(x) and the chain rule for diﬀerentiation, we find that dy dx = p0 (x) . dt dt Thus, r

dx 2 ) (1 + (p0 (x))2 ) = v dt v dx = ±p . dt 1 + (p0 (x))2 (

⇒

19

From Figure 1.2, it is clear that dx/dt < 0 as the x-coordinate of the ship decreases during the pursuit of the boat. Thus, we take dx v . = −p dt 1 + (p0 (x))2 Substituting the above expression for dx/dt into the equation xp00 (x)dx/dt = −u, we obtain the 2nd order ODE in p(x) given by up xp00 (x) − 1 + (p0 (x))2 = 0 v If we can solve this ODE for p(x), we obtain a formula for the path taken by the navy ship when it chases after the small boat. The two arbitrary constants in the general solution of the ODE may be determined from the conditions p(a) = 0 and p0 (a) = 0. (Can you see how the conditions come about?) The ODE may be converted into a 1st order ODE and solved as explained in Chapter 2 (on page 30).

1.5

Exercise II

1. An extremely thin wire lies along the x-axis from x = 0 to x = ` (` is the length of the wire). The temperature in the wire varies from one point to the next along the wire, that us, it is a function of x. If it is denoted by T (x), then that under certain conditions T (x) satisfies the simple 2nd order ODE d2 T = 0. dx2 If temperature is given by T0 and T` at x = 0 and x = ` respectively, find the temperature throughout the whole wire. 2. The tangent to a curve at the point (x, y) has gradient given by x3 + 2x + 1. If the equation of the curve is given by y = p(x), write down an ODE in p(x). Find the curve given that (1, 2) is a point on it. 20

3. The tangent to a curve at the point (x, y) has gradient given by (x − y)/x. If the equation of the curve is given by y = p(x), write down an ODE in p(x). Find the curve given that (1, 1) is a point on it. (Hint. Use the formula d dp (xp(x)) = x + p(x) to help you to solve the ODE.) dx dx

4. A body is moving to the right along a horizontal line. Its speed at time t ≥ 0 is given by 8t3 + 3t2 + 1. If w(t) is the distance of the body from a fixed point P on the line, write down an ODE in w(t) for the motion of the body. Where is the body at time t = 2 given that it is a unit distance away from P at time t = 0? 5. A body of 1 kilogramme is moving to the right along a horizontal line. It is acted upon by a rightward force given by F (t) = 1 + e−2t (in newton) at time t (in second). If s(t) is the rightward displacement (in meter) of the body from a fixed point P on the line, use Newton’s law of motion to write down an ODE in s(t). Find the displacement of the body at time t = 1 given that it is at rest at the point P at time t = 0. 6. The uniform temperature of a body at time t is given by T (t). The rate of change of the body temperature per unit time is given by c(T (t) − Tambient ), where Tambient is the constant temperature of the surrounding atmosphere and c is a constant. Express the statement in the last sentence by writing down an ODE in T (t). What can you say about the constant c? (Hint. To find out something about c, ask whether the body should be gaining or losing heat to its surrounding when T (t) > Tambient . What about when T (t) < Tambient ?)

21

1.6

Solutions to Exercise I

1. From y = ex + 2e−x , we have y 0 = ex − 2e−x , y00 = ex + 2e−x = y and y 000 = ex − 2e−x = y0 . Thus: (a)

y 00 − y = 0 (from y 00 = y).

Yes. (y = ex + 2e−x is a solution of the ODE.) (b)

y 000 + 3y 00 − 2y0 = 2ex + 8e−x 6= 2ex − 4e−x .

No. (y = ex + 2e−x is not a solution of the ODE.) (c) (d)

2y00 + 2y 0 − 3y = ex − 6e−x

6= e2x + 5e−x . No.

y 000 + 3y 00 − y0 − 3y

= y 0 + 3y − y 0 − 3y = 0. Yes. 2. (a) p = 2 sinh(3x), p0 = 6 cosh(3x), p00 = 18 sinh(3x) = 9p, p000 = 9p0 and p0000 = 9p00 = 81p. Thus, p0000 − 5p00 − 36p = 81p − 45p − 36p = 0. Yes, p = 2 sinh(3x) is a solution of the given ODE. (b) p = cosh(2x), p0 = 2 sinh(2x), p00 = 4 cosh(2x) = 4p, p000 = 4p0 and p0000 = 16p. Thus, p0000 − 5p00 − 36p = 16p − 20p − 36p 6= 0. No, p = 2 cosh(2x) is not a solution of the given ODE. (c) p(x) = cos(3x) − sin(3x) is not a solution. (d) p(x) = 2 sin(2x) + 3 cos(2x) is a solution. 3. Substitute y = α sin(2x) + β cos(2x) into ODE, we obtain (−α − 8β) sin(2x) + (−β + 8α) cos(2x) = 3 sin(2x). Thus, for the ODE to be satisfied for all x, we require α + 8β = −3 and 8α − β = 0. This gives α = −3/65 and β = −24/65. 22

4. Substitute y = α + βx + γx2 into ODE to obtain −α − βx + γx2 = x2 + 2x + 1. This gives α = −1, β = −2 and γ = 1. 5. (a) Rewrite ODE as dy = x11/2 + 5x1/2 dx and integrate directly once to obtain y=

2 13/2 10 3/2 + x + C. x 13 3

(b) Rewrite ODE as y 0 = 5e−x and integrate to obtain y = −5e−x + C. (c) Rewrite ODE as d2 y = −6x2 + 2 dx2 and integrate twice to obtain 1 y = − x4 + x2 + Cx + D. 2 (d) Rewrite ODE as d (xy) = 4x3 + 2x. dx Integrate once to obtain xy

=

x4 + x2 + C

⇒ y = x3 + x + 23

C . x

6. It may be very tedious to proceed by directly diﬀerentiating s 5 y(x) = . 2 cos(x) + 4 sin(x) + 10e2x Instead, rewrite the above as y2

5 (2 cos(x) + 4 sin(x) + 10e2x ) ⇒ y 2 (2 cos(x) + 4 sin(x) + 10e2x ) = 5 =

and diﬀerentiate both sides with respect to x to obtain 5 ) + y2 (−2 sin(x) + 4 cos(x) + 20e2x ) = 0 y2 ⇒ 10y0 = −y3 (−2 sin(x) + 4 cos(x) + 20e2x ) 2yy 0 (

⇒ 10y0 + 10y − 10y3 sin(x) 5(2 sin(x) − 4 cos(x) − 20e2x ) = y[ + 10 2 cos(x) + 4 sin(x) + 10e2x −50 sin(x) + ] 2 cos(x) + 4 sin(x) + 10e2x = 0 ⇒ y 0 + y − y 3 sin(x) = 0 (as required).

1.7

Solutions to Exercise II

1. The conditions are T (0) = T0 and T (`) = T` . Solving the ODE, we find that d2 T dT =0⇒ = C ⇒ T = Cx + D. 2 dx dx Applying the conditions, we obtain T (0) = T0 = D T (`) = T` = C` + T0 ⇒ C = Thus, the required temperature is T (x) =

T` − T0 x + T0 . ` 24

T` − T0 . `

2. The ODE giving the tangent of the curve is dp = x3 + 2x + 1. dx Integrating, we obtain 1 p(x) = x4 + x2 + x + C. 4 From the point (1, 2), we know p(1) = 2. Thus, 1 1 +1+1+C =2⇒C =− . 4 4 The required curve is 1 1 y = x4 + x2 + x − . 4 4 3. The ODE is dp dx

=

x−p x

⇒ xp0 + p = x ⇒

d (xp) = x. dx

Integrating, we obtain 1 1 C xp = x2 + C ⇒ p = x + . 2 2 x From the point (1, 1), we have p(1) = 1, that is, C = −1/2. Thus, the required curve is 1 1 y = x− (for x > 0). 2 2x 4. The ODE is dw = 8t3 + 3t2 + 1 dt which can be integrated to obtain w = 2t4 + t3 + t + C. From w(0) = 1, we find that C = 1. Thus, w(2) = 42+1 = 43. 25

5. The ODE is d2 s = 1 + e−2t . dt2 Integrating, we find that 1 1 1 ds = t − e−2t + C ⇒ s = t2 + e−2t + Ct + D. dt 2 2 4 Now, the body is at rest at t = 0. This gives s0 (0) = 0, hence C = 1/2. Also, s(0) = 0. This gives D = −1/4. Thus, the rightward displacement at t = 1 is s(1) =

1 1 −2 1 1 3 1 + e + − = e−2 + . 2 4 2 4 4 4

6. We have dT = c(T (t) − Tambient ) dt If T (t) > Tambient , the body should be losing heat, that is, dT /dt < 0. This implies c < 0. Likewise, if T (t) < Tambient , the body should be gaining heat, that is, dT /dt > 0 and hence c < 0 (as deduced earlier). (Note. This ODE can be solved as a 1st order ODE in separable form as explained in Chapter 2.)

26

Chapter 2

First order ODEs 2.1

Preamble

We will consider 1st order ODEs in y(x) which are of the form dy = G(x, y) dx where G(x, y) denotes a given mathematical expression involving x and y. For cases in which G(x, y) assumes certain specific forms, methods for solving the ODEs may be obtained. We will look at some of these cases in this chapter.

2.2

First order ODEs in separable form

A 1st order ODE given by dy = P (x) · Q(y) dx is said to be in separable form. Here P (x) and Q(y) are given functions. The ODE may be rewritten as 1 dy · = P (x). Q(y) dx 27

If we perform an indefinite integration with respect to x on both sides, we obtain Z Z dy 1 · dx = P (x)dx. Q(y) dx This may be rewritten as Z Z dy = P (x)dx. Q(y) The formula above provides us with a method for solving the ODE, if we can work out the indefinite integrals on both sides of the equation. Note that the integration on the left hand side is carried out with respect to y and that on the right hand side with respect to x. Examples 1. Solve the ODE y 0 (x) = (y + 1)2 (2x + 1) subject to y(0) = 1. We are required to find the particular solution y(x) of the ODE which is such that y(0) = 1. The ODE is 1st order and in separable form. We rewrite it as Z Z dy = (2x + 1)dx. (y + 1)2 This leads to the general solution of the form −(y + 1)−1 = x2 + x + C where C is an arbitrary constant. The general solution as given above is said to be in implicit form1 . 1 If we can rearrange the implicit solution to express y explicitly in terms of x, we obtain a general solution in explicit form. An explicit solution (if it can be easily obtained) may be preferable to an implicit one as the former may be readily used to evaluate y at any suitably chosen value of x.

28

We are given y(0) = 1, that is, y = 1 when x = 0. Applying this in the general solution above, we find that 1 −(1 + 1)−1 = 02 + 0 + C ⇒ C = − . 2 Thus, the required particular solution in implicit form is 1 −(y + 1)−1 = x2 + x − . 2 Solving for y, we obtain the required particular solution in explicit form given by y=−

−1 − 2x2 − 2x . −2x2 − 2x + 1

2. Find the general solution of the ODE x3 (7y6 + 3y 2 )y0 (x) = (y 7 + y 3 + 3).

This 1st order ODE is in separable form. It may be rewritten as Z Z (7y 6 + 3y2 ) dy = x−3 dx. (y7 + y 3 + 3) Finding the integrals on both sides, we obtain the general solution 1 ln |y7 + y 3 + 3| = − x−2 + C 2 7 3 −1/(2x2 ) ⇒ |y + y + 3| = Be (B = eC ) 2)

⇒ y 7 + y 3 + 3 = Ae−1/(2x

(A = ±B)

where A, B and C are arbitrary constants. The constant B which is related to C by B = eC is positive. The constant A which is related to B by A = ±B can either be positive or negative. The general solution as given above is in implicit form. In this case, it is diﬃcult, if not impossible, to find the general solution in explicit form. 29

3. Consider now the ODE for the pursuit problem on page 18 (Chapter 1), that is, up 1 + (p0 (x))2 = 0. xp00 (x) − v Use the substitution q(x) = p0 (x) to convert it into a 1st order separable ODE. Hence, solve the ODE to determine p(x) for the pursuit problem.

With q(x) = p0 (x), the 2nd order ODE in p(x) becomes x which is separable. Thus,

dq up 1 + q2 = dx v

Z

Z u dx p = 2 v x 1+q Z u dq p = ln |x| + E. ⇒ 2 v 1+q dq

Note that u and v are positive constants giving the speed of the boat and the navy ship respectively and E is an arbitrary constant of integration. To deal with the integral on the left hand side, we use the substitution q = sinh θ and dq = cosh θ · dθ. Noting that 1 + sinh2 θ = cosh2 θ, we find that Z Z dq p = dθ = θ + D. 1 + q2 Note that D is an arbitrary constant of integration. It follows that θ=

u ln |x| + C v

where C = E − D is an arbitrary constant. From q = sinh θ, we find that 30

u q = sinh( ln |x| + C) v C (u/v) ln |x| − e−C e−(u/v) ln |x| ⇒ 2q = e e 1 1 ⇒ q = A|x|u/v − A−1 |x|−u/v 2 2 where A = eC is an arbitrary constant. In the pursuit problem, we note that 0 < x ≤ a when the navy ship is chasing the small boat. Thus, |x| = x and 1 1 −u/v p0 (x) = Axu/v − x 2 2A x1−γ u Axγ+1 − + B if γ = 6= 1, ⇒ p(x) = 2(γ + 1) 2A(1 − γ) v γ+1 Ax 1 u or p(x) = − ln(x) + B if γ = = 1, 2(γ + 1) 2A v where B is an arbitrary constant. To determine the arbitrary constants A and B in the general solution, we apply the conditions p(a) = 0 and p0 (a) = 0. Using p0 (a) = 0, we obtain Aau/v −

1 1 = 0 ⇒ A = ± u/v . u/v Aa a

From Figure 1.2 on page 18, it is clear that p0 (x) ≤ 0 for 0 < x ≤ a as the curve giving the path of the navy ship has a non-positive slope. If we take A = −1/au/v , we find that p0 (x) =

1 − (x/a)2u/v ≥ 0 for 0 < x ≤ a. 2(x/a)u/v

Thus, for the problem under consideration, we reject A = −1/au/v . We take A = 1/au/v . The value of B can then be determined from p(a) = 0. (We leave this to readers who are interested in working out the algebraic details.)

31

4. Certain 1st order ODEs may be reduced to separable form through the use of suitable substitutions. Consider 1st order ODEs of the form y dy = G( ), dx x where G(y/x) denotes a well defined expression containing y/x in it. (The variables x and y cannot appear in the expression separately but always together in the ratio y/x.) If y(x) = x · u(x), show that the ODE above can be reduced to the following 1st order separable ODE in u(x): du 1 = (G(u) − u). dx x If we diﬀerentiate the substitution y = x · u(x) with respect to x, we obtain dy = x · u0 (x) + u(x). dx Substituiting the above into the given ODE in y(x) and noting that u = y/x, we find that x · u0 (x) + u = G(u) ⇒

du 1 = (G(u) − u). dx x

5. Solve the ODE y0 =

x y + ( )3 x y

subject to y(1) = 1. We let y = x·u(x). This gives y 0 = x·u0 (x)+u. Substituting into the given ODE, we find that 1 du 1 = ( )3 . x · u0 (x) + u = u + ( )3 ⇒ x u dx u 32

We obtain a separable ODE in u(x) which may be rewritten as Z

u3 du =

Z

dx . x

The above can be integrated to give u4 = ln |x| + C 4 where C is an arbitrary constant. Replacing u back by y/x, we obtain 1 y 4 ( ) = ln |x| + C ⇒ y 4 = 4x4 ln |x| + 4Cx4 . 4 x The above gives rise to two general solutions, namely y(x) = ±(4x4 ln |x| + 4Cx4 )1/4 . Of the two general solutions, which one should we use? Since we are given y(1) = 1 > 0, we reject the general solution y(x) = −(4x4 ln |x| + 4Cx4 )1/4 , as it implies that y < 0. Substituting y(1) = 1 into the appropriate general solution given by y(x) = (4x4 ln |x| + 4Cx4 )1/4 , we find that 1 1 = (4(1)4 ln |1| + 4C(1)4 )1/4 ⇒ C = . 4 Thus, the required particular solution is given by y(x) = (4x4 ln |x| + x4 )1/4 .

2.3

Linear 1st order ODEs

A linear 1st order ODE in y(x) has the form dy + f (x)y = r(x). dx Here f (x) and r(x) are given functions. 33

2.3.1

Homogeneous linear 1st order ODEs

The above linear ODE is said to be homogeneous if r(x) ≡ 0 (that is, r(x) is identically zero over the interval of interest). Thus, a homogeneous 1st order linear ODE is of the form dy + f(x)y = 0. dx This ODE is separable and may be rewritten as Z Z dy = − f (x)dx. y It follows that the general solution of the homogeneous linear 1st order ODE is given by Z R ln |y| = − f (x)dx + C ⇒ y = Be− f (x)dx . Here B = ±eC and C are arbitrary constants.

2.3.2

Nonhomogeneous linear 1st order ODEs

If the linear ODE dy + f (x)y = r(x) dx is such that the function r(x) has non-zero values for at least some values of x (within the interval of interest) then the ODE is said to be nonhomogeneous. Here we discuss a method known as the variation of parameter for solving the nonhomogeneous ODE. The method involves two steps as described below. Firstly, we solve the corresponding homogeneous ODE, that is, dY + f (x)Y = 0 dx for any particular solution Y (x) (except the trivial solution Y (x) = 0). 34

Secondly, to solve the nonhomogeneous ODE, we make the substitution y(x) = Y (x) · v(x). Here v(x) is an unknown function. We substitute y(x) = Y (x) · v(x) into the nonhomogeneous ODE to obtain a separable 1st order ODE in v(x). If v(x) can be determined with an arbitrary constant in it, the general solution of the nonhomogeneous ODE is given by y(x) = Y (x) · v(x). Example Solve y 0 + xy = (x + 1)ex subject to y(0) = 2.

The ODE is nonhomogeneous, 1st order and linear. Firstly, let us find a particular solution of the homogeneous ODE Y 0 + xY = 0, The homogeneous ODE is separable and may be solved to give Z Z dY 1 2 = − xdx ⇒ ln |Y | = − x2 ⇒ Y = e−x /2 . Y 2 In deriving Y above, we ignore the arbitrary constant of integration as we are only interested in finding a particular solution of the homogeneous ODE. To find the general solution of the original nonhomogeneous ODE, we let 2 /2

y = e−x

35

· v(x).

It follows that 2 /2

y 0 = e−x

2 /2

v 0 − xe−x

· v.

Substituting into the original ODE, we obtain 2 /2

e−x

2 /2

v 0 − xe−x

2 /2

· v + x · e−x

· v(x) = (x + 1)ex

which simplifies to give 2

e−x /2 v 0 = (x + 1)ex dv 2 ⇒ = (x + 1)e(x /2)+x Zdx Z ⇒

2 /2)+x

(x + 1)e(x

dv =

2 /2)+x

⇒ v(x) = e(x

dx

+C

where C is an arbitrary constant. We obtain 2 /2

y(x) = e−x

2 /2)+x

· (e(x

2 /2

+ C) = ex + Ce−x

as the required general solution. Now, since y(0) = 2, we find that 2 = 1 + C ⇒ C = 1. Thus, the required particular solution is given by 2 /2

y(x) = ex + e−x

2.4

.

Bernoulli diﬀerential equation

An example of a nonlinear 1st order ODE is the Bernoulli differential equation dy + f(x)y = r(x)y p (p constant, p 6= 0, 1). dx

36

The Bernoulli diﬀerential equation can be converted into a 1st order linear ODE if we use the substituition u(x) = (y(x))1−p . Diﬀerentiating the above with respect to x, we find that u0 = (1 − p)y −p y 0 ⇒ y 0 =

yp 0 u. 1−p

The Bernoulli diﬀerential equation now becomes yp 0 1 du u + f (x)y p u = r(x)y p ⇒ + f (x)u = r(x). 1−p 1 − p dx The linear 1st order ODE may be solved as described in earlier sections for u(x) (and hence y(x)). Example Solve y 0 + 9y = y5 e32x .

Let u = y−4 . This gives u0 = −4y −5 y 0 . From the given ODE, we find that 1 − y 5 u0 + 9y5 u = y5 e32x ⇒ u0 − 36u = −4e32x . 4 The method of variation of parameter may be used to solve the above ODE in u(x). Firstly, we look for a particular solution of the corresponding homogeneous ODE dU − 36U = 0. dx

37

Separating the variable U and x, we find that a particular solution is given by U = e36x . Secondly, for the solution of the linear ODE in u(x), we proceed as follows: u(x) = e36x v(x) ⇒ u0 = e36x v + 36e36x v 0

⇒ e36x v 0 + 36e36x v − 36e36x v = −4e32x dv = −4e−4x ⇒ v(x) = e−4x + C (C arbitrary constant). ⇒ dx Thus, the solution of y 0 + 9y = y 5 e32x is given implicitly by y −4 = (e−4x + C)e36x .

2.5

Population dynamics

First order ODEs in separable form may be found in theories which are concerned with predicting the growth of a population. The population may be a colony of bacteria thriving on an agar culture or cancerous cells in a biological organ or people living in a particular community.

2.5.1

Malthus theory of unlimited growth

According to Malthus theory, the rate of change of a growing population is proportional to its size. If the population size P can be described using a continuous and diﬀerentiable function of time t then Malthus theory can be expressed using the 1st order ODE dP = kP (t), dt where the coeﬃcient k is a positive constant. If k has a larger magnitude, the population grows faster. 38

The above ODE is separable and can be easily solved as follows. Z Z dP = k dt P ⇒ ln |P | = kt + C (C is an arbitrary constant) ⇒ |P | = Bekt (B = eC is an arbitrary constant)

⇒ P (t) = Aekt (A = ±B).

If we let t = 0 in the solution P (t) = Aekt , we obtain A = P (0), that is, A can be interpreted as the population size at time t = 0. Thus, we can write P (t) = P (0)ekt . From the above formula for the population size for k > 0, it is obvious that Malthus theory predicts a “population explosion”, that is, an unlimited growth at an exponential rate.

2.5.2

Verhulst theory of limited growth

At first glance, the assumption behind Malthus theory may appear plausible. It seems not unreasonable to think that a larger population is favourable to the rate of a growing population. On careful thought, however, we realize that a larger population is favorable to growth only if the population does not have to compete with one another for resources (such as food) to survive, that is, if the resources for survival are inexhaustible. In fact, if the resources are limited, a larger population leads to a stiﬀer competition for survival which does not favor population growth. To take the competitive factor into account, mathematical biologists modify the ODE in Malthus theory by taking the coeﬃcient k as a decreasing function of the population size. In Verhulst theory, k is taken to be of the form k = α − βP (t), 39

where α and β are positive constants. With k as given above, the ODE for population dynamics is modified to the nonlinear ODE dP = αP (t) − β(P (t))2 . dt The modified ODE is known as Verhulst equation or sometimes the logistic equation of population dynamics. The positive constant β gives an indication of the intensity of competition among members of the population for resources. In an ideal situation, such as a paradise where food and other resources are abundant and living population does not have to compete with one another for survival, the constant β has the value zero and Malthus theory is applicable. The more severe the competition for survival is, the larger the value of β is. Can we solve Verhulst equation? Rewriting it as dP − αP (t) = −β(P (t))2 , dt we recognise that it is really a Bernoulli diﬀerential equation. To solve it, we let u(t) = (P (t))−1 . This gives du + αu = β dt which can be easily solved as a separable 1st order ODE, that is, Z Z du = − dt αu − β 1 ⇒ ln |αu − β| = −t + C α ⇒ αu − β = De−αt (D = ±eαC ) α − βP = De−αt ⇒ P F αeαt 1 ⇒ P (t) = (F = ). αt 1 + F βe D Here C, D and F are arbitary constants. 40

We may express F in terms of the initial population size as follows: P (0) =

Fα P (0) ⇒F = . 1 + Fβ α − βP (0)

From the solution of the Verhulst equation, we can see that P → α/β as t → ∞. As discussed above, a larger value of β indicates that the population has to compete harder for survival. On the other hand, a higher value of α implies that the population has a high rate of reproduction. Thus, it is interesting to note that the population size in the long run is limited by the ratio α/β.

2.6

Exercise III

1. Find general solutions of the following ODEs in y(x): (a) (b) (c) (d)

2x y 0 y = (y − 2)(x + 1)

y0 =

y 0 = xy − 3y

y 0 = (1 − y)−1 cot(x)

2. Water is pumped into a tank at a constant rate of 10 meter3 per second. It is sucked out of the tank through a small hole at a rate given by 2V (t) meter3 per second, where V (t) is the volume of water (in meter3 ) present in the tank at time t ≥ 0 (in second). If the volume of water in the tank at time t = 0 second is 20 meter3 , how much water does the tank hold at time t = 1 second? If the situation described is allowed to go on indefinitely, will the tank eventually dry up? What will happen? Explain. 3. The gradient of the tangent to a curve at the point (x, y) is given by (x3 + 1)/(y 4 + 1). Express this information using a 1st order ODE. Find the equation of the curve if it is known that (1, 1) is a point lying on the curve. 41

4. The gradient of the normal to a curve at the point (x, y) is given by y/(2x). Express this information using a 1st order ODE. Find the equation of the curve if it is known that (0, 1) is a point lying on the curve. 5. With reference to a Cartesian coordinate frame, the shape of a hanging cable is given by yp = s(x), where s(x) is found to satisfy the ODE s00 (x) = 1 + (s0 (x))2 . By making an appropriate substitution, convert the 2nd order ODE into a lower order one. Given that the cable has endpoints at (0, 0) and (1, 0), find its shape. (Note. The working on page 30 may be of help here.) 6. Solve the ODE y 0 (x) =

x−y . x+y

(Hint. Rewrite the ODE as x(1 − (y/x)) 1 − (y/x) dy = = .) dx x(1 + (y/x)) 1 + (y/x)

7. Find general solutions of of the following ODEs in y(x): (a)

y 0 + (2x + 1)y = 2x + 1

(b)

xy 0 + y = 2x6 (Hint. Rewrite in the form y 0 + f(x)y = r(x).)

(c) − 2xy 0 + y = 2x6 y 3

(Hint. Bernoulli diﬀerential equation.)

8. The growth of a tumor may be described using the Gompertz equation which is a 1st order separable ODE of the form dV = (α − β ln(V ))V dt where α and β are positive constants and V (t) is the volume of the tumor at time t. Solve the Gompertz equation. Find the limiting volume of the tumor as t → ∞. 42

2.7

Solutions to Exercise III

1. (a) Rewrite ODE as ydy = 2xdx and integrate to obtain y 2 = 2x2 + C. (b) Separate the variables and integrate as follows. dy y−2

=

(x + 1)dx Z Z dy ⇒ = (x + 1)dx y−2 1 ⇒ ln |y − 2| = x2 + x + C. 2 1 2 ⇒ y = 2 + Ae 2 x +x .

(c) Rewrite ODE as y0 = y(x − 3). It follows that Z Z dy = (x − 3)dx y 1 2 −3x

⇒ y = Ae 2 x

.

(d) Separate the variables and integrate as follows. Z Z cos(x) dx (1 − y)dy = sin(x) 1 ⇒ y − y2 = ln | sin(x)| + C. 2 2. For the volume of water in the tank, we find that ‘Total rate of change of volume per unit time’ =

‘Rate of volume into the tank per unit time’

−‘Rate of volume out of the tank per unit time’ dV dV ⇒ = 10 − 2V ⇒ + 2V = 10. dt dt If we regard the ODE as a nonhomogeneous 1st order linear ODE, we first solve the corresponding homogeneous ODE, that is, dW + 2W = 0 ⇒ W (t) = e−2t . dt 43

To solve for V (t), we let V (t) = u(t)e−2t . Substitute into the nonhomogeneous ODE, we obtain du = 10e2t ⇒ u = 5e2t + C. dt This gives V (t) = 5 + Ce−2t . We are given V (0) = 20. Thus, C = 15. So, volume at time t = 1 is given by V (1) = 5 + 15e−2 ' 7. 03 meter3 . As t → ∞, we find that V (t) → 5 (since e−2t → 0 as t → ∞). The tank will not dry up if the current situation prevails. (Note. The ODE V 0 +2V = 10 may also be treated as a 1st order separable ODE.) 3. The ODE is given by dy x3 + 1 = 4 dx y +1 which can be easily solved as a separable ODE to obtain 1 5 1 y + y = x4 + x + C. 5 4 To work out C, substitute the point (x, y) = (1, 1) into the equation above. We find that C = −1/20. Thus, the required curve is 1 5 1 1 y + y = x4 + x − . 5 4 20 4. Since (gradient of normal) × (gradient of tangent) = −1, the ODE to solve is dy 2x =− . dx y The above ODE is separable and can be integrated to obtain 1 2 y = −x2 + C. 2 44

Use the point (0, 1) to obtain C = 1/2. The required curve is thus given by y 2 = −2x2 + 1. 5. To convert the 2nd order ODE to a 1st order one, let s0 (x) = p(x). Thus, dp p = 1 + p2 . dx The ODE above is separable and can be solved as follows. Z Z dp p = dx. 1 + p2 The integral on the left hand side can be evaluated as explained on page 30 by letting p = sinh θ. This leads to θ = x + C. From p = sinh θ and s0 (x) = p(x), it follows that p(x)

=

sinh(x + C)

⇒ s0 (x) = sinh(x + C)

⇒ s(x) = cosh(x + C) + F. From the endpoints (0, 0) and (1, 0), we know that s(0) = 0 and s(1) = 0. This gives eC + e−C + 2F C

−1 −C

e·e +e

e

+ 2F

= 0 = 0.

Eliminating F, we obtain (1 − e)eC +

(1 − e−1 ) =0 eC

which gives e2C

e−1 − 1 1 e−1 − 1 ⇒ C = ln( ) 1−e 2 1−e 1 ⇒ F = − (eC + e−C ) 2 1 e−1 − 1 1/2 e−1 − 1 −1/2 ⇒ F = − [( ]. ) +( ) 2 1−e 1−e =

45

Thus, the shape of the cable is given by y = s(x), that is, y =

1 e−1 − 1 1/2 x e−1 − 1 −1/2 −x e ] [( ) e +( ) 2 1−e 1−e 1 e−1 − 1 1/2 e−1 − 1 −1/2 − [( ]. ) +( ) 2 1−e 1−e

Note. In the above, we use eln(x) = x. 6. Rewrite ODE as 1 − (y/x) dy = . dx 1 + (y/x)

Let y = xu(x). (Refer to page 32.) This gives du 1−u +u = dx 1+u 1−u du = −u x dx 1+u du 1 − 2u − u2 x = (separable ODE) dx 1+u Z Z dx (1 + u)du = 2 1 − 2u − u x 1 − ln |u2 + 2u − 1| = ln |x| + ln |C| = ln(|Cx|) 2 (C is an arbitrary constant) x

⇒ ⇒ ⇒ ⇒

⇒ |u2 + 2u − 1| = C −2 x−2 y y ⇒ ( )2 + 2( ) − 1 = Ex−2 (E = ±C 2 ). x x 7. (a) Regard ODE as 1st order linear. Solve corresponding homogeneous ODE first, that is, Y 0 + (2x + 1)Y = 0. 2

A particular solution is Y = e−x −x . To solve the original 2 ODE, let y = e−x −x v(x). This gives dv dx

=

2 +x

(2x + 1)ex

2 +x

⇒ v(x) = ex 46

+ C.

Thus, the required general solution is 2 −x

y = 1 + Ce−x

.

Note. Alternatively, treat the ODE as separable by writing it as dy = (2x + 1)(1 − y). dx (b) This is a 1st order linear ODE. First solve Y0+

1 Y =0 x

for a particular solution. We obtain Y = x−1 . For the general solution of the original ODE, let y = x−1 v(x). This gives dv 2 = 2x6 ⇒ v(x) = x7 + C. dx 7 Thus, the required general solution is 2 y = x6 + Cx−1 . 7 (c) To solve this Bernoulli equation, let u = y−2 . Thus, u0

−2y−3 y 0 1 ⇒ y 0 = − y 3 u0 . 2 If we subsitute the above into the Bernoulli equation, we obtain du + u = 2x6 x dx which may be solved as a 1st order linear ODE (see part (b) above) to obtain =

2 u(x) = x6 + Cx−1 . 7 The required general solution is 2 y −2 = x6 + Cx−1 . 7 47

8. The Gompertz equation is 1st order separable. Separating the variables, we find that Z Z dV = dt V (α − β ln(V )) 1 ⇒ − ln |α − β ln(V )| = t + C β ⇒ α − β ln(V ) = Ae−βt (A = ±e−βC ) A ⇒ V (t) = exp(α/β) exp(− e−βt ) β → exp(α/β) exp(0) (as t → ∞) (since β > 0). Thus, V (t) → eα/β as t → ∞. (Note. To work out the indefinite integral involving V, use the substitution W = α − β ln(V ).)

48

Chapter 3

Second order linear ODEs 3.1

Preamble

Here we will look at 2nd order linear ODEs. These are ODEs which can be written in the form y 00 + f (x)y 0 + g(x)y = r(x) where f (x), g(x) and r(x) are given functions of x. The ODE above is said to be homogeneous if r(x) ≡ 0. Otherwise, it is nonhomogeneous. Below are some examples of 2nd order linear ODEs in y(x): (1)

y 00 (x) + 2y 0 (x) + 3y(x) = 0

(2)

y 00 (x) + xy(x) = 0

(3)

y 00 (x) + 2xy 0 (x) + (x + 1)y(x) = x2 − 2

(4)

y 00 (x) + x3 sin(x)y 0 (x) + 2x5 y(x) = x2 cos(x)

The first two ODEs in the examples above are homogeneous, while the other two are nonhomogeneous. We will examine how general solutions can be constructed for 2nd order linear ODEs. Methods of solution for solving certain 2nd order linear ODEs are then given. As we will see in Chapter 4, 2nd order linear ODEs are of interest in practical 49

engineering problems involving electric circuits and spring-mass systems.

3.2

General solution of homogeneous 2nd order linear ODE

If two linearly independent solutions of a homogeneous 2nd order linear ODE can be found, they can be used to construct the general solution of the ODE.

3.2.1

Linearly independent functions

Two functions f (x) and g(x) are said to be linearly independent over the interval [a, b] (that is, over the interval a ≤ x ≤ b), if and only if we cannot find constants α and β, other than α and β both being zero, such that αf (x) + βg(x) = 0 for all x in [a, b]. Examples 1. If f (x) = 2x2 and g(x) = x, then αf(x) + βg(x) = 2αx2 + βx. Now if α and β are both not zero, we can find at most two values of x such that 2αx2 + βx = 0. We cannot find constants α and β, other than α and β both being zero, such that 2αx2 + βx = 0 for all values of x. Thus, the functions f (x) = 2x2 and g(x) = x are linearly independent over the interval (−∞, ∞). 2. The functions f (x) = 3ex and g(x) = ex are not linearly independent as f (x) = 3g(x) (hence f (x) − 3g(x) = 0 for all x). In general, for any two functions f(x) and g(x), if f(x)/g(x) is a constant (over a given interval), then the functions are not linearly independent (over the interval). Otherwise, if f (x)/g(x) is not a constant, then the functions are linearly independent. 3. The functions f(x) = 0 and g(x) are not linearly independent as we can write f (x) = 0 · g(x), that is, we can write f (x) − 0 · g(x) = 0 (no matter what g(x) is). 50

Theorem 1 Let f (x) and g(x) be diﬀerentiable functions over the interval [a, b]. The functions f(x) and g(x) are not linearly independent over [a, b] if and only if f (x)g 0 (x) − g(x)f 0 (x) = 0 for all x in [a, b]. The above theorem can be proven as follows. Let αf (x) + βg(x) = 0 for all x in [a, b],

(1)

where α and β are constants. It follows that αf 0 (x) + βg 0 (x) = 0 for all x in [a, b].

(2)

If we multiply (1) and (2) above by g 0 (x) and g(x) respectively and take the diﬀerence between the two resulting equations, we find that [f (x)g 0 (x) − g(x)f 0 (x)]α = 0 for all x in [a, b].

(3)

We may proceed in a similar manner to obtain [f(x)g 0 (x) − g(x)f 0 (x)]β = 0 for all x in [a, b].

(4)

If the functions f (x) and g(x) are not linearly independent, then it is possible to find either α or β (or both) which is not zero. From (3) and (4) above, it immediately follows that f(x)g 0 (x)−g(x)f 0 (x) = 0 for all x in [a, b]. Thus, if the functions f(x) and g(x) are not linearly independent, then f (x)g 0 (x) − g(x)f 0 (x)α = 0 for all x in [a, b]. If f (x)g 0 (x) − g(x)f 0 (x) = 0 for all x in [a, b], then it is obvious that α and β do not have to be zero in (3) and (4). This implies that f (x) and g(x) are not linearly independent.

51

3.2.2

Construction of general solution

Consider the homogeneous 2nd order linear ODE in y(x) given by y 00 + f (x)y 0 + g(x)y = 0.

(5)

Lemma 1 If y1 (x) and y2 (x) are solutions of (5) then y(x) = Ay1 (x) + By2 (x), where A and B are any arbitrary constants, is also a solution of (5). The lemma above can be proven as follows. If y1 (x) and y2 (x) are solutions of (5) then y100 + f (x)y10 + g(x)y1 = 0

(6)

y200

(7)

+ f (x)y20

+ g(x)y2 = 0.

If we substitute y = Ay1 + By2 into the left hand side of (5), we find that y 00 + f (x)y 0 + g(x)y d2 d = (Ay1 (x) + By2 (x)) + f (x) (Ay1 (x) + By2 (x)) dx2 dx +g(x) (Ay1 (x) + By2 (x)) ¤ £ ¤ £ = A y100 + f (x)y10 + g(x)y1 + B y200 + f (x)y20 + g(x)y2 = A · 0 + B · 0 = 0 (that is, the right hand side of (5)).

Thus, y(x) = Ay1 (x) + By2 (x) is also a solution of (5). Lemma 2 If y1 (x) and y2 (x) R are any two solutions of (5) then y2 y10 − y1 y20 = De− f (x)dx , where D is a constant. Lemma 2 can be proven as follows. If y1 (x) and y2 (x) are solutions of (5) then (6) and (7) above hold. 52

Multiplying (6) by y2 and (7) by y1 , we obtain y2 y100 + f (x)y2 y10 + g(x)y2 y1 = 0 y1 y200 + f (x)y1 y20 + g(x)y1 y2 = 0. The diﬀerence between the two equations gives £ 00 ¤ £ ¤ y2 y1 − y1 y200 + f(x) y2 y10 − y1 y20 = 0.

If we define z(x) = y2 y10 − y1 y20 , then dz dx

¤ d £ 0 y2 y1 − y1 y20 dx = y2 y100 + y20 y10 − y1 y200 − y20 y10 =

= y2 y100 − y1 y200 .

Thus, we obtain the 1st order linear ODE dz + f (x)z = 0. dx The 1st order ODE is a separable one and may be rewritten as Z

dz =− z

Z

f (x)dx

to give ln |z| + C = −

Z

f (x)dx, that is, z = De−

R

f (x)dx

where C and D = ±e−C are constants. R Thus, we show that y2 y10 − y1 y20 = De− f (x)dx . Theorem 2 If y1 (x) and y2 (x) are two linearly independent solutions of the homogeneous 2nd order linear ODE in (5) then the general solution of (5) is given by y(x) = Ay1 (x) + By2 (x), where A and B are arbitrary constants. 53

From Lemma 1 above, we know that y(x) = Ay1 (x)+By2 (x) is a solution of (5), but is it the general solution? To prove the theorem, we have to show that any solution Y (x) of (5) can be written in the form Ay1 (x)+By2 (x), if y1 (x) and y2 (x) are linearly independent solutions of (5). Since y1 (x) and y2 (x) are solutions of (5), Lemma 2 tells us that y2 y10 − y1 y20 = De−

R

f (x)dx

where D is a constant. Similarly, since y1 (x) and Y (x) are solutions of (5), Lemma 2 gives Y y10 − y1 Y 0 = Ee−

R

f (x)dx

where E is a constant. Note that we should allow for the possibility that E and D are diﬀerent constants. Also, as y2 (x) and Y (x) are solutions of (5), we may write Y y20 − y2 Y 0 = F e−

R

f (x)dx

where F is a constant. The last two equations above may be used to obtain Y y10 y2 − y1 y2 Y 0 = y2 Ee−

Y y20 y1 − y2 y1 Y 0 = y1 F e−

R

R

f (x)dx f (x)dx

.

If we take the diﬀerence of the two equations above, we obtain Y y10 y2 − Y y20 y1 = −y1 F e− ⇒ Y =

−y1 F e−

R

f (x)dx + y2 Ee− R Ee− f (x)dx

f (x)dx

y2 y10

R

+ y2 − y1 y20

Using y2 y10 − y1 y20 = De− written in the form

R

f (x)dx ,

Y = Ay1 + By2 54

R

f (x)dx

we find that Y can be

where A = −F/D and B = E/D are constants, provided that D 6= 0. Can we guarantee that D 6= 0? If D =R 0 then y2 y10 − y1 y20 is identically zero, since y2 y10 − y1 y20 = De− f (x)dx . According to Theorem 1, this will then imply that y1 and y2 are not linearly independent. Now, in Theorem 2, we are given that solutions y1 and y2 are linearly independent. Thus, the constant D cannot be zero. Summarizing, we show that if y1 (x) and y2 (x) are linearly independent solutions of the homogeneous 2nd order linear ODE y 00 + f (x)y 0 + g(x)y = 0, then any solution Y (x) of the ODE can be written in the form Ay1 + By2 , where A and B are constants, that is, we can construct the general solution of the ODE to be given by y(x) = Ay1 (x) + By2 (x).

3.3

Homogeneous 2nd order linear ODEs with constant coeﬃcients

A homogeneous 2nd order linear ODE with constant coeﬃcients is given by ay 00 (x) + by 0 (x) + cy(x) = 0.

(8)

Here a 6= 0, b and c are given real constants. We can rewrite (8) in the form of (5) (on page 52) as b c y00 (x) + y0 (x) + y(x) = 0. a a According to Theorem 2, we can construct the general solution of (8) by finding two linearly independent solutions. To find a solution of (8), let us try y(x) = eλx where λ is a constant. 55

Diﬀerentiating, we obtain y 0 (x) = λeλx y 00 (x) = λ · λeλx = λ2 eλx . Substituting into (8), we find that aλ2 eλx + bλeλx + ceλx = 0 £ ¤ ⇒ eλx aλ2 + bλ + c = 0

The ODE in (8) is satisfied for all x if aλ2 + bλ + c = 0. We consider the following cases.

Case I: b2 − 4ac > 0 Now if b2 − 4ac > 0, then the quadratic equation in λ has two distinct real solutions given by √ −b + b2 − 4ac λ = λ1 ≡ √2a −b − b2 − 4ac λ = λ2 ≡ 2a and we obtain two solutions of (8) as given by y1 = eλ1 x

and

y2 = eλ2 x .

Since y2 /y1 = eλ2 x /eλ1 x = e[λ2 −λ1 ]x is not a constant as λ1 6= λ2 , the two solutions y1 and y2 are linearly independent. If b2 − 4ac > 0, then Theorem 2 tells us the general solution of (8) is given by y = Aeλ1 x + Beλ2 x where A and B are arbitrary constants and λ1 and λ2 are the two distinct real solutions of the quadratic equation aλ2 + bλ + c = 0.

56

Case II: b2 − 4ac = 0 For this particular case, the quadratic equation aλ2 + bλ + c = 0 has only one real solution given by λ = λ1 ≡ −b/(2a). Hence, in seeking a solution of the form y = eλx , we obtain only one solution of (8), that is, y1 = e−bx/(2a) . To construct the general solution of (8), another solution which is linearly independent to y1 is needed. To find another solution, we let y = u(x) · eλ1 x where u(x) is a function to be determined. Diﬀerentiating, we obtain y 0 = λ1 u(x) · eλ1 x + u0 (x) · eλ1 x

y 00 = λ21 u(x) · eλ1 x + 2λ1 u0 (x) · eλ1 x + u00 (x)eλ1 x . Substituting into (8), we find that a[λ21 u(x) · eλ1 x + 2λ1 u0 (x) · eλ1 x + u00 (x)eλ1 x ] ´ ³ +b λ1 u(x) · eλ1 x + u0 (x) · eλ1 x + cu(x)eλ1 x = 0

which may be rewritten as

[(aλ21 + bλ1 + c)u(x) + (2λ1 a + b)u0 (x) + au00 (x)]eλ1 x = 0. Since aλ21 + bλ1 + c = 0 and λ1 = −b/(2a), the equation above reduces to au00 (x)eλ1 x = 0. Since a 6= 0 and eλ1 x 6= 0, we obtain the ODE u00 (x) = 0. A solution of this simple ODE is u(x) = x. Summarizing, if b2 − 4ac = 0, two particular solutions of (8) are given by y1 = e−bx/(2a) and y2 = x · e−bx/(2a) . 57

Since y2 /y1 = x (not a constant), the two solutions above are linearly independent. From Theorem 2, the required general solution of the ODE is y = Ae−bx/(2a) + Bx · e−bx/(2a) where A and B are arbitrary constants. Case III: b2 − 4ac < 0 For this case, the quadratic equation aλ2 + bλ + c = 0 does not have any real solutions. It has two distinct complex solutions given by p |b2 − 4ac| b λ = λ1 ≡ − + i 2a p 2a |b2 − 4ac| b λ = λ2 ≡ − − i 2a 2a √ where i = −1. If we ignore the fact that λ1 and λ2 are complex and proceed as in Case I above, the general solution of (8) is given by y = Aeλ1 x + Beλ2 x where A and B are arbitrary constants. The only question that arises here is, “How can we evaluate eλx for complex λ? The general solution given above is useless if we cannot evaluate eλ1 x and eλ2 x . A proper interpretation of eλx for complex λ may be found in the theory of complex functions as explained below. In calculus of real functions, ex can be represented by its Taylor series about x = 0, that is, we can write ex =

∞ X xn

n=0

n!

.

A brief review of power series is given in Chapter 5 (Section 5.2 on page 128).

58

We extend the above to define ex+iy (for real x and y) as x+iy

e

=

∞ X (x + iy)n

n!

n=0

.

(9)

If we let x = 0 in (9) and formally regroup the terms in the series for eiy into even and odd powers of iy, we find that eiy = =

∞ X (iy)n

n=0 ∞ X k=0

= =

n!

∞

(iy)2k X (iy)2k+1 + (2k)! (2k + 1)! k=0

∞ X (i2 )k y 2k k=0 ∞ X k=0

(2k)!

+i

∞ X (i2 )k y 2k+1 k=0 ∞ X

(−1)k y 2k +i (2k)!

(2k + 1)!

k=0

(−1)k y2k+1 . (2k + 1)!

Knowing that cos(x) and sin(x) can be given by their Taylor series about x = 0 as cos(x) =

∞ X (−1)k x2k

(2k)!

k=0

and sin(x) =

∞ X (−1)k x2k+1 k=0

(2k + 1)!

,

we obtain the so called Euler’s formula. eiy = cos(y) + i sin(y) for real y. Substituting the binomial formula (x + iy)n =

n X

n! xm (iy)n−m m!(n − m)! m=0

into (9), we find that x+iy

e

= =

∞ X (x + iy)n

n=0 ∞ X

n!

n X

xm (iy)n−m . m!(n − m)! n=0 m=0 59

(10)

In the double summation above, for a fixed n which runs from 0 to ∞, we sum over m from 0 to n. All the values of n and m which we carry out the double summation are plotted as points (n, m) in Figure 3.1.

Figure 3.1 Let us now interchange the order of the double summation, that is, we fix m first and sum over n. To cover all the points in Figure 3.1, we find that we have to let m run from 0 to ∞, and for a fixed m, we sum over n from m to ∞. (For example, when m = 3, we sum over n from 3 to ∞.) Interchanging the order of the double summation, we find that e

x+iy

= = =

∞ X ∞ X xm (iy)n−m m!(n − m)! n=m

m=0 ∞ ∞ X X

xm (iy)k (if we let k = n − m) m!k!

m=0 k=0 ∞ ∞ m X X m=0

x m!

(iy)k k!

k=0 x+iy

⇒e

= ex · eiy .

60

(11)

With (10) and (11), ex+iy can be evaluated in terms of real exponential, cosine and sine functions, that is, ex+iy = ex · eiy = ex (cos(y) + i sin(y)). Examples 1. Solve the ODE y00 + y0 − 6y = 0 subject to the conditions y(0) = 1 and y 0 (0) = 7.

This is a homogeneous 2nd order linear ODE with constant coeﬃcients. Thus, we use y = eλx ,

y 0 = λeλx

and y 00 = λ2 eλx .

Substituting into the ODE, we find that λ2 eλx + λeλx − 6eλx = 0 £ ¤ ⇒ eλx λ2 + λ − 6 = 0

⇒ λ2 + λ − 6 = 0

⇒ (λ + 3)(λ − 2) = 0 ⇒ λ = −3, λ = 2.

Two linearly independent solutions of the ODE are given by y1 = e−3x

and

y2 = e2x .

The general solution is y = Ae−3x + Be2x where A and B are arbitrary constants. We use the given conditions y(0) = 1 and y0 (0) = 7 to determine A and B. Diﬀerentiating the general solution, we obtain y 0 = −3Ae−3x + 2Be2x .

61

It follows that y(0) = 1 ⇒ A + B = 1

y0 (0) = 7 ⇒ −3A + 2B = 7. Solving for A and B, we obtain A = −1 and B = 2. Thus, the required particular solution of the ODE is y = −e−3x + 2e2x . 2. Solve the ODE y 00 + 6y 0 + 9y = 0 subject to the conditions y(0) = 1 and y0 (0) = 1. This is a homogeneous 2nd order linear ODE with constant coeﬃcients. Thus, we use y = eλx ,

y 0 = λeλx

and y 00 = λ2 eλx .

Substituting into the ODE, we obtain λ2 + 6λ + 9 = 0 ⇒ (λ + 3)2 = 0 ⇒ λ = −3. Since λ = −3 is the only solution of the quadratic equation, the general solution of the ODE is given by y = Ae−3x + Bx · e−3x where A and B are arbitrary constants. Diﬀerentiating the general solution, we obtain y0 = −3Ae−3x − 3Bx · e−3x + Be−3x . Using the given conditions, we find that y(0) = 1 ⇒ A = 1

y 0 (0) = 1 ⇒ −3 + B = 1, that is, B = 4. Hence the required particular solution is y = e−3x + 4x · e−3x .

62

3. Solve the ODE y 00 −4y0 +13y = 0 subject to the conditions y(0) = 1 and y0 (0) = 2. What is the value of y at x = π/2? This is a 2nd order linear homogeneous ODE with constant coeﬃcients. We use y = eλx ,

y 0 = λeλx

and y 00 = λ2 eλx .

Substituting into the ODE, we find that λ2 − 4λ + 13 = 0

⇒ λ = 2 + 3i, λ = 2 − 3i. Thus, the general solution is given by y = Ae(2+3i)x + Be(2−3i)x where A and B are arbitrary constants. It is useful to rewrite the general solution as y = Ae(2+3i)x + Be(2−3i)x = e2x (Aei(3x) + Be−i(3x) ) = e2x (A [cos(3x) + i sin(3x)] + B [cos(3x) − i sin(3x)])

= e2x ([A + B] cos(3x) + i [A − B] sin(3x)) . Thus, we obtain y = e2x [C cos(3x) + D sin(3x)]

where C = A + B and D = i[A − B] are arbitrary constants. Diﬀerentiating, we obtain y 0 = e2x [−3C sin(3x) + 3D cos(3x)] +2e2x [C cos(3x) + D sin(3x)] . Using the given conditions, we find that y(0) = 1 ⇒ C = 1

y 0 (0) = 2 ⇒ 3D + 2C = 2, that is, D = 0. 63

Thus, the required particular solution is y = e2x cos(3x). The value of y at x = π/2 is given by 3π π y( ) = eπ cos( ) = 0. 2 2

3.4

Euler-Cauchy equations

An Euler-Cauchy equation is given by ax2 y 00 (x) + bxy 0 (x) + cy(x) = 0

(12)

where a 6= 0, b and c are given constants. Note that (12) can be rewritten in the form of (5) (on page 52) as y 00 (x) +

b 0 c y (x) + 2 y(x) = 0. ax ax

We can use Theorem 2 to construct the general solution of (12) if we can find two linearly independent solutions. To find linearly independent solutions of (12), let y = xλ where λ is a constant yet to be determined. Diﬀerentiating, we obtain y0 = λxλ−1 and y 00 = λ(λ − 1)xλ−2 . Substitution into (12) gives ax2 · λ(λ − 1)xλ−2 + bx · λxλ−1 + cxλ = 0

⇒ xλ · (aλ(λ − 1) + bλ + c) = 0 (for all x)

⇒ aλ(λ − 1) + bλ + c = 0

⇒ aλ2 + (b − a)λ + c = 0. We consider the following cases. 64

Case I: (b − a)2 − 4ac > 0 For this particular case, we find two distinct real values for λ as given by p −(b − a) + (b − a)2 − 4ac λ = λ1 ≡ p2a −(b − a) − (b − a)2 − 4ac . λ = λ2 ≡ 2a Thus, we obtain two solutions of (12) as given by y1 = xλ1 ,

y2 = xλ2 .

These solutions are linearly independent as y2 /y1 = xλ2 /xλ1 = xλ2 −λ1 is not a constant, since λ1 6= λ2 . For this particular case, the general solution of the EulerCauchy ODE in (12) is given by y = Axλ1 + Bxλ2 where A and B are arbitrary constants. Case II: (b − a)2 − 4ac = 0 For this case, the quadratic equation aλ2 + (b − a)λ + c = 0 has only one solution, that is, λ = λ1 ≡

−(b − a) . 2a

Thus, only one solution is found for (12). Two linearly independent solutions are needed to construct the general solution of the ODE. To find another solution, we let y(x) = xλ1 · u(x) where u(x) is a function yet to be determined. Diﬀerentiating, we obtain y 0 (x) = λ1 xλ1 −1 · u(x) + xλ1 · u0 (x)

y 00 (x) = λ1 (λ1 − 1)xλ1 −2 · u(x)

+2λ1 xλ1 −1 · u0 (x) + xλ1 · u00 (x). 65

Substitution into (12) gives

+

xλ1 u(x) [aλ1 (λ1 − 1) + bλ1 + c]

xλ1 +1 u0 (x) [2aλ1 + b] + axλ1 +2 · u00 (x) = 0.

Since aλ1 (λ1 − 1) + bλ1 + c = 0 and λ1 = −(b − a)/(2a), the above equation reduces to u0 (x) + xu00 (x) = 0. To solve for u(x), let v(x) = u0 (x) to obtain a 1st order ODE in separable form, that is, v+x

dv = 0. dx

Solving the 1st order ODE, we find that Z Z dx dv =− v x

1 ⇒ ln(v) = − ln(x) = ln( ) x 1 ⇒ v= x 1 du = ⇒ dx x ⇒ u = ln(x).

Note that we ignore the arbitrary constants in the integrations as we are only looking for a particular solution. Another solution of the Euler-Cauchy equation is thus given by y = xλ1 · ln(x). The solutions y = xλ1 and y = xλ1 · ln(x) are linearly independent. To summarize, if the quadratic equation aλ2 +(b−a)λ+c = 0 has only one solution given by λ1 = −(b − a)/(2a), the EulerCauchy equation in (12) has the general solution y = Axλ1 + Bxλ1 ln(x) where A and B are arbitrary constants.

66

Case III: (b − a)2 − 4ac < 0 For this case, the quadratic equation aλ2 + (b − a)λ + c = 0 has two complex solutions given by p |(b − a)2 − 4ac| (b − a) +i λ = λ1 ≡ − 2a 2a p |(b − a)2 − 4ac| (b − a) −i . λ = λ2 ≡ − 2a 2a The general solution of the Euler-Cauchy equation is y = Axλ1 + Bxλ2 where A and B are arbitrary constants. To compute xλ for complex λ, we turn to the theory of complex functions again. We use the results xz+w = xz · xw for complex z and w and xiy = eiy ln(x) = cos(y ln(x)) + i sin(y ln(x)) for x > 0 and real y.

Examples 1. Solve the ODE x2 y 00 + 4xy 0 + 2y = 0 subject to y(1) = y 0 (1) = 1. This is an Euler-Cauchy equation. We use y = xλ ,

y 0 = λxλ−1 ,

y 00 = λ(λ − 1)xλ−2 .

Substituting into the ODE, we find that λ(λ − 1) + 4λ + 2 = 0

⇒ λ2 + 3λ + 2 = 0

⇒ (λ + 1)(λ + 2) = 0 ⇒ λ = −1, 67

λ = −2.

Thus, the general solution of the ODE is y = Ax−1 + Bx−2 where A and B are arbitrary constant. Diﬀerentiating the general solution gives y0 = −Ax−2 − 2Bx−3 . Use of y(1) = y 0 (1) = 1 gives y(1) = 1 ⇒ A + B = 1

y 0 (1) = 1 ⇒ −A − 2B = 1. Solving for A and B, we obtain A = 3 and B = −2. The required particular solution is y = 3x−1 − 2x−2 .

2. Solve the ODE x2 y 00 + 3xy 0 + y = 0 subject to y(1) = 1 and y(e) = 2.

This is an Euler-Cauchy equation. We use y = xλ ,

y 0 = λxλ−1 ,

y 00 = λ(λ − 1)xλ−2 .

Substitution into the ODE gives λ(λ − 1) + 3λ + 1 = 0

⇒ λ2 + 2λ + 1 = 0

⇒ (λ + 1)2 = 0 ⇒ λ = −1 is the only solution.

Thus, the general solution of the ODE is y = Ax−1 + Bx−1 ln(x) where A and B are arbitrary constants. 68

Using the given conditions, we find that y(1) = 1 ⇒ A = 1

y(e) = 2 ⇒ e−1 + Be−1 ln(e) = 2 ⇒ B = 2e − 1.

The required particular solution is given by y = x−1 + (2e − 1) x−1 ln(x). 3. Solve the ODE x2 y 00 (x) + 3xy 0 (x) + 2y(x) = 0 subject to y(1) = y0 (1) = 1.

This is an Euler-Cauchy equation. We use y = xλ ,

y 0 = λxλ−1 ,

y 00 = λ(λ − 1)xλ−2 .

Substituting into the ODE, we find that λ(λ − 1) + 3λ + 2 = 0

⇒ λ2 + 2λ + 2 = 0 ⇒ λ = −1 + i, −1 − i. The general solution of the ODE is y = Ax−1+i + Bx−1−i where A and B are arbitrary constants. We may rewrite the above general solution as y = Ax−1+i + Bx−1−i = x−1 [Axi + Bx−i ] = x−1 [Aei ln(x) + Be−i ln(x) ] = x−1 [A (cos(ln(x)) + i sin(ln(x))) +B (cos(ln(x)) − i sin(ln(x)))]

= x−1 [C cos(ln(x)) + D sin(ln(x))] where C = A + B and D = i(A − B) are arbitrary constants.

69

Diﬀerentiating, we obtain £ ¤ y 0 = x−1 −Cx−1 sin(ln(x)) + Dx−1 cos(ln(x)) −x−2 [C cos(ln(x)) + D sin(ln(x))] .

Using y(1) = y 0 (1) = 1, we find that y(1) = 1 ⇒ C = 1

y 0 (1) = 1 ⇒ D − C = 1 ⇒ D = 2. Thus, the required particular solution is y = x−1 [cos(ln(x)) + 2 sin(ln(x))] .

3.5

Exercise IV

1. Find the general solution of each of the following ODEs in y(x): (a) y 00 − y 0 − 30y = 0 (b) 9y 00 − 30y0 + 25y = 0 (c) y 00 − 6y0 + 25y = 0

(d) x2 y 00 + 5xy0 + 3y = 0

(e) x2 y00 − 5xy 0 + 9y = 0 (f) x2 y00 + 5xy 0 + 8y = 0 2. Solve each of the following ODEs in y(x) subject to the given conditions: (a) 3y 00 + 5y0 + 2y = 0, y(0) = −1, y 0 (0) = 3 (b) x2 y 00 + 5xy0 + 4y = 0, y(1) = 1, y(e) = 2 3. If y = yp is a non-trivial solution of the 2nd order linear ODE y 00 + f (x)y 0 + g(x)y = 0 70

verify that another solution is given by Z −F (x) Z e dx with F (x) = f (x)dx. y = yp (x) · [yp (x)]2 4. Find the constant p such that y = xp is a solution of the ODE 1 3 y 00 + xy0 + ( − 2 )y = 0. 2 4x Use the result in Problem 3 above to find another solution of the ODE. Write down the general solution of the ODE. 5. Show that Riccati equation in u(x) given by the 1st order nonlinear ODE u0 = p(x)u2 + q(x)u + r(x) can be converted into the homogeneous 2nd order linear ODE in y(x) given by y00 − (

p0 (x) + q(x))y 0 + p(x)r(x)y = 0 p(x)

by letting u(x) = −

3.6

y0 (x) . y(x)p(x)

Solving nonhomogeneous ODEs

Consider the nonhomogeneous 2nd order linear ODE y 00 + f (x)y 0 + g(x)y = r(x)

(13)

where r(x) is not identically zero. Let yp (x) be any particular solution of the nonhomogeneous ODE in (13). Thus, yp00 + f (x)yp0 + g(x)yp = r(x). 71

To solve (13), let y = yp (x) + Y (x). Substituting into (13), we find that ¤ £ 00 yp + f (x)yp0 + g(x)yp + Y 00 + f (x)Y 0 + g(x)Y 00

= r(x)

0

⇒ r(x) + Y + f (x)Y + g(x)Y = r(x) ⇒ Y 00 + f (x)Y 0 + g(x)Y = 0.

(14)

In (14), we obtain a homogeneous 2nd order linear ODE in Y (x). If the general solution of the homogeneous ODE in Y (x) is given by Y = Yg then the general solution of the nonhomogeneous ODE is given by y = yp (x) + Yg (x). If two linearly independent solutions of the homogeneous ODE in Y (x) can be obtained, the method of variation of parameters may be used to construct a particular solution of the nonhomogeneous ODE. Particular solutions of nonhomogeneous ODE may, however, be also found by guesswork, especially for cases in which the coeﬃcients f (x), g(x) and r(x) assume relatively simple forms.

3.6.1

Finding a particular solution by guesswork

Below are examples of how guesswork may be used to find particular solutions of some nonhomogeneous ODEs. Examples 1. Solve the ODE y00 + 3y0 + 2y = 2e5x subject to y(0) = y 0 (0) = 0. Firstly, let us solve the corresponding homogeneous ODE, that is, Y 00 + 3Y 0 + 2Y = 0. 72

This is a 2nd order linear ODE with constant coeﬃcients. Let us use Y = eλx ,

Y 0 = λeλx

and Y 00 = λ2 eλx .

Substitution into the homogeneous ODE gives λ2 + 3λ + 2 = 0 ⇒ (λ + 1)(λ + 2) = 0 ⇒ λ = −1, −2. The general solution of the homogeneous ODE is Yg = Ae−x + Be−2x where A and B are arbitrary constants. To write down the general solution of the nonhomogeneous ODE y 00 + 3y 0 + 2y = 2e5x , a particular solution of the ODE is required. The right hand side of the ODE given by the term 2e5x suggests that we seek a particular solution of the form yp = αe5x where α is a constant to be determined. Diﬀerentiating, we obtain yp0 = 5αe5x ,

yp00 = 25αe5x .

Substitution into the ODE gives 25αe5x + 15αe5x + 2αe5x = 2e5x 1 ⇒ 42αe5x = 2e5x ⇒ α = . 21 Hence, we obtain yp =

1 5x e 21

as a particular solution of the nonhomogeneous ODE.

73

The general solution of the nonhomogeneous ODE is y=

1 5x e + Ae−x + Be−2x 21

where A and B are arbitrary constants. Diﬀerentiating, we obtain y0 =

5 5x e − Ae−x − 2Be−2x . 21

Using y(0) = y 0 (0) = 0, we find that 1 21 5 0 y (0) = 0 ⇒ A + 2B = . 21

y(0) = 0 ⇒ A + B = −

Solving for A and B, we obtain B = 2/7 and A = −1/3. Thus, the required particular solution is y=

1 5x 1 −x 2 −2x e − e + e . 21 3 7

2. Find the general solution of the ODE y00 −y0 = − cos(2x). As before, let us first solve the corresponding homogeneous ODE, that is, Y 00 − Y 0 = 0. To find the general solution, use Y = eλx ,

Y 0 = λeλx

and Y 00 = λ2 eλx .

This gives λ2 − λ = 0 ⇒ λ = 0, 1. The general solution of the homogeneous ODE is Yg = A + Bex 74

where A and B are arbitrary constants. A particular solution of the nonhomogeneous ODE y 00 −y 0 = − cos(2x) is required. The right hand side of the ODE given by the term − cos(2x) suggests that we seek a particular solution of the form1 yp = α sin(2x) + β cos(2x) where α and β are constants to be determined. Diﬀerentiating, we obtain yp0 = 2α cos(2x) − 2β sin(2x)

yp00 = −4α sin(2x) − 4β cos(2x). Substituting into the nonhomogeneous ODE, we obtain −4α sin(2x) − 4β cos(2x) − 2α cos(2x) + 2β sin(2x) = − cos(2x) which may be rewritten as (−4α + 2β) sin(2x) − (2α + 4β) cos(2x) = − cos(2x). The equation above is satisfied for all x if −4α + 2β = 0

2α + 4β = 1.

Solving for α and β, we obtain α = 1/10 and β = 1/5. The required general solution of the nonhomogeneous ODE is thus given by y=

1 1 sin(2x) + cos(2x) + A + Bex 10 5

where A and B are arbitrary constants. 1

If we seek a particular solution having the simpler form of either yp = α sin(2x) or yp = α cos(2x), we find that the nonhomogeneous ODE cannot be satisfied, no matter how we choose the constant α.

75

3. Find the general solution of the ODE y 00 − y = 3ex . Firstly, let us solve the corresponding homogeneous ODE Y 00 − Y = 0. Proceeding as usual, we find that the general solution of the homogeneous ODE is given by Yg = Ae−x + Bex where A and B are arbitrary constants. A particular solution of the nonhomogeneous ODE is required. We seek one of the form2 yp = αxex where α is a constant to be determined. Diﬀerentiating, we obtain yp0 = αex + αxex yp00 = 2αex + αxex . Substituting into the nonhomogeneous ODE, we find that 2αex + αxex − αxex = 3ex 3 ⇒ 2αex = 3ex ⇒ α = . 2 Thus, a particular solution of the nonhomogeneous ODE is 3 yp = xex . 2 The required general solution of the nonhomogeneous ODE is 3 y = xex + Ae−x + Bex 2 where A and B are arbitrary constants. 2 Looking at the term on the right hand side of the ODE, that is, 3ex , we may be tempted to seek a particular solution of the simpler form yp = αex . However, this will not lead to success, as y = αex is a particular solution of the corresponding homogeneous ODE.

76

4. Find the general solution of the ODE x2 y 00 + 4xy 0 + 2y = 3x2 + 1. Firstly, let us solve the corresponding homogeneous ODE x2 Y 00 + 4xY 0 + 2Y = 0. We use Y = xλ , Y 0 = λxλ−1 , Y 00 = λ(λ − 1)xλ−2 . Substituting into the homogeneous ODE, we find that λ2 + 3λ + 2 = 0 ⇒ (λ + 1)(λ + 2) = 0 ⇒ λ = −1, −2. Thus, the general solution of the homogeneous ODE is Yg = Ax−1 + Bx−2 where A and B are arbitrary constants. A particular solution of the nonhomogeneous ODE is required. The right hand side of the ODE suggests that we seek one of the form yp = αx2 + βx + γ, where α, β and γ are constants to be determined. It follows that yp0 = 2αx + β, yp00 = 2α. Substituting into the nonhomogeneous ODE, we find that 12αx2 + 6βx + 2γ = 3x2 + 1. For the nonhomogeneous ODE to be satisfied, we take 12α = 3, 6β = 0 and 2γ = 1, that is, α = 1/4, β = 0 and γ = 1/2. Thus, a particular solution of the nonhomogeneous ODE is 1 1 yp = x2 + . 4 2 The required general solution is 1 1 y = x2 + + Ax−1 + Bx−2 4 2 where A and B are arbitrary constants. 77

3.6.2

Method of variation of parameters

The method of variation of parameters gives a systematic approach for finding a particular solution of the nonhomogeneous 2nd order linear ODE in (13). Let us examine the method from a general view point before applying it to a specific example. Assume that we can find two linearly independent solutions Y1 (x) and Y2 (x) of the corresponding homogeneous ODE in (14). The two solutions may be used to construct a particular solution yp of (13) as follows. Let yp take the form yp = u(x)Y1 (x) + v(x)Y2 (x) where u(x) and v(x) are functions to be determined. Diﬀerentiating, we obtain yp0 = uY10 + u0 Y1 + vY20 + v 0 Y2 yp00 = uY100 + 2u0 Y10 + u00 Y1 + vY200 + 2v0 Y20 + v00 Y2 . Substitution into (13) gives ¢ ¡ ¢ ¡ 0 0 2u Y1 + u00 Y1 + 2v 0 Y20 + v 00 Y2 + f (x) u0 Y1 + v 0 Y2 = r(x) since Y1 and Y2 are solutions of (14). The ODE above may be rewritten as ¢ ¡ ¢ ¡ ¢ d ¡ 0 u Y1 + v0 Y2 + u0 Y10 + v0 Y20 + f (x) u0 Y1 + v0 Y2 = r(x) dx which is satisfied if u0 Y1 + v0 Y2 = 0 u0 Y10 + v0 Y20 = r(x). Multiplying the first equation above by Y20 and the second by Y2 gives u0 Y1 Y20 + v0 Y2 Y20 = 0 u0 Y10 Y2 + v0 Y2 Y20 = r(x)Y2 (x). 78

The diﬀerence between the two equations above yields du r(x)Y2 (x) =− dx (Y1 Y20 − Y10 Y2 ) which may be integrated directly to give Z r(x)Y2 (x) dx. u(x) = − (Y1 Y20 − Y10 Y2 ) Similarly, we obtain dv r(x)Y1 (x) = dx (Y1 Y20 − Y10 Y2 ) and v(x) =

Z

r(x)Y1 (x) dx. (Y1 Y20 − Y10 Y2 )

The formulae above for u and v break down if Y1 Y20 − Y10 Y2 is identically zero. Theorem 1 (page 51), however, guarantees us that Y1 Y20 − Y10 Y2 is not identically zero as Y1 and Y2 are linearly independent. As Y1 and Y2 are assumed known, the functions u and v required in finding a particular solution of the nonhomogeneous ODE can be determined at least in theory. Example Solve the nonhomogeneous ODE y 00 − y = 3ex using the method of variation of parameters.

In the previous section, through guesswork, we find that the general solution of the ODE is given by 3 y = xex + Ae−x + Bex 2 where A and B are arbitrary constants. 79

Here we use the method of variation of parameters to find a particular solution for the ODE. Note that the corresponding homogeneous ODE Y 00 −Y = 0 has two linearly independent solutions given by Y1 = e−x and Y2 = ex . Hence Y1 Y20 − Y10 Y2 = e−x ex − (−e−x )ex = 2. According to the method of variation of parameters, a particular solution of the nonhomogeneous ODE is given by yp = u(x)e−x + v(x)ex with Z 3 3ex ex dx = − e2x u(x) = − 2 4 and v(x) =

Z

3ex e−x 3 dx = x. 2 2

Note that the constants of integration are ignored as we are only interested in a particular solution. Thus, yp = u(x)e−x + v(x)ex 3 3 3 3 = − e2x e−x + xex = − ex + xex . 4 2 4 2 The required general solution of the nonhomogeneous ODE is 3 3 y = − ex + xex + Ce−x + Dex 4 2 where C and D are arbitrary constants. The general solution above may be rewritten as 3 y = xex + Ae−x + Bex (as before) 2 where A = C and B = D − 3/4 are arbitrary constants. 80

3.7

Extension to higher order linear ODEs

Much of the discussion above for 2nd order linear ODEs can be extended to higher order linear ODEs. The extension is indicated below.

3.7.1

General N-th order linear ODEs

An N -th order linear ODE in y(x) may be written in the form y(N) + fN−1 (x)y (N −1) + · · · + f1 (x)y (1) + f0 (x)y = r(x) (15) where f0 , f1 , · · · , fN−2 , fN−1 and r are given real functions of x. If r(x) ≡ 0, the N -th order linear ODE is said to be homogeneous. Otherwise, it is nonhomogeneous.

3.7.2

General solution of a homogeneous ODE

Theorem 2 (page 53) for the general solution of 2nd order linear ODEs can be generalized as follows. If y1 , y2 , · · · , yN−1 and yN are linearly independent solutions of the homogeneous N-th order linear ODE y (N) + fN −1 (x)y (N−1) + · · · + f1 (x)y(1) + f0 (x)y = 0 (16) then the general solution of the ODE is y = A1 y1 + A2 y2 + · · · + AN −1 yN −1 + AN yN where A1 , A2 , · · · , AN−1 and AN are arbitrary constants. The functions y1 , y2 , · · · , yN−1 and yN are linearly independent if we cannot find any constants k1 , k2 , · · · , kN −1 and kN , other than k1 , k2 , · · · , kN−1 and kN all being zero, such that k1 y1 + k2 y2 + · · · + kN −1 yN −1 + kN yN ≡ 0. 81

The functions y1 = x2 , y2 = x and y3 = 1 are linearly independent because k1 x2 + k2 x + k3 cannot be identically zero for any constants k1 , k2 and k3 other than k1 = k2 = k3 = 0. The functions y1 = x2 + x + 1, y2 = x + 1, y3 = 2x2 and y4 = 3x5 are not linearly independent as we can find constants k1 , k2 , k3 and k4 (other than k1 = k2 = k3 = k4 = 0) such that k1 y1 + k2 y2 + k3 y3 + k4 y4 ≡ 0. For example, k1 = 2, k2 = −2, k3 = −1 and k4 = 0.

3.7.3

General solution of a nonhomogeneous linear ODE

If yp is a particular solution of the nonhomogeneous linear ODE in (15) and Yg is the general solution of the corresponding homogeneous linear ODE in (16), then the general solution of (15) is given by y = yp + Yg . Examples 1. Find the general solution of the ODE in y(x) given by y 000 − 2y 00 − 5y0 + 6y = 0.

This is a homogeneous linear ODE with constant coeﬃcients. Let y = eλx , y 0 = λeλx , y 00 = λ2 eλx , y000 = λ3 eλx . Substituting into the ODE, we obtain ¡ ¢ eλx λ3 − 2λ2 − 5λ + 6 = 0

that is,

(λ − 1)(λ + 2)(λ − 3) = 0 ⇒ λ = 1, λ = −2, λ = 3. 82

We obtain three linearly independent solutions, namely y1 = ex , y2 = e−2x and y3 = e3x . The required general solution is y = Aex + Be−2x + Ce3x where A, B and C are arbitrary constants.

2. Find the general solution of the ODE in y(x) given by y000 − 2y 00 − 5y 0 + 6y = 12.

It is easy to see that a particular solution of this nonhomogeneous ODE is yp = 2. The general solution of the corresponding homogeneous ODE is found in the example above. Thus the required general solution is y = 2 + Aex + Be−2x + Ce3x where A, B and C are arbitrary constants.

3.8

Exercise V

1. Find the general solution of each of the following ODEs in y(x): (a) y 00 − y 0 − 30y = 3e−3x (b) 9y 00 − 30y0 + 25y = x3 + 3x + 2 (c) y 00 − y0 − 30y = 3e−5x (d) y 00 − y 0 = − cos(2x) 2. Solve each of the following ODEs in y(x) subject to the given conditions: 83

(a) y 00 + 3y = 3x2 + 2x − 1, y(0) = 1, y0 (0) = 2 (b) y 00 − y 0 − 20y = x + ex , y(0) = 1, y0 (0) = 0 (c) x2 y00 + 5xy 0 + 8y = 5, y(1) = 0, y0 (1) = 1 (d) y 000 − 2y 00 − y 0 + 2y = 0, y(0) = y0 (0) = y00 (0) = 1 (e) y 000 − 2y 00 − y0 + 2y = e−3x , y(0) = y 0 (0) = y 00 (0) = 1 (f) y 0000 − y = 0, y(0) = 0, y0 (0) = y00 (0) = y 000 (0) = 1 3. In Chapter 1 (Figure 1.1, page 15), we show that the motion of an object falling under the influence of gravity with air resistance is governed by the 2nd order ODE s00 (t) +

k 0 s (t) = g m

where s(t) is the vertical distance of the object at time t from a fixed point P along the path of the motion, m is the mass of the object, k is a positive coeﬃcient associated with air resistance and g is the acceleration due to gravity. If k/m = 1/5 per second and g = 10 meter per second per second, the object is released from rest at the point P at time t = 0, and it takes 5 seconds to hit the ground, find the height of P above the ground. If there is no air resistance, how long will it take the object to hit the ground? Is your answer what you would expect to get? Explain. (Note. There are two ways to solve the ODE above. One way is to solve it directly as an nonhomogeneous 2nd order linear ODE with constant coeﬃcients using the appropriate method given in this chapter; the other is to convert it into a 1st order ODE by making an appropriate substitution and proceed as in Chapter 2.) 4. Consider the N -th order ODE in y(x) given by a0 y + a1 y (1) + · · · + aN−1 y (N−1) + aN y (N) = 0 84

where an are constants such that aN 6= 0 and such that the polynomial function defined by P (ξ) = a0 + a1 ξ + a2 ξ 2 + · · · + aN−1 ξ N−1 + aN ξ N can be written in the form P (ξ) = (ξ − c)2 Q(ξ) where Q(ξ) is a polynomial function of order N − 2 and c is a given real constant. (Assume that N ≥ 2.) (a) Show that y = ecx is a solution of the ODE above. (b) Show that y = xecx is also a solution of the ODE. (c) Use parts (a) and (b) above together with the generalization of Theorem 2 (as given on page 81) to solve y 0000 − 7y000 + 16y 00 − 12y 0 = 0 subject to y(0) = y0 (0) = y00 (0) = y 000 (0) = 1.

3.9

Solutions to Exercise IV

1. In parts (a), (b) and (c), we have homogeneous 2nd order linear ODEs with constant coeﬃcients. To find two linearly independent solutions, try y = eλx . For the EulerCauchy equations in parts (d), (e) and (f), try y = xλ . In each case, find value(s) of λ from a quadratic equation in λ. Final general solutions are given below. (The arbitrary constants are A and B.) (a) λ = 5 and λ = −6. General solution is y = Ae5x + Be−6x . (b) Only λ = 5/3. General solution is y = Ae5x/3 + Bxe5x/3 . 85

(c) λ = 3 ± 4i. General solution is y = e3x (A cos(4x) + B sin(4x)). (d) λ = −1 and λ = −3. General solution is y = Ax−1 + Bx−3 . (e) Only λ = 3. General solution is y = Ax3 + Bx3 ln(x). (f) λ = −2 ± 2i. General solution is y = x−2 [A cos(2 ln(x)) + B sin(2 ln(x))]. 2. (a) General solution of the ODE is y(x) = Ae−2x/3 + Be−x . Applying given conditions gives A + B = −1 2 − A−B = 3 3 which may be solved to give A = 6 and B = −7. Thus, required solution is y(x) = 6e−2x/3 − 7e−x . (b) General solution of the ODE is y(x) = x−2 (A + B ln(x)). Applying given conditions gives y(1)

=

A=1

y(e)

=

e−2 (A + B) = 2

⇒ B = 2e2 − 1. Thus, required solution is y(x) = x−2 (1 + [2e2 − 1] ln(x)). 86

3. Since yp is a solution, we can write: yp00 + f (x)yp0 + g(x)yp = 0. We have: y = yp y

0

=

yp0

Z

e−F (x) dx yp2

Z

e−F (x) e−F (x) dx + yp0 2 yp yp2

Z

y 00 = yp00

e−F (x) e−F (x) dx + yp2 yp

f (x)e−F (x) yp + yp0 e−F (x) yp2 Z −F (x) e e−F (x) = yp00 dx − f(x) . yp2 yp −

It follows that y 00 + f (x)y 0 + g(x)y =

(yp00

+ f (x)yp0

−f(x) = 0

Z

+ g(x)yp )

Z

e−F (x) dx yp2

e−F (x) e−F (x) + f (x) yp yp

e−F (x) dx + 0 = 0. yp2

Thus, we verify that y = yp

Z

e−F (x) dx yp2

is another solution of y 00 + f(x)y 0 + g(x)y = 0. 4. Substituting y = xp into the ODE, we find that 3 1 (p + )[xp + (p − )xp−2 ] = 0. 2 2 87

Thus, y = xp is a solution of the ODE for p = −1/2. (The above equation is true for all x only if p = −1/2.) According to Problem 3, another solution is given by y

=

−1/2

x

Z

2

e−x /2 dx x−1 1 2

⇒ y = −x−1/2 e− 2 x . Since the given ODE is homogeneous 2nd order linear ODE, we may use Theorem 2 to write down the general solution 1 2

y = Ax−1/2 − Bx−1/2 e− 2 x . 5. Let u(x) = −

y0 (x) . y(x)p(x)

It follows that u0 =

−ypy00 + y0 (y 0 p + yp0 ) . (yp)2

Substituting into the Ricatti equation, we find that −ypy 00 + y 0 (y 0 p + yp0 ) y0 2 y0 − p( + q ) −r =0 (yp)2 yp yp Multiplying the above by py, we obtain −ypy 00 + y 0 (y 0 p + yp0 ) − p(y 0 )2 + qy0 − rpy = 0 yp which may be simplified to give p0 0 y + qy0 − rpy = 0 p p0 ⇒ y 00 − ( + q)y 0 + pry = 0. p −y00 +

88

3.10

Solutions to Exercise V

1. (a) The corresponding homogeneous ODE has general solution Y (x) = Ae−5x + Be6x . For a particular solution of the nonhomogeneous ODE, let yp = αe−3x . Substituting yp into the nonhomogeneous ODE, we obtain 1 −18α = 3 ⇒ α = − . 6 Thus, required general solution is 1 y = − e−3x + Ae−5x + Be6x . 6 (b) The corresponding homogeneous ODE has general solution 5

Y (x) = (A + Bx)e 3 x . For a particular solution of the nonhomogeneous ODE, let yp = αx3 + βx2 + γx + δ. Substituting yp into nonhomogeneous ODE, we obtain 25αx3 + (−90α + 25β)x2 +(54α − 60β + 25γ)x + (18β − 30γ + 25δ)

= x3 + 3x + 2 Thus,

1 , − 90α + 25β = 0 25 54α − 60β + 25γ = 3, 18β − 30γ + 25δ = 2 α =

which can be solved to obtain β = 18/125, γ = 237/625 and δ = 1348/3125. 89

Required general solution is y (x) =

1348 237 18 2 + x+ x 3125 625 125 5 1 + x3 + (A + Bx)e 3 x . 25

(c) The corresponding homogeneous ODE has general solution Y (x) = Ae−5x + Be6x . For a particular solution of the nonhomogeneous ODE, let yp = cxe−5x . Substituting yp into nonhomogeneous ODE gives c = −3/11. Thus, required general solution is y=−

3 −5x xe + Ae−5x + Be6x . 11

(d) The corresponding homogeneous ODE has general solution Y (x) = A + Bex . For a particular solution of the nonhomogeneous ODE, let yp = α sin(2x) + β cos(2x). Substituting yp into nonhomogeneous ODE gives α = 1/10 and β = 1/5. Thus, required general solution is y=

1 1 sin(2x) + cos(2x) + A + Bex . 10 5

2. (a) The corresponding homogeneous ODE has general solution √ √ Y (x) = A cos( 3x) + B sin( 3x). 90

For a particular solution of the nonhomogneous ODE, let yp = αx2 + βx + γ. Substituting yp into nonhomogeneous ODE, we find that α = 1, β = 2/3 and γ = −1. Thus, general solution of the nonhomogeneous ODE is √ √ 2 y = x2 + x − 1 + A cos( 3x) + B sin( 3x). 3 Applying the condition y(0) = 1, we obtain A − 1 = 1, that is, y 0 (0) = 2 gives √ √ A = 2. The other condition 2/3 + B 3 = 2, that is, B = 4 3/9. Thus, required soliution is √ √ √ 2 4 3 2 sin( 3x). y = x + x − 1 + 2 cos( 3x) + 3 9 (b) The corresponding homogeneous solution has general solution Y (x) = Ae−4x + Be5x . For a particular solution, let yp = αx + β + γex . We find that α = −1/20, β = 1/400 and γ = −1/20. General solution of the nonhomogeneous ODE is y=−

1 1 1 x+ − ex + Ae−4x + Be5x . 20 400 20

Applying the given conditions y(0) = 1 and y 0 (0) = 0, we obtain A + B = 419/400 and −4A + 5B = 1/10 which can be solved to obtain A = 137/240 and B = 143/300. Thus, required solution is y (x) = −

1 137 −4x 143 5x 1 1 + x+ − ex + e e . 20 400 20 240 300

91

(c) The corresponding homogeneous solution has general solution Y (x) = [A cos(2 ln(x)) + B sin(2 ln(x))]x−2 . It is easy to see that a particular solution of the nonhomogeneous ODE is y = 5/8. General solution of nonhomogeneous ODE is y(x) =

5 + [A cos(2 ln(x)) + B sin(2 ln(x))]x−2 . 8

Applying the given conditions y(1) = 0 and y 0 (1) = 1, we obtain y(x) =

5 1 5 + [− cos(2 ln(x)) − sin(2 ln(x))]x−2 . 8 8 8

(d) Try y = eλx . Substituting into the given ODE, we obtain λ3 − 2λ2 − λ + 2 = 0 ⇒ λ = 1, −1, 2. General solution of ODE is y(x) = Aex + Be−x + Ce2x . Applying the given conditions y(0) = y 0 (0) = y 00 (0) = 1, we obtain A = 1 and B = C = 0. Thus, required solution y = ex . (e) Try Y = eλx to obtain the general solution of the corresponding homogeneous ODE as in part (d) above. We obtain Y (x) = Aex + Be−x + Ce2x . For a particular solution of the nonhomogeneous ODE, try y = αe−3x . Substituting into ODE, we find that α = −1/40. The general solution of the nonhomogeneous ODE is y(x) = −

1 −3x + Aex + Be−x + Ce2x . e 40

Applying the given conditions, we obtain A = 7/8, B = 1/12 and C = 1/15. Required solution of the nonhomogeneous ODE is y(x) = −

1 1 1 −3x 7 x + e + e−x + e2x . e 40 8 12 15 92

(f) Try y = eλx . This gives λ4 − 1

=

0 ⇒ λ2 = 1, λ2 = −1

⇒ λ = ±1, λ = ±i.

It follows that the general solution can be written as y(x) = Aex + Be−x + C cos(x) + D sin(x). Applying the given conditions gives the required solution 1 1 3 y(x) = ex − e−x − cos(x). 4 4 2 3. The ODE to solve is 1 s00 (t) + s0 (t) = 10. 5 The corresponding homogeneous ODE is 1 S 00 (t) + S 0 (t) = 0. 5 Try S(t) = eλt . This gives λ2 +λ/5 = 0, that λ = 0, −1/5. Thus, the general solution of the homogeneous ODE is S(t) = A + Be−t/5 . For a particular solution of the original nonhomogeneous ODE, try s(t) = αt + β. This gives α = 50. We can let β = 0. Thus, the general solution of the nonhomogeneous ODE is s(t) = 50t + A + Be−t/5 . The conditions to use are s(0) = 0 and s0 (0) = 0 (the body is at rest at the point P at time t = 0 and s(t) is the distance from P ). Applying the conditions gives A = −250 and B = 250. Thus, the distance s(t) is given by s(t) = 50t − 250 + 250e−t/5 . 93

The height of P from the ground is given by s(5) = 250e−1 meter. If there is no air resistance, k/m = 0. The ODE becomes s00 (t) = 10. This can be easily solved to give the general solution s(t) = 5t2 + Ct + D. Applying the conditions s(0) = 0 and s0 (0) = 0, we obtain s(t) = 5t2 . If there is no air resistance, the time t taken to reach the ground is obtained by solving 5t2 = 250e−1 ,that is, t ' 4.3 second. The body hits the ground quicker than when there is air resistance. This is expected. 4. (a) If y(x) = ecx then y 0 (x) = cecx , y 00 (x) = c2 ecx , · · · , y (N) (x) = cN ecx . The LHS of the given ODE can then be written as LHS of ODE = a0 y + a1 y(1) + a2 y (2) + · · · + aN−1 y (N−1) + aN y (N) = (a0 + a1 c + a2 c2 + · · · + aN−1 cN −1 + aN cN )ecx = P (c)ecx

= (c − c)2 Q(c)ecx = 0 = RHS of ODE. Thus, y = ecx is a solution of the given ODE. (b) From y = xecx , we find that y 0 = cxecx + ecx = ecx (1 + cx) y 00 = c2 xecx + cecx + cecx = ecx (2c + c2 x) y000 = c3 xecx + c2 ecx + 2c2 ecx = ecx (3c2 + c3 x) .. . y (N) = ecx (N cN−1 + cN x). Substitute into the LHS of ODE gives LHS of ODE = xecx (a0 + a1 c + a2 c2 + · · · + aN xN )

+ecx (a1 + 2ca2 + 3c2 a3 + · · · + N cN−1 aN )

= xecx P (c) + ecx P 0 (c) 94

From P (ξ) = (ξ − c)2 Q(ξ), we find that P (c) = 0 and P 0 (ξ)

=

2(ξ − c)Q(ξ) + (ξ − c)2 Q0 (ξ)

⇒ P 0 (c) = 0. It follows that LHS of ODE

= xecx P (c) + ecx P 0 (c) = 0 = RHS of ODE.

Thus, y = xecx is also a solution of the ODE. (c) To solve y 0000 − 7y000 + 16y 00 − 12y 0 = 0, try y = eλx . This gives λ4 − 7λ3 + 16λ2 − 12λ = 0

⇒ λ(λ3 − 7λ2 + 16λ − 12) = 0 ⇒ λ (λ − 3) (λ − 2)2 = 0

⇒ λ = 0, λ = 2 (twice), λ = 3.

Thus, we obtain 4 particular solutions: y = e0x (or y = 1), y = e2x , y = xe2x and y = e3x . We form the general solution y = A + Be2x + Cxe2x + De3x . Applying the conditions y(0) = y0 (0) = y 00 (0) = y000 (0) = 1, we find that A = 1/6, B = 1/2, C = −1 and D = 1/3. Thus, required solution is y (x) =

1 1 2x 1 + e − xe2x + e3x . 6 2 3

95

Chapter 4

Circuits and springs 4.1

Preamble

In earlier chapters, we show how ODEs arise in the formulation of various problems, such as the motion of bodies and population growth. In this chapter, we examine two specific problems in physical and engineering sciences which are governed by linear ODEs with constant coeﬃcients. One concerns the flow of electric current in circuits and the other vibrational motion in spring-mass systems. In both problems, we state the laws of physics governing the behaviors of the systems and show how they may be mathematically formulated in terms of ODEs.

4.2 4.2.1

Electric circuits Basic electrical components

Basic components which may be found in an electric circuit include resistor, inductor and capacitor. The resistor is a device which restricts the flow of electric current to within a safe level in the circuit. The inductor is essentially a coil of wire (usually copper) which stores energy in a magnetic field. The capacitor stores energy in an electric field in between two oppositely charged plates. In addition to the components just mentioned, 96

an external source voltage (for example, a battery) may also be present in the circuit. The external source voltage provides an input voltage which helps to initiate the flow of electric current in the circuit.

Figure 4.1 The symbols which we use to represent external source voltage, resistor, inductor and capacitor are shown in Figure 4.1. A simple circuit which contains each of the basic components mentioned, that is, the so called LCR circuit, is shown at the bottom of Figure 4.1. Circuits may contain less components than the LCR circuit shown in Figure 4.1 or they may be more complicated being made up of many simple circuits joined together to form a network.

97

4.2.2

Voltage across an electric component

The voltage across a resistor or inductor or capacitor may be measured by using a voltmeter. Relations between the voltage and the electric current flowing across these components are well established. If there is a current I (in ampere) flowing through a resistor, then the voltage across the resistor, VR (in volt), is related to the current by VR = RI where R ≥ 0 is the resistance (in ohm). The coeﬃcient R is a measure of the resistance put up by the resistor against the flow of current. If a resistor is made up of a material which conducts electricity better, it has a lower value of R. In theory, it is possible for R to be zero. In practice, however, all materials show some resistance (even if only an extremely small amount) to the flow of current. In general, the current I flowing in a circuit varies with time t, that is, I is a function of t. The voltage across the inductor, VL , is related to the time rate of change of current flowing through it by VL = L

dI dt

where L ≥ 0 is the inductance (in henry). The voltage across a capacitor, VC , is related to the current passing through it by 1 dVC = I(t) dt C where C > 0 is the capacitance (in farad). It is useful to rewrite the above relation as Z 1 t VC (t) = I(τ )dτ + VC (0). C 0 Note that VC (0) is the voltage across the capacitor at time t = 0. If the capacitor is not charged at time t = 0 then VC (0) = 0. For our purpose here, we take L, C and R as constants. 98

4.2.3

ODEs in electric circuits

The flow of current in an electric circuit may be described in terms of ODEs. ODEs for electric circuits are obtained from the Kirchoﬀ’s voltage law which may be stated as follows. For any closed loop of circuit (such as the simple LCR circuit shown in Figures 4.1) where the current flows in only one direction (that is, either clockwise or counter-clockwise direction), the total sum of the voltages across all resistors, inductors and capacitors present is equal to the total input voltages supplied by external sources. Kirchoﬀ’s voltage law is applicable for any closed loop of circuits, no matter how many components they contain. We now apply the law to some electric networks. Another law known as Kirchoﬀ’s current law is also useful in circuit analysis. This law implies the conservation of charge and may be stated as follows. Let P be any point or node in an electric circuit. If Iin and Iout denote respectively the electric current entering and leaving the point P then Iin = Iout . Thus, according to the Kirchoﬀ’s current law, the electric current in a simple circuit (such as the ones in Figures 4.1 and 4.2) is the same at all points (in every part of the circuit). Examples 1. Consider the simple LR circuit in Figure 4.2. The resistance of the resistor is R = 2 ohm and the inductance of the inductor is 3 henry. The input voltage supplied by the external source at time t (in second) is V (t) = 2 + e−t volt. If the current at time t = 0 is 1 ampere, find the current at t = 2 second (correct to 3 significant places). Predict what happens to the current in the circuit in the long run. 99

Figure 4.2 As mentioned earlier on, the electric current is the same in every part of the circuit. If we denote the electric current by I(t) then dI . dt According to Kirchoﬀ’s voltage law, VR + VL equals the input voltage supplied by the external source. Thus, VR = 2I

and VL = 3

dI + 2I = 2 + e−t . dt This is a 1st order nonhomogeneous linear ODE. We solve it now by the method of variation of parameter. Firstly, we find a particular solution of the corresponding homogeneous ODE 3

dJ + 2J = 0. dt Solving the ODE, we obtain Z Z 2 dJ =− dt J 3 2 ⇒ ln(J) = − t 3 ⇒ J = e−2t/3 . 3

100

Note that the constant of integration is ignored as we are only interested in finding a particular solution. A particular solution of the homogeneous ODE is J = e−2t/3 . To find the general solution of the original nonhomogeneous ODE, we use I = u(t)e−2t/3 dI 2 ⇒ = − u(t)e−2t/3 + u0 (t)e−2t/3 . dt 3 Substituting into the nonhomogeneous ODE, we obtain −2u(t)e−2t/3 + 3u0 (t)e−2t/3 + 2u(t)e−2t/3 = 2 + e−t . Rearranging, we find that Z Z 3 du = [2e2t/3 + e−t/3 ]dt ⇒ u = e2t/3 − e−t/3 + C

where C is an arbitrary constant. The required general solution is I(t) = (e2t/3 − e−t/3 + C)e−2t/3 = 1 − e−t + Ce−2t/3 .

We are given I(0) = 1. Thus, 1 = 1 − e0 + Ce0 ⇒ C = 1. The current flowing in the circuit is thus given by I(t) = 1 − e−t + e−2t/3 . The current at t = 2 second is given by I(2) = 1 − e−2 + e−4/3 ' 1.128 ampere. To find out what happens to the current in the long run, we let t → ∞ in I(t), that is, we examine the limit lim I(t) = lim [1 − e−t + e−2t/3 ] = 1.

t→∞

t→∞

Thus, in the long run, we expect the current to get closer and closer to a constant value of 1 ampere.

101

2. Consider the LCR circuit in Figure 4.1. If the resistance, inductance and capacitance are respectively R = 3 ohm, L = 1 henry and C = 12 farad and the input voltage supplied by the external source is V (t) = sin(t) volt, write down a 2nd order ODE to describe the flow of current in the circuit. At time t = 0, the current is 2 ampere and the capacitor is not charged (that is, VC (0) = 0). Find the current at any time t ≥ 0. The voltages across the resistor, inductor and capacitor are respectively Z t dI VR = 3I, VL = I(τ )dτ . and VC = 2 dt 0 According to Kirchoﬀ’s voltage law, VR + VL + VC = sin(t). Thus Z t dI I(τ )dτ = sin(t). + 3I + 2 dt 0 Such an equation is called an integro-diﬀerential equation. The integro-diﬀerential equation may be easily converted into a 2nd order ODE by diﬀerentiating it with respect to t once and using Liebnitz’s rule1 , that is, Z t d d dI [ + 3I + 2 [sin(t)] I(τ )dτ ] = dt dt dt 0 ⇒ I 00 (t) + 3I 0 (t) + 2I(t) = cos(t). The 2nd order ODE is nonhomogeneous. To solve it, let us solve the corresponding homogeneous ODE first, that is, J 00 (t) + 3J 0 (t) + 2J(t) = 0. 1

In a nutshell, the Leibnitz’s rule says: Z x d f (u)du = f (x). dx 0

102

Let us use J = eλt , J 0 = λeλt , J 00 = λ2 eλt . This leads to λ2 + 3λ + 2 = 0 ⇒ (λ + 1)(λ + 2) = 0 ⇒ λ = −1, −2. The general solution of the homogeneous ODE is J = Ae−t + Be−2t where A and B are arbitrary constants. Let us now find a particular solution of the original nonhomogeneous ODE. The right hand side of the ODE suggests that we try I = α sin(t) + β cos(t) where α and β are constants to be determined. Diﬀerentiating, we obtain I 0 = α cos(t) − β sin(t)

I 00 = −α sin(t) − β cos(t). Substituting into the nonhomogeneous ODE, we obtain −α sin(t) − β cos(t) + 3α cos(t) − 3β sin(t)

+2α sin(t) + 2β cos(t) ≡ cos(t)

which leads to [α − 3β] sin(t) + [3α + β] cos(t) ≡ cos(t). The nonhomogeneous ODE is satisfied if we choose α and β satisfying α − 3β = 0 and 3α + β = 1, 103

that is, α=

1 3 and β = . 10 10

Thus, a particular solution of the nonhomogeneous ODE is I=

3 1 sin(t) + cos(t), 10 10

and the required general solution is I(t) =

3 1 sin(t) + cos(t) + Ae−t + Be−2t 10 10

where A and B are arbitrary constants. We are given I(0) = 2. To determine the two arbitrary constants A and B, another condition is required. This may be obtained by letting t = 0 in the original integro-diﬀerential equation. We obtain the other condition as I 0 (0) = −6 as follows. ¯ Z 0 dI ¯¯ + 3I(0) + 2 I(τ )dτ = sin(0) dt ¯t=0 0 ⇒ I 0 (0) = −3I(0) = −6 Diﬀerentiating the general solution, I 0 (t) =

3 1 cos(t) − sin(t) − Ae−t − 2Be−2t . 10 10

We may now determine A and B. We find that 1 19 = 10 10 3 63 0 I (0) = −6 ⇒ −A − 2B = −6 − =− 10 10 I(0) = 2 ⇒ A + B = 2 −

which leads to A=−

5 22 and B = . 2 5

Hence, the current flowing in the circuit is given by I(t) =

3 22 1 5 sin(t) + cos(t) − e−t + e−2t . 10 10 2 5 104

3. Consider the electric network in Figure 4.3. Write down a system of 3 equations for the electric currents I1 , I2 and I3 . Explain briefly how these equations can be solved for I1 , I2 and I3 , without actually solving them.

Figure 4.3 To analyze the network, we break it up into two simple circuits as shown in Figure 4.4 and apply Kirchoﬀ’s voltage law separately on each of the circuits. Applying Kirchoﬀ’s voltage law on the each of the circuits in Figure 4.4, we obtain Z t 2I3 + I1 (τ )dτ + VC (0) = 0 0

dI2 − 2I3 = 9 dt

where VC (0) is the voltage across the capacitor at time t = 0. We reverse the direction of the current I3 in the circuit at the bottom of Figure 4.4, as Kirchoﬀ’s voltage is applicable to a closed loop of a circuit where the current flows in only one direction only. (Note that −I3 gives the current having the same magnitude as I3 but flowing in the opposite direction.) 105

Figure 4.4 The first equation above is an integro-diﬀerential equation in 2 unknowns, namely I1 and I3 . The second equation is a 1st order ODE in I2 and I3 . Since there are three unknowns I1 , I2 and I3 , a third equation is required to completely describe the flow of current in the electric network. Another equation may be obtained by applying Kirchoﬀ’s curent law at the point (node) P in Figure 4.3. The total amount of electric current entering P is I2 +I3 , while the electric current leaving P is I1 . Thus, I1 = I2 + I3 . This is the required third equation. How do we solve this system of 3 equations for I1 , I2 and I3 ? Diﬀerentiating the integro-diﬀerential equation with respect to t, we obtain 2

dI3 + I1 (t) = 0. dt 106

This may be rewritten as dI3 + I2 (t) + I3 (t) = 0 dt

2

since I1 = I2 + I3 . Diﬀerentiating once more with respect to t, we obtain 2

d2 I3 dI2 dI3 + + = 0. dt2 dt dt

Now, from the ODE dI2 /dt − 2I3 = 9, that is, dI2 /dt = 9 + 2I3 , we obtain the 2nd order ODE 2

dI3 d2 I3 + 9 + 2I3 + =0 dt2 dt

which may be rearranged to give 2

d2 I3 dI3 + 2I3 = −9 + dt2 dt

Once I3 is determined from the 2nd order ODE, we may determine I1 from the ODE I1 (t) = −2dI3 /dt. The remaining unknown I2 may then be found from I1 = I2 + I3 . 4. Consider once again the simple LR circuit in Figure 4.2. Take the external source voltage V (t) to be given by the piece-wise function ⎧ ⎨ t for 0 ≤ t < 1 1 for 1 ≤ t < 2 V (t) = ⎩ 0 for t ≥ 2. If the electric current I is zero at time t = 0, find I(t) for t ≥ 0.

According to Kirchoﬀ’s voltage law, the ODE for the electric current in the circuit is given by 3

dI + 2I = V (t). dt 107

From the given external source voltage V (t), it is obvious that we have to solve diﬀerent ODEs over the time intervals 0 ≤ t < 1, 1 ≤ t < 2 and t ≥ 2. The first ODE to solve is 3

dI + 2I = t for 0 ≤ t < 1. dt

Solving the above ODE subject to the condition I(0) = 0, we obtain 1 3 3 I(t) = t − + e−2t/3 . 2 4 4 This gives 1 3 I(1) = − + e−2/3 . 4 4 The second ODE 3

dI + 2I = 1 for 1 ≤ t < 2 dt

is to be solved subject to I(1) as computed above. We obtain I(t) =

1 3 + (1 − e2/3 )e−2t/3 for 1 ≤ t < 2 2 4

and I(2) =

1 3 + (1 − e2/3 )e−4/3 . 2 4

Lastly, we solve 3

dI + 2I = 0 for t ≥ 2 dt

subject to I(2) as computed above. We obtain 1 3 I(t) = [ e4/3 + (1 − e2/3 )]e−2t/3 for t ≥ 2. 2 4

108

4.3

Exercise VI

1. Consider the LCR circuit in Figure 4.1. If L = 1 (henry), C = 1/13 (farad) and R = 6 (ohm), the input (external source) voltage is V (t) = 2 sin(2t) (volt), the current at time t = 0 is 0 (ampere) and the capacitor is not charged at t = 0, find the current at time t = 2 (second). 2. Repeat Problem 1 with the input voltage given by ½ t for 0 ≤ t < π/4 V (t) = 0 for t ≥ π/4. 3. Find the currents I1 , I2 and I3 in Figure 4.3 (page 105), if it is known that I1 (0) = I2 (0) = I3 (0) = 0 and the capacitor is not charged at time t = 0. 4. A simple RC circuit is made up of a resistor, a capacitor and an external source voltage. If the external source voltage provides an input voltage which does not change with time t and if the resistance R of the resistor and the capacitance C of the capacitor are such that the RC = 1 (ohm farad), find the time taken for the magnitude of the current to be reduced to 20% of its magnitude at t = 0. 5. A simple LC circuit is made up of an inductor, a capacitor and an external source voltage. If the external source voltage provides a constant input voltage V0 and if the capacitance C of the capacitor and the inductance L of the inductor are such that LC = 1 (ohm farad) and V0 C = 1 (volt farad), find the current flowing in the circuit, given that at t = 0 there is no current and the capacitor is not charged.

4.4 4.4.1

Spring-mass systems A simple spring-mass system

We now look at the motion of a body (mass) attached to a spring. The spring is hung vertically from a ceiling. A sketch 109

of the situation is given in Figure 4.5. We assume that the body moves vertically up and down. There is no sideway motion. We measure the position of the body at time t by y(t). According to Figure 4.5, y is the downward displacement of the body from a fixed reference position, that is, y > 0 if the body is below the reference position and y < 0 if it is above.

Figure 4.5 Hooke’s law The spring is said to be in its ‘natural’ condition when it does not exert any force on the body. For convenience, we assume that the spring is in its ‘natural’ condition when the body is at the fixed reference position, that is, when the downward displacement y of the body is zero. Let the vertical length of the spring in its ‘natural’ condition be `natural . If the vertical length of the spring is changed to `, the spring exerts a force (which we call the ‘spring force’) on the body. According to Hooke’s law, the magnitude of the spring force is proportional to |` − `natural |. What is the direction of the spring force? For the spring-mass system in Figure 4.5, when ` > `natural (the spring extends in length), that is, when the body is below the fixed reference position (y > 0), the 110

spring force acts upward on the body. On the other hand, when ` < `natural (the spring compresses in length), that is, when y < 0, the spring force acts downward on the body. In short, for the system in Figure 4.5, the direction of the spring force on the body is always opposite to that of the displacement y. Mathematically, the downward spring force2 on the body in the spring-mass system in Figure 4.5 is given by ½ μ|` − `natural | if ` < `natural Fspring = −μ|` − `natural | if ` ≥ `natural where μ > 0 is the spring coeﬃcient (assumed to be a constant here). The constant μ depends on the strength of the spring. It is larger in magnitude for stronger springs. Since y is the downward displacement of the body, we find that y = −|` − `natural | when ` < `natural , and y = |`−`natural | when ` ≥ `natural . Thus, for the spring-mass system in Figure 4.5, if the fixed reference position is where the body is when the spring does not exert any force, then Hooke’s law tells us that the downward force exerted by the spring on the body is given by Fspring = −μy. ODE for the spring-mass system If we can identify all the forces acting on the body attached to the spring (the spring force is one of them), we may use Newton’s law of motion to form an ODE to describe the motion of the body. Apart from the spring force, the body may experience a damping force as it moves through a medium such as air or viscous fluid. The downward force due to damping by the medium, Fdamping , may be given by Fdamping = −k 2

dy dt

When we say that a force F is “downward,” we mean that the force is pointing downward if F > 0 and pointing upward if F < 0.

111

where k > 0 is the damping coeﬃcient. (If the spring-mass system is placed in vacuum, k = 0.) Note when the body is moving downward, dy/dt > 0 and Fdamping < 0, that is, the damping force is in the upward direction. When the body moves upward, dy/dt < 0 and Fdamping > 0, that is, the damping force is acting downward. As we will see, the presence of damping force helps to kill the motion of the body. Another force acting on the body is the gravitational force. This always acts downward. The downward gravitational force acting on the body, Fgravity , is given by Fgravity = mg where g is the acceleration due to gravity (g ' 9.81 meter per second per second around planet earth) and m is the mass of the body. We assume that there are no other types of forces acting on the body. Thus, the total downward force acting on the body is given by Fspring + Fdamping + Fgravity = −μy − k

dy + mg. dt

Newton’s law of motion tells us that (total downward force acting on body) = (mass of body) × (downward acceleration of body) d2 y = m 2. dt It follows that d2 y dy + mg = m 2 dt dt d2 y k dy μ ⇒ 2 + + y = g. dt m dt m −μy − k

We obtain a 2nd order linear ODE in y(t). The ODE is nonhomogeneous if g 6= 0. 112

Solution of the ODE We now solve the ODE k dy μ d2 y + + y=g 2 dt m dt m for the spring-mass system in Figure 4.5. We first solve the corresponding homogeneous ODE, that is, d2 Y μ k dY + Y = 0. + 2 dt m dt m It is a homogeneous 2nd order linear ODE with constant coefficients. Let us use Y = eλt , Y 0 = λeλt , Y 00 = λ2 eλt . Substituting into the homogeneous ODE, we find that k μ λ+ =0 m pm −k ± k2 − 4mμ . ⇒ λ= 2m λ2 +

The general solution for the homogeneous ODE is √ 2 √ 2 Y (t) = Ae(−k+ k −4mμ)t/(2m) + Be(−k− k −4mμ)t/(2m) where A and B are arbitrary constants. A particular solution of the original nonhomogeneous ODE is mg . y(t) = μ This may be easily verified by substitution into the ODE. Thus, the required general solution of the nonhomogeneous ODE is √ 2 √ 2 mg y(t) = + Ae(−k+ k −4mμ)t/(2m) + Be(−k− k −4mμ)t/(2m) μ

113

where A and B are arbitrary constants. If the displacement y(t) and the velocity y 0 (t) of the body are known at time t = 0, we may determine the constants A and B. Even without knowing A and B, we may make certain deductions about the motion of the body attached to the spring. If k = 0, that is, in a vacuum (absence of damping), the displacement becomes √ √ mg + Aeit μ/m + Be−it μ/m y(t) = μ r r μ μ mg = + C cos( t) + D sin( t). μ m m In this case, y does not tend to any fixed value as t → ∞. Because of the real sine and cosine terms in time t, the body keeps on oscillating forever if there is no damping to the motion. If k2 < 4mμ, we find that the displacement may be rewritten as p |k 2 − 4mμ| t mg y(t) = + Ce−kt/(2m) cos( ) μ 2m p |k 2 − 4mμ| t ). +De−kt/(2m) sin( 2m

For the case k2 < 4mμ, if k 6= 0, we find thar y → mg/μ (a constant) as t → ∞. The presence of the real sine and cosine terms (in time) shows that the body does oscillate up and down but the oscillation becomes smaller and smaller as time goes by. The spring-mass system is said to be damped. p For k2 ≥ 4mμ, we observe that the numbers −k± k 2 − 4mμ are real and negative. Hence, for this case, the displacement as given by √ √ mg (−k+ k2 −4mμ)t/(2m) (−k− k2 −4mμ)t/(2m) y(t) = + Be + Ae μ also tend to mg/μ as t → ∞. There is, however, no sine and cosine term in the solution. The body does not oscillate. The motion dies down rapidly without any oscillation. The springmass system is said to be overdamped. 114

4.4.2

A more complicated spring-mass system

Consider now a more complicated spring-mass system which consists of two vertical springs and two bodies connected as shown in Figure 4.6. Let us assume that the only forces acting on the masses are due to the springs. Gravitational and damping forces are ignored. It is not diﬃcult, however, to incorporate them into the analysis, if necessary.

Figure 4.6

The two bodies are in motion here. We expect a system of two ODEs to describe the motion. Let y1 (t) and y2 (t) be the downward displacement of body 1 and 2 respectively from selected fixed reference positions. Refer to Figure 4.6. For convenience, the fixed reference position for each body is chosen to be where it (the body) is when the springs do not exert any force (that is, when the springs are in their ‘natural’ condition).

115

Newton’s law of motion gives d2 y1 dt2 d2 y2 (total downward spring forces on body 2) = m2 2 . dt

(total downward spring forces on body 1) = m1

Here m1 and m2 are the mass of body 1 and 2 respectively. What are the total spring forces acting on the masses? For our analysis, let us assume that y1 > 0 and y2 > 0, that is, both bodies are below their fixed positions. (This assumption is made only for the purpose of providing a clear explanation. The final outcome of the analysis is the same whether the assumption holds or not.) It is clear that as body 1 displaces itself below its fixed reference position, it extends spring 1 but compresses spring 2. On the other hand, as mass 2 displaces itself below its fixed reference position, it extends spring 2. Thus, spring 1 increases in length by y1 . The net increase in length of spring 2 is y2 − y1 . (If y2 > y1 then there is a positive increase in the length of spring 2. If y2 < y1 , there is a negative increase in length, that is, spring 2 shortens in length.) Let the spring coeﬃcient of body 1 and 2 be denoted by μ1 and μ2 respectively. The downward force due to spring 1 acting on body 1 is −μ1 y1 . (For y1 > 0, this is negative, that is, upward, as expected.) The downward spring force due to spring 2 on body 1 is μ2 (y2 −y1 ). (To see this, let us consider the case y2 > y1 . For this case, spring 2 extends in length. When the length of spring 2 increases, it exerts a downward force on body 1. Downward force is positive, hence μ2 (y2 − y1 ) > 0 for y2 > y1 .) According to Newton’s law of motion, body 1 gives −μ1 y1 + μ2 (y2 − y1 ) = m1

d2 y1 dt2

(1)

an ODE with 2 unknowns y1 and y2 . The only force acting on body 2 is due to spring 2. The downward spring force due to spring 2 on body 2 is given by −μ2 (y2 − y1 ). (For y2 > y1 , spring 2 increases in length and we expect the spring force due to spring 2 to point upward. 116

This is correct, as −μ2 (y2 − y1 ) is negative for y2 > y1 .) Thus, Newton’s law for body 2 gives −μ2 (y2 − y1 ) = m2

d2 y2 dt2

(2)

another ODE in y1 and y2 . How do we solve this system of ODEs? One way is to convert it to a single higher order ODE in only one unknown as follows. If we divide (2) by m2 and (1) by m1 , the diﬀerence between the two resulting ODEs gives −μ2 (

1 1 μ d2 Y + )Y + 1 y1 = 2 m2 m1 m1 dt

(3)

where Y = y2 − y1 . If we diﬀerentiate (3) twice with respect to t, we obtain −μ2 (

1 1 d2 Y μ d2 y1 d4 Y + ) 2 + 1 = m2 m1 dt m1 dt2 dt4

which, by the use of (1), may be converted to −μ2 (

1 1 d2 Y μ μ μ d4 Y + ) 2 + 1 (− 1 y1 + 2 Y ) = 4 . m2 m1 dt m1 m1 m1 dt

Further use of the (3) to eliminate y1 gives a homogeneous 4th order linear ODE in Y (t), that is, Y 0000 + αY 00 + βY = 0 where 1 1 μ + )+ 1 m2 m1 m1 μ1 μ2 1 1 μ μ ( + ) − 1 22 . m1 m2 m1 m1

α = μ2 ( β =

The general solution of the ODE in Y may be obtained by letting Y = eλt 117

where λ is a constant to be determined from λ4 + αλ2 + β = 0. If the quartic equation in λ gives 4 distinct solutions, that is, λ = λ1 , λ = λ2 , λ = λ3 and λ = λ4 , then the required general solution is given by Y (t) = Aeλ1 t + Beλ2 t + Ceλ3 t + Deλ4 t where A, B, C and D are arbitrary constants. For further details, refer to Chapter 3. Once Y is found, y1 may be determined directly out from (3). Subsequently, y2 = Y + y1 may be obtained. The 4 arbitrary constants in y1 and y2 can be found if we know, say, y1 (0), y10 (0), y2 (0) and y20 (0) which gives the state of motion of the two bodies at t = 0.

4.5

Exercise VII

1. Consider the spring-mass system in Figure 4.5. The only forces acting on the body are the spring force, damping force and gravity, and the spring force is 0 when the displacement y = 0, as discussed. The damping coeﬃcient, the spring coeﬃcient and the acceleration due to gravity are denoted by k, μ and g respectively. If k/m = 2 (per second), μ/m = 10 (per second per second), g = 10 (meter per second per second) and the displacement and velocity are both zero at time t = 0, find the displacement and velocity at time t = 1 (second). 2. An electric motor is attached to the body in Figure 4.5. The motor exerts a vertical force on the body. The downward force due to the motor is given by Fmotor = 10m cos(t) (newton), where m is the mass (kilogram) of the body. The only other force acting on the body is the spring force. (Gravity and damping are ignored.) The spring force is 0 when the displacement y is 0. The spring coeﬃcient is denoted by μ. Write down an ODE to describe the 118

motion of the body. If μ/m = 1 (per second per second) and the displacement and velocity of the body are both 0 at time t = 0, find the displacement and velocity of the body at time t = 1 (second). 3. As in Problem 2, an electric motor is attached to the body in Figure 4.5. The downward force exerted by the motor on the body is given by Fmotor = −4m cos(2t) (newton). The spring force is 0 when the displacement y is 0. Take the damping coeﬃcient k and the spring coeﬃcient μ to be given by k/m = 6 (per second) and μ/m = 13 (per second per second) respectively. Ignore the eﬀect of gravity on the motion of the body. If the displacement and velocity of the body are both 0 at t = 0, find the displacement at t = 2 (second). 4. Taking m1 = 1, m2 = 1, μ1 = 6 and μ2 = 4, solve the ODEs for the spring-mass system in Figure 4.6 subject to y1 (0) = y2 (0) = 1 and y10 (0) = y20 (0) = 0.

4.6

Solutions to Exercise VI

1. The integro-diﬀerential equation for the LCR circuit is Z t dI 6I + I(τ )dτ = 2 sin(2t) + 13 dt 0 which may be diﬀerentiated to obtain the nonhomogeneous 2nd order linear ODE 6

d2 I dI + 2 + 13I = 4 cos(2t). dt dt

We are given I(0) = 0. From the integro-diﬀerential equation, we obtain I 0 (0) = 0. Thus, the nonhomogeneous 2nd order linear ODE is to be solved subject to I(0) = I 0 (0) = 0. The corresponding homogeneous ODE is 6

dJ d2 J + 2 + 13J = 0. dt dt 119

The general solution of the homogeneous ODE (obtained by trying J = eλt ) is J(t) = e−3t (A cos(2t) + B sin(2t)). To obtain a particular solution of the nonhomogeneous ODE, try I(t) = α sin(2t) + β cos(2t). Substituting into the ODE gives α = 16/75 and β = 4/25. Thus, the general solution of the ODE is I(t) =

16 4 sin(2t) + cos(2t) 75 25 +e−3t (A cos(2t) + B sin(2t)).

Applying the conditions I(0) = I 0 (0) = 0 gives A = −4/25 and B = −34/25. The current is 4 16 sin(2t) + cos(2t) 75 25 4 34 +e−3t (− cos(2t) − sin(2t)). 25 75

I(t) =

The current at time t = 2 second is 16 4 sin(4) + cos(4) 75 25 4 34 +e−6 (− cos(4) − sin(4)) 25 75 ' 0. 264 9 ampere.

I(2) =

2. The integro-diﬀerential equation for the LCR circuit is Z t dI 6I + + 13 I(τ )dτ = V (t) dt 0 which may be diﬀerentiated to give 6

d d2 I dI + 2 + 13I = [V (t)]. dt dt dt

Since the voltage is given by ½ t for 0 ≤ t < π/4 V (t) = 0 for t ≥ π/4 120

and hence d [V (t)] = dt

½

1 for 0 ≤ t < π/4 0 for t ≥ π/4

the task of finding the current has to be divided into two parts as follows. In the first part, the ODE to solve is 6

dI d2 I + 2 + 13I = 1. dt dt

The above ODE is valid for 0 ≤ t < π/4. At time t = 0, we are given I(0) = 0. From the above integro-diﬀerential equation with V (t) = t, we find that I 0 (0) = 0. Solving the ODE subject to these conditions, we obtain I (t) =

3 π 1 1 − e−3t cos(2t) − e−3t sin(2t) for 0 ≤ t < . 13 13 26 4

In the second part which is for t ≥ π/4, the ODE to solve is 6

dI d2 I + 2 + 13I = 0. dt dt

We need to have two conditions. We can calculate the current I(t) at t = π/4 using the solution above for 0 ≤ t < π/4. (Due to the law of conservation of charge, the current has to be a continuous function of time t. Hence we expect the current for 0 ≤ t < π/4 to match with that for t ≥ π/4 at time t = π/4.) We find that π I( ) = 4 =

1 1 3 π π − e−3π/4 cos( ) − e−3π/4 sin( ) 13 13 2 26 2 1 3 −3π/4 . − e 13 26

The value of I 0 (π/4) must be obtained from the integrodiﬀerential equation with V (t) = 0 (that is, the integrodiﬀerential equation which is valid for t ≥ π/4). We find 121

that π π 6I( ) + I 0 ( ) + 13 4 4

Z

π/4

I(τ )dτ = 0.

0

Using integration by parts twice, after a bit of tedious work, we find that Z π/4 Z π/4 1 1 I(τ )dτ = ( − e−3t cos(2t) 13 13 0 0 3 −3t − e sin(2t))dt 26 5 −3π/4 6 1 π+ e − = 52 338 169 and hence π 1 1 I 0 ( ) = e−3π/4 − π. 4 2 4 For t ≥ π/4 , we have to solve 6

dI d2 I + 2 + 13I = 0 dt dt

subject to π π 1 1 3 1 I( ) = − e−3π/4 and I 0 ( ) = e−3π/4 − π. 4 13 26 4 2 4 We obtain I(t) =

e3π/4 (−12 − 8e−3π/4 + 13π)e−3t cos(2t) 104 e3π/4 − (−2 + 3e−3π/4 )e−3t sin(2t) for t ≥ π. 26

3. Refer to page 105 (and a few pages thereafter) for equations for I1 , I2 amd I3 . We start with 2

d2 I3 dI3 + + 2I3 = −9. dt2 dt 122

To solve this, we need two conditions on I3 . One condition is given, that is, I3 (0) = 0. Another condition can be obtained from I1 (0) = 0 and the ODE 2I30 (t) + I1 (t) = 0, that is, we obtain I30 (0) = 0. Solving the ODE in I3 (t) subject to I3 (0) = 0 and I30 (0) = 0 (standard procedure), we obtain 9 9 1√ I3 (t) = − + e−t/4 cos( 15t) 2 2 4 1√ 3 √ −t/4 15e sin( 15t). + 10 4 The current I1 (t) can be obtained from 2I30 (t) + I1 (t) = 0, that is, dI3 dt d 1√ 9 9 = −2 (− + e−t/4 cos( 15t) dt 2 2 4 1√ 3 √ −t/4 15e sin( 15t)) + 10 4 12 √ −t/4 1√ = 15e sin( 15t). 5 4

I1 (t) = −2

Lastly, I2 (t) is given by I2 (t) = I1 (t) − I3 (t) 21 √ −t/4 1√ 9 = 15e sin( 15t) + 10 4 2 9 −t/4 1√ − e cos( 15t). 2 4 4. Let the external source voltage be given by V (t) = K (cosntant). The integral equation for the RC circuit is Z 1 t RI + I(τ )dτ = K C 0 which may be diﬀerentiated to give dI + I = 0. dt 123

Note that RC = 1. The 1st order ODE can be easily solved to give I(t) = I(0)e−t . To find the time when the time t is 20% of I(0), we set I(t) = I(0)/5. This gives 1 1 = e−t ⇒ t = − ln( ) = ln(5) ' 1. 609 second. 5 5 5. We are given LC = 1 and V0 C = 1. This implies V0 /L = 1. For the LC circuit, the integro-diﬀerential equation is Z dI 1 t L + I(τ )dτ = V0 dt C 0 Z t dI ⇒ I(τ )dτ = 1 + dt 0 V0 = 1) (after using LC = L d2 I ⇒ + I(t) = 0 . dt2 The 2nd order ODE can be solve to give the general solution I(t) = A cos(t) + B sin(t). We are told I(0) = 0. From the integro-diﬀerential equation, I 0 (0) = 1. Using I(0) = 0 and I 0 (0) = 1, we obtain A = 0 and N = 1, that is, the required solution is I(t) = sin(t).

4.7

Solutions to Exercise VII

1. The ODE for the spring-mass system is y 00 (t) + 2y 0 (t) + 10y(t) = 10. To solve the corresponding homogeneous ODE, that is, Y 00 (t) + 2Y 0 (t) + 10Y (t) = 0, we let Y = eλt to look for two linearly independent solutions. We obtain the general solution Y (t) = e−t (A cos(3t) + B sin(3t)). It is easy to see that a particular solution of the nonhomogeneous ODE for the spring-mass system is y = 1. Thus, the general solution of the nonhomogeneous ODE is y(t) = 1 + e−t (A cos(3t) + B sin(3t)). 124

Fitting the given conditions y(0) = 0 and y 0 (0) = 0 into the general solution above, we obtain A = −1 and B = −1/3. Thus, y(t) = 1 − e−t (cos(3t) +

1 sin(3t)). 3

The velocity of the mass is given by y 0 (t) = =

d 1 (1 − e−t (cos(3t) + sin(3t))) dt 3 10 −t e sin(3t) 3

The displacement at t = 1 is given by y(1) = 1 − e−1 (cos(3) +

1 sin(3)) 3

' 1. 347 meter. The velocity at t = 1 is y 0 (1) = per second.

10 −1 sin(3) 3 e

' 0. 173 meter

2. Newton’s law gives d2 y dt2 d2 y ⇒ −μy + 10m cos(t) = m 2 dt μ 00 ⇒ y (t) + y(t) = 10 cos(t) (after using = 1). m Fspring + Fmotor = m

The corresponding homogeneous ODE Y 00 (t) + Y (t) = 0 has general solution Y (t) = A cos(t) + B sin(t). To look for a particular solution of the nonhomogeneous ODE, we let y(t) = t(C cos(t) + D sin(t)) (can you see why y(t) = C cos(t) + D sin(t) would not work?), substitute it into the ODE and choose C and D to ensure that the ODE is satisfied. We obtain y(t) = 5t sin(t) as a particular solution. It follows that the general solution of the nonhomogeneous ODE is y(t) = 5t sin(t) + A cos(t) + B sin(t). 125

Applying the given conditions y(0) = 0 and y 0 (0) = 0, we find that A = B = 0. Thus, the displacement and velocity are given by y(t) = 5t sin(t) d y 0 (t) = 5 (t sin(t)) = 5t cos(t) + 5 sin(t). dt At time t = 1, the displacement is y(1) = 5 sin(1) ' 4. 207 meter and the velocity is y 0 (1) = 5 cos(1) + 5 sin(1) ' 6. 909 meter per second. 3. Newton’s law gives d2 y dt2 2 d y dy − 4m cos(2t) = m 2 ⇒ −μy − k dt dt 00 0 ⇒ y (t) + 6y (t) + 13y(t) = −4 cos(2t) k μ = 13 and = 6). (after using m m The general solution of the corresponding homogeneous ODE Y 00 (t)+6Y 0 (t)+13Y (t) = 0 is Y (t) = e−3t (A cos(2t)+ B sin(2t)). To find a particular solution of the nonhomogeneous ODE, we let y(t) = C cos(2t) + D sin(2t) and work out C and D. We obtain C = −4/25 and D = −16/75. Thus, the displacement is given generally by Fspring + Fdamping + Fmotor = m

16 4 cos(2t) − sin(2t) 25 75 +e−3t (A cos(2t) + B sin(2t))

y(t) = −

Using the conditions y(0) = 0 and y 0 (0) = 0, we find that A = 4/25 and B = 34/75. Thus, the required displacement is 4 16 y(t) = − cos(2t) − sin(2t) 25 75 4 34 sin(2t)). +e−3t ( cos(2t) + 25 75 The displacement at time t = 2 is y(2) ' 0. 265 meter. 126

4. Refer to the analysis on page 116 (and a few pages thereafter). Now y1 and y2 denote the displacement of body 1 and 2 respectively. We first solve for Y = y2 − y1 . The ODE to solve is Y 0000 + 14Y 00 + 24Y = 0. If we try Y =√eλt , we √ obtain λ2 = −2 or −12. It follows that λ = ± 2i, ±2 3i. This gives the general solution √ √ Y (t) = A cos( 2t) + B sin( 2t) √ √ +C cos(2 3t) + D sin(2 3t). We can now work out y1 as follows. 6y1

=

Y 00 (t) + 8Y

√ √ ⇒ y1 = A cos( 2t) + B sin( 2t) √ √ 2 2 − C cos(2 3t) − D sin(2 3t). 3 3

It follows that y2 = Y + y1

√ √ = 2A cos( 2t) + 2B sin( 2t) √ √ 1 1 + C cos(2 3t) + D sin(2 3t). 3 3

We are given y1 (0) = y2 (0) = 1. This gives A − 2C/3 = 1 and 2A + C/3 = 1, that is, A = 3/5 and C = −3/5. To work out B and√D, we use y10 (0) = y20 (0)√= 0, that is, we √ √ obtain 2B − 4 3 3 D = 0 and 2 2B + 2 3 3 D = 0 to give B = D = 0. Thus, the required displacements are y1 (t) = y2 (t) =

√ 3 cos( 2t) + 5 √ 6 cos( 2t) − 5

127

√ 2 cos(2 3t) 5 √ 1 cos(2 3t). 5

Chapter 5

Series solutions 5.1

Preamble

As seen in Chapter 3, if we can find two linearly independent solutions of a homogeneous 2nd order linear ODE, we can use Theorem 2 (page 53) to construct the general solution of the ODE. So far, however, we have learnt how to systematically obtain solutions for only two specific types of homogeneous 2nd order linear ODEs: those with constant coeﬃcients and the Euler-Cauchy equations. In this chapter, we will look at the power series method and the Frobenius method for deriving series solutions of rather general homogeneous 2nd order linear ODEs. The methods can be applied to treat well known ODEs in mathematical physics, such as the Legendre’s equation and Bessel’s equation, giving rise to particular special functions, but here we will not go into the details of those ODEs and the associated special functions.

5.2

Review of power series

A power series of x centered about the constant α is a series of the form ∞ X

n=0

cn (x − α)n = c0 + c1 (x − α) + c2 (x − α)2 + · · · 128

where c0 , c1 , c2 , · · · are coeﬃcients of the powers of (x − α). Note that c0 , c1 , c2 , · · · are independent of x. The radius of convergence of the above power series is the largest non-negative number R such that the series converges for α − R < x < α + R. It can be found by using the ratio test which states that the power series converges if ¯ ¯ ¯ (m + 1)th non-zero term in the series ¯ ¯ < 1. lim ¯ m→∞ ¯ mth non-zero term in the series ¯

If ck 6= 0 for k greater than a positive integer (for example, ck 6= 0 for k > 10) then the ratio test above gives ¯ ¯ ¯ cm+1 (x − α)m+1 ¯ ¯ ¯<1 lim ¯ m→∞ ¯ cm (x − α)m ¯ ¯ ¯ ¯ ¯ cm ¯ ¯ cm ¯ ¯ ¯ ¯ ¯. ⇒ R = lim ¯ ⇒ |x − α| < lim ¯ m→∞ cm+1 ¯ m→∞ cm+1 ¯

Power series may be used to represent functions. If a function F (x) can be represented by a power series of x centered about α , that is, if we can write ∞ X cm (x − α)m for α − R < x < α + R F (x) = m=0

then we can diﬀerentiate the power series term by term to obtain a power series representation for F 0 (x) as follows. F 0 (x) =

∞ X

m=0

cm

d (x − α)m dx

d d d (1) + c1 (x − α) + c2 (x − α)2 dx dx dx d +c3 (x − α)3 + · · · dx = c1 + 2c2 (x − α) + 3c3 (x − α)2 + · · · ∞ X = kck (x − α)k−1 (note that k starts from 1) = c0

k=1

∞ X = (n + 1)cn+1 (x − α)n (if we let k = n + 1) n=0

for α − R < x < α + R. 129

Similarly, if we start from 0

F (x) =

∞ X

(m + 1)cm+1 (x − α)m for α − R < x < α + R

m=0

then ∞ X

F 00 (x) =

(m + 1)cm+1

m=0 ∞ X

=

k=1

(k + 1)kck+1 (x − α)k−1

∞ X

=

d (x − α)m dx

(n + 2)(n + 1)cn+2 (x − α)n

n=0

for α − R < x < α + R.

If F (x) is a function which can be diﬀerentiated as many times as we like at x = α, its Taylor series about x = α is given by the power series ∞ X F (n) (α)

n=0

n!

(x − α)n = F (α) + F 0 (α)(x − α) +

F 00 (α) (x − α)2 + · · · . 2!

If the Taylor series of F (x) about x = α has a positive radius of convergence R and if the function F (x) is such that1 F (n) (γ) (x − α)n = 0 n→∞ n! lim

for all γ between α and x and α − R < x < α + R then we can represent F (x) by its Taylor series about x = α for α − R < x < α + R, that is, we can write F (x) =

∞ X F (n) (α)

n=0 1

n!

(x − α)n for α − R < x < α + R.

Check out on the Lagrange remainder term for Taylor series.

130

5.3

Power series method for ODEs

Consider the homogeneous 2nd order linear ODE y 00 + f (x)y 0 + g(x)y = 0.

(1)

If f (x) and g(x) can be represented by their Taylor series about x = α then the power series method may be used to find a general solution (in series form) for (1). The steps involved are described in a general manner as follows. (i) Write f (x) =

∞ X f (m) (α) (x − α)m m! m=0

∞ X g (m) (α) g(x) = (x − α)m . m! m=0

(2)

(ii) For a solution of (1), let y(x) be given by a power series of x centered about α, that is, let y(x) =

∞ X

n=0

cn (x − α)n .

(3)

Here c0 , c1 , c2 , · · · are coeﬃcients to be determined using (1). If the two series in (2) are valid for α−R < x < α+R (R is a positive number) then so is the series in (3). (iii) Diﬀerentiate (3) to obtain2 ∞ X y (x) = (n + 1)cn+1 (x − α)n . 0

(4)

n=0

Similarly, diﬀerentiate (4) to obtain y 00 (x) =

∞ X (n + 2)(n + 1)cn+2 (x − α)n .

n=0 2

The diﬀerentiation of power series is explained on page 129.

131

(5)

(iv) Substitute (2), (3), (4) and (5) into the ODE in (1) to obtain ∞ X (n + 2)(n + 1)cn+2 (x − α)n

n=0 ∞ X

+ +

∞ X f (m) (α) (n + 1)cn+1 (x − α)n (x − α)m m! m=0 n=0 ∞ ∞ X X g (m) (α) cn (x − α)n = 0. (x − α)m m! m=0 n=0

(6)

(v) Rewrite (6) in the form ∞ X p=0

bp (x − α)p = 0.

(7)

To work out bp , we note that ∞ ∞ X X f (m) (α) (n + 1)cn+1 (x − α)n (x − α)m m! m=0 n=0

= =

∞ X

(n + 1)cn+1

n=0 p ∞ X X

∞ X f (m) (α) (x − α)m+n m! m=0

(n + 1)cn+1

p=0 n=0

f (p−n) (α) (x − α)p (p − n)!

(if we let m + n = p).

(8)

Similarly, we also note that ∞ ∞ X X g (m) (α) cn (x − α)n (x − α)m m! m=0 n=0

=

p ∞ X X

p=0 n=0

cn

g (p−n) (α) (x − α)p . (p − n)!

(9)

If we rename the summation (dummy) index n as p (that is, let p = n) in the series on the first line of (6) and 132

replace the series on the second and third lines using (8) and (9) respectively, we obtain bp = (p + 2)(p + 1)cp+2 p X f (p−n) (α) g (p−n) (α) + cn } + {(n + 1)cn+1 (p − n)! (p − n)! n=0 for p = 0, 1, 2, · · · .

(10)

If the series solution (3) is valid for α − R < x < α + R then, to ensure that (7) holds for all the values of x within the given range, we need to set bp = 0 for p = 0, 1, 2, · · · , that is, we have to ensure that (p + 2)(p + 1)cp+2 p X f (p−n) (α) g (p−n) (α) + cn } = − {(n + 1)cn+1 (p − n)! (p − n)! n=0

for p = 0, 1, 2, · · · .

(11)

(vi) Use (11) to determine the coeﬃcients c2 , c3 , c4 · · · in terms of c0 and c1 . The constants c0 and c1 can be regarded as arbitrary. No matter what values are given to c0 and c1 , we can always work out c2 , c3 , c4 · · · from (11). (Note that the right hand side of (11) does not contain cp+2 and the factor (p + 2)(p + 1) multiplied to cp+2 is never zero.) We will now apply the steps above to specific examples of ODEs. In the first three examples below, we consider 2nd order linear ODEs which can also be solved as explained in Chapter 3. In those examples, we can check that the series solutions agree with the solutions obtained as described in Chapter 3. Examples 1. Use the power series method to solve the ODE y00 − 2y 0 = 0, taking y to be a power series of x centered about 0.

133

In this example, the coeﬃcients f (x) and g(x) (refer to the general ODE (1) on page 131) are given by f (x) = −2 and g(x) = 0. They are already in the form of their Taylor series about x = 0. To solve the given ODE, let y(x) =

∞ X

cn xn .

n=0

It follows that y0 = y00 =

∞ X

(n + 1)cn+1 xn

n=0 ∞ X

(n + 2)(n + 1)cn+2 xn .

n=0

Substituting the above into the given ODE, we obtain ∞ X {(p + 2)(p + 1)cp+2 − 2(p + 1)cp+1 }xp = 0. p=0

Setting the coeﬃcient of xp in the series above to 0, we find that 2 cp+2 = cp+1 for p = 0, 1, 2, · · · . (p + 2) Note that the above relation is a specific case of (11). If we let p = 0, we obtain 2 c2 = c1 . 2 Subsequent values of p give 2 22 c2 = c1 3 3! 2 23 = c3 = c1 4 4! 2 24 = c4 = c1 5 5! .. .

c3 = c4 c5

134

From the above, we deduce that cn =

2n−1 c1 for n = 2, 3, 4, · · · . n!

We can easily check whether the deduced formula for cn is correct as follows. 2n−1 c1 n! 2p+1−1 2p ⇒ cp+1 = c1 = c1 (p + 1)! (p + 1)! 2p+2−1 cp+2 = c1 (p + 2)! 2 2p = c1 (p + 2) (p + 1)! 2 = cp+1 (verified). (p + 2) cn =

The required series solution is given by y(x) =

∞ X

cn xn

n=0

= c0 +

∞ X 2n−1

n=1

n!

c1 xn

∞ c1 X (2x)n = c0 + . 2 n=1 n!

There are two observations we may make concerning the series solution above. Firstly, if we solve the ODE y 00 − 2y 0 = 0 as described in Chapter 3 (that is, by letting y = eλx to find two linearly independent solutions and then applying Theorem 2 on page 53), we obtain the general solution y = A + Be2x , where A and B are arbitrary constants. Thus, we may like to check if we can recover this general solution from the series solution. We

135

can rewrite the series solution as ∞ c1 X (2x)n y(x) = c0 + 2 n=1 n!

∞ c1 c1 c1 X (2x)n = c0 + (− + ) + 2 2 2 n=1 n!

∞ c1 X (2x)n c1 c1 c1 = (c0 − ) + e2x = (c0 − ) + 2 2 n! 2 2 n=0

since from the Taylor series of ex about x = 0 we know that ex

=

∞ X xn

n=0

n!

⇒ e2x =

∞ X (2x)n

n=0

n!

.

As there is no restriction on the coeﬃcients c0 and c1 , they are arbitrary. Thus, if we let A = c0 − 12 c1 and B = 12 c1 , A and B are arbitrary constants and we recover the general solution y = A + Be2x . Secondly, we note that the series solution above converges for all x (as the Taylor series for e2x is valid for −∞ < x < ∞). This is as expected since the Taylor series of the coeﬃcients, that is, f (x) = −2 and g(x) = 0, used in the derivation of the series solution converges for all x. 2. Derive a solution for the ODE y 00 + x−1 y0 − x−2 y = 0 by taking y to be a power of x centered about 1. Determine the range of x for which the series solution converges. (For this ODE, it is not possible to find a solution in the form of a power series centered about 0.) To express the coeﬃcient of y 0 , that is f (x) = x−1 , in terms

136

of its Taylor series about x = 1, we note that f (x) = x−1 ⇒ f (1) = 1

f 0 (x) = −x−2 ⇒ f 0 (1) = −1

f 00 (x) = 2x−3 ⇒ f 00 (1) = 2

f 000 (x) = −3 · 2x−4 ⇒ f 000 (1) = −3!

f 0000 (x) = 4 · 3 · 2x−5 ⇒ f 0000 (1) = 4! .. . From the above, we deduce that f (m) (1) = m!(−1)m .

If the Taylor series of f (x) = x−1 about x = 1 converges, then we can write ∞ X 1 = (−1)m (x − 1)m . x m=0

Whether the Taylor series of f (x) = x−1 about x = 1 converges or not depends on the value of x. For example, if x = 2, the series diverges as the n-th term in the series does not tend to 0 as n increases. The range of x for which the series converges can be determined by calculating the radius of convergence of the series. Since the Taylor series of x−1 about x = 1 has non-zero coeﬃcients for all positive powers of (x − 1), its radius of convergence is given by ¯ ¯ ¯ (−1)m ¯ ¯ = 1. R = lim ¯¯ m→∞ (−1)m+1 ¯

Thus, the power series of x−1 about x = 1 converges for 0 < x < 2. In a similar way, we can work out the Taylor series of g(x) = −x−2 about x = 1. If the Taylor series converges then we can write ∞ X 1 − 2 =− (m + 1)(−1)m (x − 1)m . x m=0

137

The series also converges for 0 < x < 2. Note that g (m) (1) = −(m + 1)!(−1)m An alternative way to derive the power series representing g(x) = −x2 is to diﬀerentiate the Taylor series of f (x) = x−1 about x = 1 as follows. d 1 ( ) dx x

=

∞ ∞ X X n d n (−1) n(−1)n (x − 1)n−1 (x − 1) = dx n=0 n=1

∞ X 1 ⇒ − 2 =− (m + 1)(−1)m (x − 1)m x m=0

(if we let n = m + 1)

If we look for a solution of the ODE in the series form y(x) =

∞ X

n=0

cn (x − 1)n

then (11) in step (v) above of the power series method (page 133) gives (p + 2)(p + 1)cp+2 p X = − {(n + 1)cn+1 − (p − n + 1)cn }(−1)p−n n=0

for p = 0, 1, 2, · · · .

For p = 0, we obtain 2c2 = c0 − c1 . Subsequent values of p give 6c3 = 2(c1 − c0 − c2 ) ⇒ c3 = −c2

12c4 = −3c1 + 3c0 + 3c2 − 3c3 ⇒ c4 = c2

20c5 = 4c1 − 4c0 − 4c2 + 4c3 − 4c4 ⇒ c5 = −c2 .. . 138

Thus, we deduce that 1 cp = (−1)p (c0 − c1 ) for p = 2, 3, 4, · · · . 2 The required series solution is given by ∞ X 1 (−1)p (x − 1)p . y(x) = c0 + c1 (x − 1) + (c0 − c1 ) 2 p=2

We can rewrite the series solution as 1 1 y(x) = c0 + c1 (x − 1) − (c0 − c1 ) + (c0 − c1 )(x − 1) 2 2 ∞ X 1 (−1)p (x − 1)p + (c0 − c1 ) 2 p=0 which simplifies to ∞ X 1 1 y(x) = (c0 + c1 )x + (c0 − c1 ) (−1)p (x − 1)p . 2 2 p=0

The series above (with the sum from p = 0 to ∞) converges to x−1 for 0 < x < 2. Thus, the series solution is valid only for 0 < x < 2. This is as expected since the power series for the coeﬃcients of y and y 0 in the ODE converges for the same range of x. Furthermore, if we set A = 12 (c0 + c1 ) and B = 12 (c0 − c1 ) (where A and B are arbitrary constants since c0 and c1 can be given any values we like), we find the series solution of the ODE gives y(x) = Ax +

∞ X B 1 (−1)p (x − 1)p = ). (since x x p=0

This is the general solution which we obtain if we solve y 00 + x−1 y 0 − x−2 y = 0 as an Euler-Cauchy equation (that is, by rewriting and solving it as x2 y 00 + xy0 − y = 0, as described in Chapter 3). In the above, in finding the solution in terms of a power series centered about 1 directly from the ODE y 00 + x−1 y0 − 139

x−2 y = 0, we have to work out the power series centered about 1 for the coeﬃcients x−1 and −x−2 . For this particular ODE, it is possible to avoid working out power series representations for the coeﬃcients, if we use the ODE in its rewritten form x2 y 00 + xy 0 − y = 0 and proceed as follows to work out the same series solution. Rewrite x2 y 00 + xy 0 − y = 0 further as x2 y00 + xy 0 − y = 0

⇒ ([x − 1] + 1)2 y 00 + ([x − 1] + 1)y0 − y = 0

⇒ ([x − 1]2 + 2[x − 1] + 1)y00 + ([x − 1] + 1)y 0 − y = 0. Substitute the proposed power series solution given by y(x) =

∞ X

n=0

and its derivatives y

0

cn (x − 1)n

∞ X = (n + 1)cn+1 (x − 1)n n=0 ∞ X

y 00 =

(n + 2)(n + 1)cn+2 (x − 1)n .

n=0

into the rewritten ODE above to obtain ∞ X (n + 1)(n + 2)cn+2 (x − 1)n+2 n=0

∞ X +2 (n + 1)(n + 2)cn+2 (x − 1)n+1 n=0

∞ X + (n + 1)(n + 2)cn+2 (x − 1)n

+ +

n=0 ∞ X

(n + 1)cn+1 (x − 1)n+1

n=0 ∞ X

n

(n + 1)cn+1 (x − 1) −

n=0

= 0

140

∞ X

n=0

cn (x − 1)n

which may be rewritten as ∞ ∞ X X (p − 1)pcp (x − 1)p + 2 p(p + 1)cp+1 (x − 1)p p=2

+

p=1

∞ X p=0

(p + 1)(p + 2)cp+2 (x − 1)p +

∞ X p=1

pcp (x − 1)p

∞ ∞ X X p + (p + 1)cp+1 (x − 1) − cp (x − 1)p = 0 p=0

p=0

to give 2c2 + c1 − c0 + (6c2 + 6c3 )(x − 1) ∞ X (p + 1){(p + 2)cp+2 + (2p + 1)cp+1 + p=2

= 0.

+(p − 1)cp }(x − 1)p

To satisfy the ODE for all x, we set 2c2 + c1 − c0 = 0 6c2 + 6c3 = 0

and (p + 2)cp+2 + (2p + 1)cp+1 + (p − 1)cp = 0 for p = 2, 3, · · · , that is, 1 c2 = (c0 − c1 ), c3 = −c2 , c4 = c2 , c5 = −c2 and so on. 2 From the above, we deduce that 1 cp = (−1)p (c0 − c1 ) for p = 2, 3, 4, · · · 2 as before. 141

3. Use the power series method to solve y00 + 4y = 0 subject to y(0) = 1 and y0 (0) = 2. Let the series solution be given by y(x) =

∞ X

cn xn

n=0

As before, it follows that

∞ X (n + 1)cn+1 xn y (x) = 0

y 00 (x) =

n=0 ∞ X

(n + 2)(n + 1)cn+2 xn .

n=0

Substituting the above into the given ODE, we find that ∞ X {(p + 2)(p + 1)cp+2 + 4cp }xp = 0 p=0

which gives (p + 2)(p + 1)cp+2 = −4cp for p = 0, 1, 2, · · · . For p = 0 and p = 1, we obtain 4 2c2 = −4c0 ⇒ c2 = − c0 2 4 c1 3 · 2c3 = −4c1 ⇒ c3 = − 3·2 Subsequent even values of p yields 42 c0 4! 43 = −4c4 ⇒ c6 = − c0 6! 4 4 = −4c6 ⇒ c8 = c0 8! 45 = −4c8 ⇒ c10 = − c0 10! .. .

4 · 3c4 = −4c2 ⇒ c4 = 6 · 5c6 8 · 7c8 10 · 9c10

142

while odd values of p gives 42 c1 5! 43 = −4c5 ⇒ c7 = − c1 7! 4 4 = −4c7 ⇒ c9 = c1 9! 45 = −4c9 ⇒ c11 = − c1 11! .. .

5 · 4c5 = −4c3 ⇒ c5 = 7 · 6c7 9 · 8c9 11 · 10c11

From the above, we deduce that c2k =

(−1)k 4k c0 for k = 0, 1, 2, · · · (2k)!

and c2k+1 =

(−1)k 4k c1 for k = 0, 1, 2, · · · . (2k + 1)!

Now from y(0) = 1, we obtain c0 = 1. Similarly, y 0 (0) = 2 gives c1 = 2. Thus, c2k =

(−1)k 4k for k = 0, 1, 2, · · · (2k)!

and c2k+1 =

2(−1)k 4k for k = 0, 1, 2, · · · . (2k + 1)!

We break up the series for y into odd and even powers of x, that is, y(x) =

=

∞ X

cn xn +

n=0,2,4,··· ∞ X c2k x2k k=0

∞ X

cn xn

n=1,3,5,···

+

143

∞ X k=0

c2k+1 x2k+1 .

From the deduced formulae for c2k and c2k+1 for k = 0, 1, 2, · · · , it follows that the required series solution of the ODE is y(x) =

∞ X (−1)k 4k k=0

(2k)!

x2k +

∞ X 2(−1)k 4k k=0

(2k + 1)!

x2k+1 .

As a check, let us rewrite the series solution as y(x) =

∞ X (−1)k 22k k=0

=

(2k)!

x2k +

k=0

∞ X (−1)k (2x)2k k=0

∞ X (−1)k 22k+1

(2k)!

+

(2k + 1)!

x2k+1

∞ X (−1)k (2x)2k+1 k=0

(2k + 1)!

.

From the Taylor series of sin(x) and cos(x) about x = 0, we find that sin(x) = cos(x) =

∞ X (−1)k x2k+1 k=0 ∞ X k=0

(2k + 1)!

⇒ sin(2x) =

(−1)k x2k ⇒ cos(2x) = (2k)!

∞ X (−1)k (2x)2k+1

k=0 ∞ X k=0

(2k + 1)!

(−1)k (2x)2k . (2k)!

Thus, the series solution derived above can be rewritten as y(x) = cos(2x) + sin(2x) which is what we obtain if we solve y 00 + 4y = 0 subject to y(0) = 1 and y 0 (0) = 2 as described in Chapter 3. 4. Consider solving the ODE y 00 + ex y 0 + y = 0 subject to y(0) = y0 (0) = 1 using the power series method, by taking y to be a power series of x centered about 0. Truncate the series solution to obtain a quartic (4th order polynomial) approximation of y. Let the series solution be given by y(x) =

∞ X

n=0

144

cn xn .

Here c0 = y(0) = 1 and c1 = y 0 (0) = 1. Applying (11) in step (v) above of the power series method (page 133), we obtain (p + 2)(p + 1)cp+2 = −

p X

(n + 1)cn+1

n=0

for p = 0, 1, 2, · · · .

1 − cp (p − n)!

Letting p = 0, 1, 2, we find that 2c2 = −c1 − c0 = −1 − 1 = −2 ⇒ c2 = −1

6c3 = −c1 − 2c2 − c1 = 0 ⇒ c3 = 0 1 5 5 12c4 = − c1 − 2c2 − 3c3 − c2 = ⇒ c4 = 2 2 24 Thus, the required approximation is given by y(x) ' 1 + x − x2 +

5 4 x . 24

An alternative way to derive the approximation above is as follows. If we assume that the required solution of the ODE can be represented by its Taylor series about x = 0 then we can write y(x) ' y(0) + y 0 (0)x +

y 00 (0) 2 y000 (0) 3 y0000 (0) 4 x + x + x . 2! 3! 4!

It is given that y(0) = 1 and y0 (0) = 1. We can work out y 00 (0), y000 (0) and y 0000 (0) from the ODE y00 + ex y0 + y = 0 as follows. y00 + ex y0 + y = 0 ⇒ y000 + ex y 0 + ex y 00 + y0 = 0

⇒ y0000 + ex y 0 + 2ex y 00 + ex y 000 + y 00 = 0 y00 (0) = −e0 y0 (0) − y(0) = −2

y000 (0) = −e0 y 0 (0) − e0 y 00 (0) − y 0 (0) = 0

y0000 (0) = −e0 y 0 (0) − 2e0 y00 (0) − e0 y 000 (0) − y00 (0) = 5.

Thus, from the Taylor series of y about x = 0, we can also obtain the quartic approximation which we have derived earlier on. 145

5.4

Exercise VIII

1. Use the power series method to solve each of the following ODEs. Give your solution as a power series of x centered about 0. State the range of x over which the series solution converges. (a) (b) (c)

y 00 − 25y = 0

y 00 + 5y0 − 6y = 0 with y(0) = 2 and y 0 (0) = −5

y 00 + 5(x − 1)−1 y0 + 3(x − 1)−2 y = 0

with y(0) = 0 and y0 (0) = 2

(Multiply this ODE by (x − 1)2 , if you like.) 2. For each of the following ODEs and given conditions, write down the first four non-zero terms of its power series solution centered about 0. (a)

y 00 + sin(x)y = 0 with y(0) = 1 and y0 (0) = 0

(b)

y 00 − 4y0 + 13y = 0 with y(0) = y 0 (0) = 1

(c)

y 00 + sin(2x)y 0 + cos(2x)y = 0

with y(0) = 0 and y 0 (0) = 1 3. Apply the power series method directly on (1 − x2 )y 00 − 2xy 0 + 42y = 0 to find the exact solution of the ODE subject to y(0) = 2 and y 0 (0) = 0. (Begin by taking y to be in the form of a power series of x centered about 0. You will find that the coeﬃcient of xn in the series solution, that is, cn is zero if n exceeds a certain value and the required exact solution turns out to be a polynomial function.) 4. Repeat Problem 3 above using the ODE (1 − x2 )y 00 − 2xy 0 + 56y = 0 with y(0) = 0 and y0 (0) = 2. 146

5.5

Frobenius method

We will consider now the homogeneous 2nd order linear ODE in (1) (page 131) with the coeﬃcients f (x) and g(x) given by f(x) = g(x) =

p(x) x−α q(x) (x − α)2

(12)

where p(x) and q(x) are functions which can be represented by their Taylor series about x = α. In general, the coeﬃcients f (x) and g(x) in (12) are not well defined at x = α and Taylor series about x = α are not available to represent them. Thus, the power series method described above (page 131) cannot be applied directly on (1) and (12) to obtain a solution in the form of a power series centered about the constant α. An attempt may, however, be made to apply the power series method on the ODE rewritten as (x − α)2 y 00 + (x − α)p(x)y 0 + q(x)y = 0

(13)

to obtain a power series solution centered about α. Depending on p(x) and q(x), it may just end up giving us the trivial solution y = 0. There is a more general approach for solving (13) to obtain a solution in terms of a series containing powers of (x − α) (not necessarily non-negative integer powers). According to the Frobenius method, we can look for a solution of (13) in the form of a series given by y(x) = (x − α)λ

∞ X

n=0

dn (x − α)n

(14)

where λ and dn are constants to be determined from (13) as explained below.

147

We can diﬀerentiate (14) twice as follows3 . y(x)

=

∞ X

n=0

dn (x − α)n+λ ∞ X (n + λ)dn (x − α)n+λ−1

⇒ y 0 (x) = ⇒ y 00 (x) =

(15)

n=0 ∞ X

(n + λ)(n + λ − 1)dn (x − α)n+λ−2 .

n=0

Representing p(x) and q(x) by their Taylor series about x = α and substituting (15) into (13), we obtain ∞ X (n + λ)(n + λ − 1)dn (x − α)n+λ

n=0

∞ ∞ X p(m) (α) X + (n + λ)dn (x − α)m+n+λ m! m=0 n=0

∞ ∞ X q (m) (α) X + dn (x − α)m+n+λ = 0 m! m=0 n=0

which can be rewritten as (if we follow step (v) of the power series method above, as explained on page 132, that is, if we let k = m + n in the second and third lines of the equation above) ∞ X {(k + λ)(k + λ − 1)dk k=0

+

k X

(n + λ)dn

n=0

+

k X

n=0

dn

p(k−n) (α) (k − n)!

q (k−n) (α) }(x − α)k+λ = 0. (k − n)!

3

(16)

Bear in mind that λ is not necessarily 0 here. If λ = 0 (as in the power series method) then the summation in the series for y 0 and y 00 in (15) can be taken to start from n = 1 and n = 2 respectively (instead of 0).

148

To ensure that (16) is satisfied for all x, we set the coeﬃcient of (x − α)k+λ to zero for k = 0, 1, 2, · · · , that is, (k + λ)(k + λ − 1)dk

+

k X p(k−n) (α) (n + λ)dn (k − n)! n=0

+

k X

n=0

dn

q (k−n) (α) (k − n)!

= 0 for k = 0, 1, 2, · · · (17)

In (17), k = 0 yields [λ(λ − 1) + λp(α) + q(α)]d0 = 0.

(18)

To keep open the option of choosing d0 as an arbitrary constant, we select the constant λ in such a way that λ(λ − 1) + λp(α) + q(α) = 0.

(19)

With λ chosen to satisfy (19), we can work out the remaining coeﬃcients d1 , d2 , · · · from (17) with k = 1, 2, · · · , that is, from [(k + λ)(k + λ − 1) + (k + λ)p(α) + q(α)]dk = −

k−1 X

n=0

dn {(n + λ)p(k−n) (α) + q (k−n) (α)} (k − n)! for k = 1, 2, · · · .

(20)

We choose λ to satisfy (19) so that d0 is not forced to take any specific value in (18). Nevertheless, whether we can take d0 to be arbitrary or not would depend on the coeﬃcient of dk on the left hand side of (20). To see this, let us write out (20) for k = 1, that is, after using (19), we obtain (2λ + p(α))d1 = λ(λ − 1)d0 . Now if it happens that 2λ + p(α) = 0 but λ(λ − 1) 6= 0 then we are forced to take d0 = 0 and to look into the possibility of taking d1 as an arbitrary constant instead. 149

Note that if the coeﬃcient of dk on the left hand side of (20) is not zero for all k = 1, 2, · · · , then we are free to give d0 any non-zero value, for example, d0 = 1 (for convenience), in order to work out a particular solution for the ODE in (13). Note that (19) is a quadratic equation in λ. It may have up to two distinct solutions. Let λ = λ1 and λ = λ2 be solutions of (19). With λ determined, the coeﬃcients dk can then be worked out from (20). If the coeﬃcients are given by ½ dk1 if λ = λ1 dk = dk2 if λ = λ2 then the solutions corresponding to λ = λ1 and λ = λ2 are respectively y1 (x) and y2 (x) given by λ1

y1 (x) = (x − α)

y2 (x) = (x − α)λ2

∞ X

n=0 ∞ X n=0

dn1 (x − α)n dn2 (x − α)n .

(21)

Now if y1 (x) and y2 (x) in (21) are linearly independent then according to Theorem 2 in Chapter 3 (page 53) the general solution of (13) is y(x) = Ay1 (x) + By2 (x)

(22)

where A and B are arbitrary constants. If λ1 = λ2 then it is obvious that y1 (x) and y2 (x) are not linearly independent. In this case, to construct the general solution of (13), we have to look for another linearly independent solution. To examine whether y1 (x) and y2 (x) are linearly independent or not for the case in which λ1 6= λ2 , let us write either y2 (x) = y1 (x)

∞ P

n=0

dn2 (x − α)n+λ2 −λ1 ∞ P

n=0

dn1 (x − α)n

150

(23)

or ∞ P

dn2 (x − α)n y2 (x) n=0 . = ∞ P y1 (x) dn1 (x − α)n+λ1 −λ2

(24)

n=0

For the case in which λ2 − λ1 or λ1 −λ2 is a positive integer, on the right hand side of (23) or (24), we see that y2 /y1 can be given by one power series centered about α divided by another power series also centered about α. Depending on what the coeﬃcients dn1 and dn2 are, the power series in the numerator may or may not a constant multiple of the one in the denominator. Thus, for the case in which λ2 − λ1 = ±1, ±2, ±3, · · · , we find that y2 /y1 may or may not be a constant, that is, y1 (x) and y2 (x) may or may not be linearly independent. For the case in which λ2 − λ1 is not an integer, on the right hand side of (23), the denominator is a power series centered about α, but the numerator is a series with non-integer powers (n + λ2 − λ1 may be a real number which is not an integer or it may be a complex number with a non-zero imaginary part). It follows that y2 /y1 cannot be a constant, hence y1 and y2 are linearly independent. Thus, if λ2 − λ1 is not an integer (that is, if λ2 − λ1 6= 0, ±1, ±2, ±3, · · · ), then the general solution of (13) is given by (22) If y1 (x) and y2 (x) as given in (21) are not linearly independent, we have to find another linearly independent solution in order to construct the general solution of (13). This may be done by modifying the second solution y2 (x) to be given by y2 (x) = y1 (x) ln(x − α) + (x − α)min(λ1 ,λ2 )

∞ X

n=0

gn (x − α)n (25)

where min(λ1 , λ2 ) = λ1 if λ1 ≤ λ2 or min(λ1 , λ2 ) = λ2 if λ2 ≤ λ1 . In (25) above, note that y1 (x) is the same particular solution in (21) (as obtained by using λ = λ1 ) and λ = λ2 is the second solution of (19). We can work out the coeﬃcients 151

gn (n = 0, 1, 2, · · · ) by substituting (25) into (13) and equating the coeﬃcients of the resulting series to zero. The task of determining gn (n = 0, 1, 2, · · · ) for (13) in its general form will not be described here. Instead, it will be illustrated below through some specific examples. Examples 1. Use the Frobenius method to find the general solution of x2 y 00 − 2xy0 + y = 0. Comparing the given ODE to (13) (page 147), we find that α = 0, p(x) = −2 and q(x) = 1. Thus, if we take y(x) to be of the series form y(x) = xλ

∞ X

dn xn

n=0

then according to (19) (page 149) the constant λ must satisfy λ(λ − 1) − 2λ + 1 = 0. √ √ The above equation gives λ = 32 + 12 5 and λ = 32 − 12 5. Use of (20) (page 149) with p(x) = −2 and q(x) = 1 gives k(k + 2λ − 3)dk = 0 for k = 1, 2, 3, · · ·

(since p(m) (x) = 0 and q (m) (x) = 0 for m = 1, 2, · · · ). √ √ For both λ = 32 + 12 5 and λ = 32 − 12 5, we find that k(k + 2λ − 3) 6= 0 for k = 1, 2, · · · and hence the above implies that 2, · · · . dk = 0 for k = 1, √ For λ = 32 + 12 5, we may take d0 = 1 to obtain one particular solution

Similarly, for λ = tion

3 2

3

1

√

3

1

√ 5

y1 (x) = x 2 + 2 5 . √ − 12 5, we obtain another particular soluy2 (x) = x 2 − 2 152

.

The two particular solutions given above are linearly independent. Hence, the required general solution is given by 3

1

y(x) = Ax 2 + 2

√ 5

3

1

+ Bx 2 − 2

√ 5

where A and B are arbitary constants. The ODE in this example is a Cauchy-Euler equation. It is chosen here to show that the Frobenius method may be regarded as a generalization of the solution procedure in Section 3.4 for solving Cauchy-Euler equations. 2. Use the Frobenius method to find the general solution of 2x2 y 00 − xy0 + (2x + 1)y = 0. We look for a series solution of the form y(x) = xλ

∞ X

dn xn .

n=0

Substituting the proposed series solution into the given ODE, we obtain 2

∞ ∞ X X (n + λ)(n + λ − 1)dn xn+λ − (n + λ)dn xn+λ

n=0

n=0

+2

∞ X

dn xn+λ+1 +

n=0

∞ X

dn xn+λ = 0

n=0

which can be rewritten as [2λ2 − 3λ + 1]d0 xλ ∞ X + {2(k + λ)(k + λ − 1)dk − (k + λ)dk k=1

+2dk−1 + dk }xk+λ = 0.

To ensure that the ODE is satisfied, we choose λ such that 2λ2 − 3λ + 1 = 0 153

and the coeﬃcients in the series solution such that k[2k + 4λ − 3]dk = −2dk−1 for k = 1, 2, · · · . Solving the quadratic equations in λ, we obtain λ = 1/2 and λ = 1. For λ = 1/2 and λ = 1, we find that k[2k + 4λ − 3] 6= 0 for k = 1, 2, 3, · · · , hence we may take d0 = 1 to work out particular solutions of the ODE. For λ = 1/2, taking d0 = 1, we find that d1 = −2 2d1 22 = d2 = − 2·3 2·3 2d2 23 d3 = − =− 3·5 (3 · 2) · (5 · 3) .. . (−1)k 2k . dk = k!(2k − 1)(2k − 3) · · · · · 5 · 3 It follows that a particular solution of the ODE is 1/2

y(x) = x

(1 +

∞ X k=1

(−1)k 2k xk ). k!(2k − 1)(2k − 3) · · · · · 5 · 3

For λ = 1, also taking d0 = 1, we find that 2 1·3 2d1 22 = − = 2·5 (2 · 1) · (5 · 3) .. . (−1)k 2k = . k!(2k + 1)(2k − 1) · · · · · 5 · 3

d1 = − d2

dk

Thus, another particular solution of the ODE is y(x) = x(1 +

∞ X k=1

(−1)k 2k xk ). k!(2k + 1)(2k − 1) · · · · · 5 · 3 154

The first particular solution is not a constant multiple of the second one. The two particular solutions given above are linearly independent. Hence, the required general solution is given by 1/2

y(x) = Ax

(1 +

∞ X k=1

+Bx(1 +

∞ X k=1

(−1)k 2k xk ) k!(2k − 1)(2k − 3) · · · · · 5 · 3

(−1)k 2k xk ) k!(2k + 1)(2k − 1) · · · · · 5 · 3

where A and B are arbitary constants. (Note that the series in the general solution can be shown to converge for all x.) 3. Use the Frobenius method to find two linearly independent solutions of y00 +

x−3 0 1 y = 0. y + x−2 (x − 2)2

Rewrite the ODE as (x − 2)2 y 00 + (x − 2)(x − 2 − 1)y 0 + y = 0 and propose a series solution of the form y(x) = (x − 2)λ

∞ X

n=0

dn (x − 2)n .

Substitute the proposed series solution into the rewritten ODE, we find that ∞ X (n + λ)(n + λ − 1)dn (x − 2)n+λ

n=0 ∞ X

+ − = 0

(n + λ)dn (x − 2)n+λ+1

n=0 ∞ X

n+λ

(n + λ)dn (x − 2)

n=0

155

+

∞ X

n=0

dn (x − 2)n+λ

which may be rewritten as [λ(λ − 1) − λ + 1]d0 (x − 2)λ ∞ X + {(n + λ)(n + λ − 1)dn n=1

+(n − 1 + λ)dn−1 − (n + λ)dn + dn }(x − 2)n+λ

= 0.

We choose λ such that λ(λ − 1) − λ + 1 = 0

⇒ (λ − 1)2 = 0 ⇒ λ = 1. With λ = 1, the constants d1 , d2 , · · · can be determined from [(n + 1)n − (n + 1) + 1]dn = −ndn−1 for n = 1, 2, · · · . Taking d0 = 1, we find that d1 = −1 2d2 = −d1 ⇒ d2 =

1 2

3d3 = −d2 ⇒ d3 = −

dn

1 3·2

.. . (−1)n = for n = 0, 1, 2, · · · . n!

From the above analysis, we manage to obtain one particular solution, that is, y1 (x) =

∞ X (−1)n (x − 2)n+1

n!

n=0

.

For another particular solution y2 (x) such that y1 (x) and y2 (x) are linearly independent, let y2 (x) = y1 (x) ln(x − 2) + (x − 2) 156

∞ X

n=0

gn (x − 2)n

which gives y20 = y10 (x) ln(x − 2) + y200 = y100 (x) ln(x − 2) + +

∞ X

n=1

∞

y1 (x) X + gn (n + 1)(x − 2)n x−2 n=0

2y10 (x) x−2

−

y1 (x) (x − 2)2

gn (n + 1)n(x − 2)n−1 .

Substituting y2 (x) into the ODE and making use of (x − + (x − 2)(x − 2 − 1)y10 + y1 = 0 (since y1 is a particular solution), we obtain

2)2 y100

2

∞ X (−1)n (n + 1)(x − 2)n+1

n=0 ∞ X

+

+ − = 0

n=1 ∞ X n=0 ∞ X n=0

n!

gn (n + 1)n(x − 2)n+1 + gn (n + 1)(x − 2)n+2 − gn (n + 1)(x − 2)n+1 +

−

∞ X (−1)n (x − 2)n+1

n!

n=0 ∞ X

(−1)n (x − 2)n+2 n! n=0

∞ X (−1)n (x − 2)n+1

n=0 ∞ X n=0

n!

gn (x − 2)n+1

which can be tidied up to give ∞ X (−1)n + n2 gn + ngn−1 }(x − 2)n+1 = 0. { (n − 1)!

n=1

It follows that ngn = −

(−1)n − gn−1 for n = 1, 2, · · · . n!

There is no restriction on the coeﬃcient g0 here. Taking

157

g0 = 0, we find that the equation above gives g1 = 1 − g0 = 1 1 3 2g2 = − − g1 ⇒ g2 = − 2 4 1 11 3g3 = − g2 ⇒ g3 = 3! 36 1 25 4g4 = − − g3 ⇒ g4 = − 4! 288 1 137 − g4 ⇒ g5 = 5g5 = 5! 7200 1 49 6g6 = − − g5 ⇒ g6 = − 6! 14 400 .. . It appears diﬃcult here to deduce a general formula for gn by just looking at the first few coeﬃcients above. To summarize, two linearly independent solutions of the ODE are ∞ X (−1)n (x − 2)n+1 y1 (x) = n! n=0 y2 (x) = ln(x − 2) ·

∞ X (−1)n (x − 2)n+1

n! n=0 ∙ 3 11 +(x − 2) (x − 2) − (x − 2)2 + (x − 2)3 4 36 25 137 − (x − 2)4 + (x − 2)5 288 7200 ¸ 49 (x − 2)6 + · · · . − 14 400

4. Use the Frobenius method to solve 4x2 y00 + 4xy0 − (1 + x)y = 0. Take y(x) to be of the series form y(x) = xλ

∞ X

n=0

158

dn xn .

Substitute the proposed series solution into the ODE to obtain ∞ ∞ X X (n + λ)(n + λ − 1)dn xn+λ + 4 (n + λ)dn xn+λ 4 n=0 ∞ X

− = 0

n=0

n=0

dn xn+λ −

∞ X

dn xn+λ+1

n=0

which can be rearranged to give (4λ2 − 1)d0 xλ +

∞ X {[4n2 + 8nλ + 4λ2 − 1]dn − dn−1 }xn+λ = 0.

n=1

The ODE is satsified if we choose λ = ±1/2 and 4n(n + 2λ)dn = dn−1 for n = 1, 2, · · · . The diﬀerence between the two values of λ is an integer. There is a possibility that we cannot obtain two linearly independent solutions of the series form proposed above by using the two diﬀerent values of λ. For λ = −1/2, we find that

1 n(n − 1)dn = dn−1 for n = 1, 2, · · · . 4 If we let n = 1, we find that d0 = 0. Thus, d0 cannot be arbitrary. Noticing that 4n(n − 1) 6= 0 for n = 2, 3, · · · , we may regard d1 as an arbitrary constant. Let us take d1 = 1. It follows that 1 1 2d2 = ⇒ d2 = 4 4·2 1 1 3 · 2d3 = d2 ⇒ d3 = 2 4 4 ·3·2·2 1 1 4 · 3d4 = d3 ⇒ d4 = 3 4 4 ·4·3·3·2·2 .. . 1 for k = 1, 2, · · · . dk = k−1 4 k[(k − 1)!]2 159

Thus, a particular solution is given by y1 (x) =

∞ X k=1

1 4k−1 k[(k

− 1)!]2

xk−1/2

1 1 5/2 1 7/2 = x1/2 + x3/2 + x + x + ··· 8 192 9216 Now for λ = 1/2, we find that 1 n(n + 1)dn = dn−1 for n = 1, 2, · · · . 4 If we take d0 = 1, we obtain d1 =

1 ·3·2·2 1 = 3 4 ·4·3·3·2·2 .. . 1 = for k = 0, 1, 2, · · · . k 4 (k + 1)[k!]2

d2 = d3

dk

1 4·2 42

It is easy to see that the above leads to the same particular solution which we have derived earlier on. If we select d0 to be a non-zero value other than 1, we will find that we obtain a particular solution which is a constant multiple of the earlier one. Thus, for this example, the particular solutions from λ = −1/2 and λ = 1/2 are not linearly independent. Following (25) (page 151), we look for a second particular solution by letting −1/2

y2 (x) = y1 (x) ln(x) + x

∞ X

gn xn .

n=0

Substituting the proposed solution into the ODE, we obtain ∞ X { k=1

8(k − 1/2) + 4k(k − 1)gk − gk−1 }xk−1/2 = 0. 4k−1 k[(k − 1)!]2 160

To ensure the ODE is satisfied, we take 4k(k − 1)gk = −

8(k − 1/2) + gk−1 for k = 1, 2, · · · . − 1)!]2

4k−1 k[(k

For k = 1, the above gives g0 = 4. The constant g1 can be aribtrary (as there is restriction on it). If we take g1 = 0, subsequent values of k give 3 3 8g2 = − + g1 ⇒ g2 = − 2 16 5 7 24g3 = − + g2 ⇒ g3 = − 48 576 7 35 48g4 = − + g3 ⇒ g4 = − 2304 110 592 .. . Thus, the general solution of the ODE is y(x) = A

∞ X k=1

"

1 4k−1 k[(k

+B ln(x)

∞ X k=1

− 1)!]2

xk−1/2

1 xk−1/2 + 4x−1/2 4k−1 k[(k − 1)!]2

3 7 5/2 35 − x3/2 − x − x7/2 − · · · 16 576 110 592

¸

where A and B are arbitrary constants.

5.6

Exercise IX For each of the following ODEs, use the Frobenius method to find two linearly independent solutions.

1. 6x2 y 00 − x(1 + x2 )y 0 − 3y = 0 2. 4y 00 + 19(x − 1)−1 y0 − (5 − x)(x − 1)−2 y = 0 3. x2 y 00 + xy0 + (x2 − 9)y = 0 161

4. 9x2 y 00 + 15xy 0 + (x + 1)y = 0 5. 16x2 y00 + 8xy 0 − (x + 3)y = 0 6. 16x2 y00 + x(8 − x)y 0 − 3y = 0

5.7

Solutions to Exercise VIII

1. Let the series solution be given by y(x) =

∞ X

cn xn

n=0

and hence y0 = y 00 =

∞ X (n + 1)cn+1 xn

n=0 ∞ X

(n + 2)(n + 1)cn+2 xn .

n=0

(a) Substituting the series solution into the given ODE y 00 − 25y = 0 gives ∞ X {(n + 2)(n + 1)cn+2 − 25cn }xn = 0.

n=0

It follows that

(n + 2)(n + 1)cn+2 = 25cn for n = 0, 1, 2, · · · . We find that 52 c0 (if we let n = 0) 2! 52 = c1 (if we let n = 1) 3! 52 54 = c2 = c0 (if we let n = 2) 4·3 4! 52 54 = c3 = c1 (if we let n = 3) 5·4 5! .. .

c2 = c3 c4 c5

162

We deduce that c2k =

52k 52k c0 and c2k+1 = c1 (2k)! (2k + 1)! for k = 0, 1, 2, · · · .

Thus, the required series solution is ∞ ∞ X X 52k 2k 52k x + c1 x2k+1 y = c0 (2k)! (2k + 1)! k=0

k=0

where c0 and c1 are arbitrary constants. Since the coeﬃcients of the ODE are constant functions, their Taylor series about x = 0 should converge for all values of x. Thus, the above series solution converges for −∞ < x < ∞. (b) From the conditions y(0) = 2 and y 0 (0) = −5, we find that y(0) = c0 = 2 and y 0 (0) = c1 = −5. If we substitute the series solution into the given ODE, we obtain (n + 2)(n + 1)cn+2 = −5(n + 1)cn+1 + 6cn for n = 0, 1, 2, · · · .

It follows that 52 + 6 · 2 1 + 62 = 2! 2! 2 −5(1 + 6 ) − 6(5) 1 − 63 = = 3! 3! 3 2 −5(1 − 6 ) + 6(1 + 6 ) 1 + 64 = = 4! 4! .. . 1 + (−1)n 6n = for n = 0, 1, 2, · · · . n!

c2 = c3 c4

cn

Thus, the required solution is y=

∞ X 1 + (−1)n 6n

n=0

163

n!

xn .

The above series solution converges for −∞ < x < ∞ (as the coeﬃcients of y 0 and y in the ODE have Taylor series about x = 0 which converge for −∞ < x < ∞). (c) Rewrite ODE as (x − 1)2 y 00 + 5(x − 1)y 0 + 3y = 0

⇒ (x2 − 2x + 1)y 00 + 5(x − 1)y 0 + 3y = 0

.

Substitute the series solution into the ODE to obtain (x2 − 2x + 1) +5(x − 1)

∞ X

∞ X

(n + 2)(n + 1)cn+2 xn

n=0 n

(n + 1)cn+1 x + 3

n=0

∞ X

cn xn = 0

n=0

which can be rewritten as 3c0 − 5c1 + 2c2 + (3c1 − 10c2 + 5c1 + 6c3 − 4c2 )x ∞ X {3cn − 5(n + 1)cn+1 + 5ncn + n=2

+(n + 2)(n + 1)cn+2

= 0.

−2(n + 1)ncn+1 + n(n − 1)cn }xn

From given conditions, y(0) = c0 = 0 and y 0 (0) = c1 = 2. Thus, 3c0 − 5c1 + 2c2 = 0 ⇒ c2 = 5

3c1 − 10c2 + 5c1 + 6c3 − 4c2 = 0 ⇒ c3 = 9 and (n + 2)(n + 1)cn+2 = (5 + 2n)(n + 1)cn+1 −3cn − 5ncn − n(n − 1)cn

for n = 2, 3, · · · . 164

It follows that c4 = 14, c5 = 20, c6 = 27, c7 = 35 and so on. We observe that cn = cn−1 + 1 + n for n = 1, 2, 3, · · · . Hence cn = cn−1 + 1 + n = cn−2 + 1 + (n − 1) + 1 + n = cn−2 + 1 + n + n

= cn−3 + 1 + (n − 2) + 1 + n + n

= cn−3 + 1 + (n − 1) + n + n

= cn−4 + 1 + (n − 3) + 1 + (n − 1) + n + n

= cn−4 + 1 + (n − 2) + (n − 1) + n + n .. . = c0 + 1 + 2 + 3 + · · · + n + n 1 = n(n + 1) + n. 2 The required series solution is ∞ X 1 y= ( n(n + 1) + n)xn . 2 n=1

The series solution converges for −1 < x < 1. 2. Let the series solution be given by y(x) =

∞ X

cn xn .

n=0

(a) From the given conditions, c0 = 1 and c1 = 0. Proceeding as explained on page 133, we find that 2 · 1c2 3 · 2c3

c4

= =

−c0 sin(0) = 0

−c0 cos(0) − c1 sin(0) = −1 1 ⇒ c3 = − 3! 1 1 = 0, c5 = and c6 = . 5! 180 165

Thus, 1 1 5 1 6 y(x) ' 1 − x3 + x + x . 6 120 180 (b) Here c0 = c1 = 0 and (n + 2)(n + 1)cn+2 = 4(n + 1)cn+1 − 13cn for n = 0, 1, 2, · · · . Form the above, we can work out c2 = −9/2 and c3 = −49/6. Thus, 9 49 y(x) ' 1 + x − x2 − x3 . 2 6 (c) Here c0 = 0 and c1 = 1. Proceeding as explained on page 133, we find that 2 · 1c2

3 · 2c3

c4

= =

−(sin(0) + 0 cos(0)) = 0

−(1 · 2 · cos(0) + cos(0)) = −3 1 ⇒ c3 = − 2! 41 923 = 0, c5 = , c7 = − . 5! 7!

Thus, y(x) ' x −

1 3 41 5 923 7 x + x − x . 2! 5! 7!

3. Let the series solution be given by y(x) =

∞ X

cn xn .

n=0

Here c0 = 2 and c1 = 0. Substituting the series solution into the given ODE gives 2c2 + 42c0 = 0 ⇒ c2 = −42

−2c1 + 6c3 + 42c1 = 0 ⇒ c3 = 0

(p + 2)(p + 1)cp+2 = (p − 6)(p + 7)cp

for p = 2, 3, · · · .

166

It follows that 4 · 3c4 = (−4) · 9c2 ⇒ c4 = 126 462 c5 = 0, c6 = − , 5 c7 = 0, c8 = 0 .. . cn = 0 for n = 9, 10, 11, · · · . Thus, the required exact solution is y(x) = 2 − 42x2 + 126x4 −

462 6 x . 5

4. Let the series solution be given by y(x) =

∞ X

cn xn .

n=0

Here c0 = 0 and c1 = 2. Substituting the series solution into the given ODE gives 2c2 + 56c0 = 0 ⇒ c2 = 0

6c3 + 54c1 = 0 ⇒ c3 = −18

(p + 2)(p + 1)cp+2 = (p − 7)(p + 8)cp

for p = 2, 3, · · · .

It folllows that c4 = c6 = c8 = 0, c5 = 198/5, c7 = −858/35, c9 = 0 and cn = 0 for n ≥ 10. Thus, the required exact solution is y(x) = 2x − 18x3 +

5.8

198 5 858 7 x − x . 5 35

Solutions to Exercise IX

1. Let the series solution be given by y=

∞ X

n=0

167

dn xn+λ .

Substituting the series solution into the given ODE gives (6λ2 − 7λ − 3)d0 = 0

(6λ2 + 5λ − 4)d1 = 0

k(6k + 12λ − 7)dk = (k − 2 + λ)dk−2

for k = 2, 3, · · · .

To allow for the possibility that d0 is an arbitrary number, we choose λ such that 3 1 6λ2 − 7λ − 3 = 0 ⇒ λ = , λ = − . 2 3 Since 6λ2 + 5λ − 4 6= 0 for the above values of λ, we have to take d1 = 0. Also, since k(6k + 12λ − 7) 6= 0 for k = 2, 3, · · · , we can take d0 to be arbitrary and we find that d3 = d5 = d7 = ... = 0 (since d1 = 0). Let us take d0 = 1 to obtain a particular solution corresponding to each of the two values of λ. The two particular solutions corresponding to λ = 3/2 and λ = −1/3 are linearly independent since the diﬀerence between 3/2 and −1/3 is not an integer. For λ = 3/2, we obtain d2 =

3 3 11 , d4 = , d6 = ,··· 92 3680 691840

while for λ = −1/3, we find that 1 5 11 , d6 = − ,··· . d2 = − , d4 = − 6 936 84240 The required series solution is 3 2 3 4 11 x + x + x6 + · · · ) 92 3680 691840 1 5 4 11 6 +Bx−1/3 (1 − x2 − x + x + · · · ). 6 936 84240

y(x) = Ax3/2 (1 +

2. Rewrite the ODE as 4(x − 1)2 y 00 + 19(x − 1)y0 − (4 − (x − 1))y = 0. 168

Let the series solution be given by y=

∞ X

n=0

dn (x − 1)n+λ .

Substituting the series solution into the ODE gives (4λ2 + 15λ − 4)d0 = 0

n(4n + 8λ + 15)dn = −dn−1 for n = 1, 2, · · · . To allow for the possibility of having an arbitrary d0 , we choose λ to satisfy 4λ2 + 15λ − 4 = 0. This gives λ = 1/4 and λ = −4. Check that n(4n + 8λ + 15) 6= 0 for these values of λ and for n = 1, 2, · · · , that is, so that we can take d0 is arbitrary. Let us take d0 = 1. For λ = 1/4, we obtain d1 = −

1 1 1 , d2 = , d3 = − 21 1050 91350

and for λ = −4, we find that d1 =

1 1 1 , d2 = , d3 = . 13 234 3510

Thus, the required general solution is 1 2 1 1 x+ x − x3 + · · · ) 21 1050 91350 1 1 3 1 2 +Bx−4 (1 + x + x + x + · · · ). 13 234 3510

y(x) = Ax1/4 (1 −

3. Let the series solution be given by y=

∞ X

dn xn+λ .

n=0

Substituting the series solution into the given ODE gives (λ2 − 9)d0 = 0 d1 = 0

n(n + 2λ)dn = −dn−2 for n = 2, 3, · · · . 169

To allow for the possibility that d0 is arbitrary, we take λ2 − 9 = 0, that is, λ = −3 or λ = 3. However, for λ = −3, we see that n(n + 2λ) is zero when n = 6. Thus, for λ = −3, letting n = 6 in n(n + 2λ)dn = −dn−2 , we obtain d4 = 0 which in turn leads to d2 = 0 and d0 = 0. From d1 = 0 and n = 3, 5, 7, · · · , we deduce that d3 = d5 = d7 = · · · = 0. From n = 8, we find that 8 · 2d8 = −d6 . Also, 10 · 4d10 = −d8 , 12 · 6d12 = −d10 and so on. We can take either one of d6 , d8 , d10 , · · · to be arbitrary here. Taking d6 = 1 (here we choose d6 to be any real number except 0-can you see the reason for not choosing g6 = 0?), we find that λ = −3 gives 1 1 1 , d10 = and d12 = − . 16 640 46080 Thus, a particular series solution corresponding to λ = −3 is given by d8 = −

1 8 1 10 1 x + x − x12 + · · · ) 16 640 46080 1 1 7 1 = x3 − x5 + x − x9 + · · · 16 640 46080 If we attempt to find another particular solution by using λ = 3, we find that it is a merely a multiple of y1 above, that we cannot obtain two linearly independent solutions from λ = −3 and λ = 3. To seek another particular solution y2 which is linearly independent of y1 , let y1 (x) = x−3 (x6 −

y2 = y1 ln(x) +

∞ X

gn xn−3 .

n=0

Substituting the above into the ODE gives 5 4 7 6 9 x + x − x8 + · · · ) 16 640 46080 ∞ X −2 {n(n − 6)gn + gn−2 }xn−3 = 0. −5g1 x + 2x(3x2 −

n=2

Note that the coeﬃcient of gn in n(n − 6)gn + gn−2 is 0 when n = 6. As we will see, this forces g0 , g2 and g4 to 170

have definite values (that is, we cannot assign an arbitrary value to any one of these three coeﬃcients). Setting the coeﬃcient of x−2 to 0, we obtain g1 = 0. Setting the coeﬃcients of x−1 , x0 , x1 , x2 , x3 , x4 and so on to 0, we obtain −8g2 + g0

=

0

g3

=

0

8g4 − g2

=

0

=

6 + g4

=

0 ⇒ g5 = 0

5g5 − g3

7g7 − g5

0 ⇒ g4 = −6

⇒ g2 = −48 ⇒ g0 = −384 =

16g8 + g6

=

27g9 + g7

=

40g10 + g8

=

0 ⇒ g7 = 0 5 8 0 ⇒ g9 = 0 7 − 320 .. .

It appears that we can choose one of the coeﬃcients from g6 , g8 , g10 , · · · to be an arbitrary constant. If we choose g6 = 0 (we can choose g6 = 0 because here it does not lead to a trivial solution for y2 ) then g8 = 5/128 and g10 = −39/25600. The solution y2 is thus given by 1 5 1 7 1 x + x − x9 + · · · ) ln(x) 16 640 46080 +x−3 (−348 − 48x2 − 6x4 5 8 39 10 + x − x + · · · ). 128 25600

y2 = (x3 −

171

The required general solution is 1 5 1 7 1 x + x − x9 + · · · ) 16 640 46080 1 1 7 1 +B[(x3 − x5 + x − x9 + · · · ) ln(x) 16 640 46080 +x−3 (−348 − 48x2 − 6x4 5 8 39 10 + x − x + · · · )]. 128 25600

y = A(x3 −

(Note that in the above we choose g6 = 0 to work out the particular solution y2 . Now if we give g6 a diﬀerent value or if we choose to assign a value to either g8 or g10 (instead of g6 ), we will obtain a diﬀerent particular solution from y2 given above, but we should be able to recover that particular solution by selecting appropriate values of A and B in the general solution above. For example, if we choose g8 = 0, we obtain g6 = 5/8 and g10 = −7/12 800 , that is, we obtain the particular solution 1 5 1 7 1 x + x − x9 + · · · ) ln(x) 16 640 46080 +x−3 (−348 − 48x2 − 6x4 5 7 + x6 − x10 + · · · ). 8 12800

y2 = (x3 −

We can easily check that the particular solution above can be recovered from the general solution by letting A = 5/8 and B = 1.) 4. Let the series solution be given by y=

∞ X

dn xn+λ .

n=0

Substituting the series solution into the given ODE gives (3λ + 1)2 d0 = 0 (9[n − λ][n − λ − 1] + 15[n − λ] + 1)dn = −dn−1

for n = 1, 2, 3, · · · .

172

From the above, we obtain only one value for λ, that is, λ = −1/3. We find that the coeﬃcient (9[n − λ][n − λ − 1] + 15[n − λ] + 1) is not 0 for λ = −1/3 and n = 1, 2, · · · . Thus, we can assign a non-zero arbitrary value to d0 . Taking λ = −1/3 and d0 = 1, we find that 1 d1 = − 2 3 1 d2 = 2 (3 )2 22 1 d3 = − 2 3 (3 ) (3 · 2)(3 · 2) 1 d4 = 2 4 (3 ) (4 · 3 · 2)(4 · 3 · 2) .. . (−1)n dn = for n = 0, 1, 2, · · · 32n (n!)2 A particular solution of the ODE is ∞ X (−1)n n−1/3 y1 (x) = x . 32n (n!)2 n=0 To find another particular solution, let ∞ ∞ X X (−1)n n−1/3 −1/3 x +x gn xn . y2 (x) = ln(x) 2n (n!)2 3 n=0 n=0

If we substitute the above into the ODE and equate the coeﬃcients of x−1/3 , x2/3 , x5/3 and so on to 0, we obtain g0 = 0, g1 = 2/9, g2 = −1/108 and so on. Thus, the required general solution is ∞ X (−1)n n−1/3 y(x) = A x 32n (n!)2 n=0 +B{ln(x)

∞ X (−1)n n−1/3 x 32n (n!)2 n=0

2 1 2 +x−1/3 ( x − x + · · · )}. 9 108 173

5. Let the series solution be given by y=

∞ X

dn xn+λ .

n=0

Substituting the series solution into the given ODE gives (16λ2 − 8λ − 3)d0 = 0

8n(2n + 4nλ − 1)dn = dn−1

for n = 1, 2, 3, · · · .

We obtain λ = 3/4 and λ = −1/4. From λ = 3/4, we obtain one particular solution given by y1 (x) = x3/4 (1 +

1 1 1 2 x+ x + x3 + · · · ). 32 3072 589 824

The two diﬀerent values of λ give linearly dependent solutions. To obtain another particular solution which is not a multiple of y1 , let y2 (x) = y1 ln(x) + x−1/4

∞ X

gn xn .

n=0

Proceeding as before, we obtain g0 = 16, g1 = 0, g2 = −3/64, g3 = −7/9216 and so on. Thus, the required general solution is 1 1 1 2 x+ x + x3 + · · · ) 32 3072 589 824 1 2 1 x +B{x3/4 (1 + x + 32 3072 1 + x3 + · · · ) ln(x) 589 824 3 7 3 +x−1/4 (16 − x2 − x + · · · )}. 64 9216

y(x) = Ax3/4 (1 +

6. Let the series solution be given by y=

∞ X

n=0

dn (x − 1)n+λ .

174

Substituting the series solution into the ODE gives (16λ2 − 8λ − 3)d0 = 0

8n(2n + 4λ − 1)dn = (n + λ − 1)dn−1 for n = 1, 2, · · · .

Setting 16λ2 − 8λ − 3 = 0, we obtain λ = 3/4 and λ = −1/4. For λ = 3/4, we find that 8n(2n + 2)dn = (n − 1/4)dn−1 for n = 1, 2, · · · . Since 8n(2n + 2) 6= 0 for n = 1, 2, · · · , we can choose d0 = 1 when working out d1 , d2 and so on for λ = 3/4. For λ = 3/4, we obtain d1 = 3/128, d2 = 7/16 384, d3 = 77/12 582 912 and so on. Thus, we obtain the particular solution 3 77 7 y1 (x) = x3/4 (1 + x+ x2 + x3 128 16 384 12 582 912 77 + x4 + · · · ). 1073 741 824 If we attempt to find another particular solution using λ = −1/4, we obtain a solution which is a multiple of y1 above. To find another particular solution which is not a multiple of y1 above, we let −1/4

y2 (x) = y1 ln(x) + x

∞ X

gn xn .

n=0

Proceeding as before, we obtain g0 = −64, g1 = −4, g2 = −25/256, g3 = −31/16 384 and so on. Thus, the required general solution is 3 77 7 y1 (x) = Ax3/4 (1 + x+ x2 + x3 128 16 384 12 582 912 77 + x4 + · · · ) 1073 741 824 3 77 7 x+ x2 + x3 +B{x3/4 (1 + 128 16 384 12 582 912 77 + x4 + · · · ) ln(x) 1073 741 824 1 25 −64x−1/4 (1 + x + x2 16 16 384 31 x3 + · · · )}. + 1048 576 175

Chapter 6

Numerical methods 6.1

Preamble

Analytical methods of solution such as those given in Chapters 2 and 3 are not available for many ODEs. If an ODE cannot be solved exactly, one may resort to numerical methods for finding approximately the required solution. In this chapter, we look at the Euler’s method for solving numerically 1st order ODEs and show how it can be extended to solve 2nd order ODEs subject to certain prescribed conditions. The Euler’s method is definitely not the best numerical method for solving ODEs. It is examined here merely for instructional purposes to give a flavor of what is involved in the numerical solution of an ODE. A somewhat better (more accurate) numerical method is the second order Runge-Kutta method described in Problems 5 and 6 in Exercise X (page 193).

6.2

Euler’s method for 1st order ODEs

We consider now solving the 1st order ODE in y(x) given by dy = F (x, y) dx subject to the condition y(a) = c. Here F (x, y) is a given function of x and y and a and c are constants. 176

Let us assume that we are interested in finding y(x) for a ≤ x ≤ b. To find y(x) approximately, we divide the interval a ≤ x ≤ b into N smaller subintervals given by x0 ≤ x ≤ x1 , x1 ≤ x ≤ x2 , · · · , xk−1 ≤ x ≤ xk , · · · , xN−2 ≤ x ≤ xN−1 and xN−1 ≤ x ≤ xN , where a = x0 < x1 < · · · < xN−1 < xN = b. Let the value of y at the point x = xk be denoted by yk for k = 0, 1, 2, · · · , N. From the condition given by the problem, that is, y(a) = c, we know that y0 = c. The values of y1 , y2 , · · · , yk , · · · , yN−1 and yN are, however, not known. A simple approximate technique for finding y1 , y2 , · · · , yk , · · · , yN−1 and yN is the Euler’s method. To derive the Euler’s method, we use the Taylor series1 of y(x) about x = xk−1 , that is, y(x) = y(xk−1 ) + y0 (xk−1 )[x − xk−1 ] 1 + y 00 (xk−1 )[x − xk−1 ]2 2 1 + y 000 (xk−1 )[x − xk−1 ]3 + · · · . 6 For x very near xk−1 , y(x) may be reasonably approximated using only the first two terms of the series, that is, y(x) ' y(xk−1 ) + y 0 (xk−1 )[x − xk−1 ] for xk−1 ≤ x ≤ xk .

If we bear in mind that y 0 (x) = F (x, y) (from the given ODE) and y(xk−1 ) = yk−1 , we can write y(x) ' yk−1 + [x − xk−1 ]F (xk−1 , yk−1 ) for xk−1 ≤ x ≤ xk .

Thus, if yk−1 is known, we may calculate y(x) approximately for xk−1 ≤ x ≤ xk using the formula above. If we let x = xk in the formula above, we obtain yk ' yk−1 + [xk − xk−1 ]F (xk−1 , yk−1 ). 1

Refer to Section 5.2 in Chapter 5 for some details on Taylor series.

177

The above formula gives the Euler’s method for calculating approximate numerical values of y1 , y2 , · · · , yk , · · · , yN−1 and yN . If we let k = 1, we obtain y1 ' y0 + [x1 − x0 ]F (x0 , y0 ). Since y0 , x0 and x1 are known, we may calculate y1 using the approximation above. With y1 now known, we may let k = 2 and compute y2 using y2 ' y1 + [x2 − x1 ]F (x1 , y1 ) and subsequently y3 ' y2 + [x3 − x2 ]F (x2 , y2 )

y4 ' y3 + [x4 − x3 ]F (x3 , y3 ) .. . yN

' yN−1 + [xN − xN−1 ]F (xN−1 , yN−1 ).

All the above calculations may be done by hand, that is, on a piece of paper with the help of a hand calculator, if N is not very large. For large N, hand calculations become extremely tedious. In this case, we may write a computer program (using, for example, Fortran 77, Pascal or C++) to execute the large number of calculations. Alternatively, it may also be possible to implement the calculations on an electronic spreadsheet (for example, Microsoft Excel), as explained in the example below. To summarize, once the numerical values of y1 , y2 , · · · , yk , · · · , yN−1 and yN are obtained from the Euler’s method, that is, from yk ' yk−1 + [xk − xk−1 ]F (xk−1 , yk−1 ), we can calculate y(x) approximately for all x in the interval a ≤ x ≤ b by using y(x) ' yk−1 + [x − xk−1 ]F (xk−1 , yk−1 ) for xk−1 ≤ x ≤ xk .

178

As shown graphically in Figure 6.1, the Euler’s method approximates the actual solution curve using a set of straight lines over the subintervals x0 ≤ x ≤ x1 , x1 ≤ x ≤ x2 , · · · , xk−1 ≤ x ≤ xk , · · · , xN−2 ≤ x ≤ xN −1 and xN −1 ≤ x ≤ xN . The straight line over the subinterval xk−1 ≤ x ≤ xk passes through the point (xk−1 , yk−1 ) and its gradient is given by F (xk−1 , yk−1 ) (from the ODE y 0 = F (x, y)). Note that the straight line over the subinterval xk−1 ≤ x ≤ xk is parallel to the tangent to the solution curve at x = xk−1 .

Figure 6.1

Example Use the Euler’s method to solve numerically the ODE y 0 (x) = 2xy subject to y(0) = 1 over the interval 0 ≤ x ≤ 1, in order to find an approximate value for y(1). Note that this ODE can be solved exactly. It is a 1st order 2 separable ODE. The exact solution is y = ex . There is really no need to use the Euler’s method to solve this problem numerically. Nevertheless, for the purpose of illustration, we will 179

use the Euler’s method here to solve the ODE and examine the accuracy of the numerical results obtained through comparison 2 with the exact solution y = ex . We divide up the interval 0 ≤ x ≤ 1 into N smaller subintervals of equal size h = (1−0)/N = 1/N. Thus, x0 = 0, x1 = h, x2 = 2h, · · · , xk = kh, · · · , xN−1 = (N − 1)h and xN = N h = 1. In this case, xk − xk−1 = h for all k = 1, 2, · · · , N. The required approximate value for y(1) is given by yN . The Euler’s method is then given by the formula yk ' yk−1 + 2 h xk−1 · yk−1

= yk−1 + 2 h 2 (k − 1) · yk−1 for k = 1, 2, · · · , N.

If N is not very large, we may carry out the calculations in the formula by hand. For large N, we may write a simple computer program to carry out the calculations in the formula. The pseudocode for the calculations is as given below. read N {input a positive integer N } h=1.00/N {calculate h} ykm1=1.00 {value of y(0)} do k=1 to N {set current value of k} yk=ykm1+2*h*h*(k-1)*ykm1 {calculate yk at current k} ykm1=yk {give yk−1 the value of yk at current k} enddo {end of loop} print yk {print out approximate value of y(1)} Alternatively, we may carry out the calculations on an electronic spreadsheet. A typical spreadsheet comprises cells like Table 6.1. We will use only columns A to E to carry out the calculations required by the Euler’s method. We define them as follows. Column A:= input N Column B:= value of h = 1/N 180

Column C:= current value of the integer k Column D:= value of yk−1 for k given in column C Coulmn E:= value of yk for k given in column C

Table 6.1 A

B

C

D

E

F

G

1 2 3 4 We set up the spreadsheet as outlined in the steps below. 1. In row 1 of column A (that is, in the cell A1), we enter the positive integer N (the number of subintervals) (for example, 10). 2. In the cell B1, we insert the formula =1.0/A1. 3. In C1, we set the initial value of k to 1. 4. In D1, we enter the value 1 for the condition y(0) = 1. 5. In E1, we insert the formula for the Euler’s method, that is, =D1+2.0*B1*B1*(C1−1.0)*D1. 6. In A2, we insert the formula =A1, since the value of N does not change throughout the calculations. 7. In B2, we insert the formula =B1. 8. In C2, we insert the formula =C1+1, since k increases by 1 in the next step of the calculations. 9. In D2, we insert the formula =E1. 181

10. Copy and paste the formula in E1 into E2. If this is correctly done, the formula in E2 should be =D2+2.0*B2*B2*(C2−1.0)*D2. All the columns A to E now contain programmed formulae in them. We are ready to perform the rest of the calculations. If we are doing the calculations with N = 10, we need to command the sheet to repeat the calculations according to the programmed formulae from rows 3 to 10. The approximate numerical value of y(1), that is, y10 , is given in cell E10. In Table 6.2, we show the results of the calculations for N = 10. For N = 10, we find that the approximate numerical value of y(1) is 2.3346. How does this compare with the exact value? 2 Recall that the exact solution is y = ex . Thus, the exact value is y(1) = e = 2.7183 (accurate to 5 significant figures). The percentage error in our approximation is about 14%. Here the percentage error is calculated according to the formula |

“exact value of y(1)” − “numerical value of y(1)” | × 100%. “exact value of y(1)”

Table 6.2

1 2 3 4 5 6 7 8 9 10 Guide

A 10 10 10 10 10 10 10 10 10 10 N

B 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 h

C 1 2 3 4 5 6 7 8 9 10 k 182

D 1 1 1.02 1.0608 1.1244 1.2144 1.3358 1.4961 1.7056 1.9785 yk−1

E 1 1.02 1.0608 1.1244 1.2144 1.3358 1.4961 1.7056 1.9785 2.3346 yk

Table 6.3 N 10 50 100 500

Approximate y(1) 2.3346 2.6309 2.6738 2.7093

% Error 14 3.2 1.6 0.3

We can use the same spreadsheet to carry out calculations for diﬀerent values of N. We only need to change the value of A1 in the sheet. All other required calculations will then be carried out automatically. We repeat the calculations for N = 50, 100 and 500. The results for the approximate numerical values of y(1) together with the percentage errors of the calculations are shown in Table 6.3. From Table 6.3, it is obvious that the accuracy of Euler’s method improves if a larger value of N is used. For N = 500, the error in our approximation of y(1) is well under 1%.

6.3

Second order ODEs

Consider now solving the 2nd order ODE d2 y = F (x, y, y 0 ) dx2 subject to y(a) = c and y0 (a) = d. Here F (x, y, y 0 ) denotes a mathematical expression containing x, y and y 0 , and a, c and d are given constants. As in the Euler’s method, to solve the ODE for x within the interval a ≤ x ≤ b, we divide the interval up into N subintervals given by x0 ≤ x ≤ x1 , x1 ≤ x ≤ x2 , · · · , xk−1 ≤ x ≤ xk , · · · , xN−2 ≤ x ≤ xN−1 and xN −1 ≤ x ≤ xN , with x0 = a and xN = b. Also, as before, let the value of y at x = xk be yk . The value of y 0 at x = xk is denoted by yk0 . From y(a) = c and y 0 (a) = d, 183

we know that y0 = c and y00 = d. We are interested in finding approximate numerical values of y1 , y2 , · · · , yk , · · · , yN−1 and yN . Recall the approximation y(x) ' y(xk−1 ) + y0 (xk−1 )[x − xk−1 ] for xk−1 ≤ x ≤ xk .

In the above approximation, if we replace y and y 0 in the approximation by y0 and y 00 respectively, we obtain y0 (x) ' y0 (xk−1 ) + y 00 (xk−1 )[x − xk−1 ] for xk−1 ≤ x ≤ xk .

If we let x = xk in both approximate formulae above and use the given ODE y 00 (x) = F (x, y(x), y 0 (x)), we obtain 0 yk ' yk−1 + (xk − xk−1 ) yk−1

0 0 yk0 ' yk−1 + (xk − xk−1 ) F (xk−1 , yk−1 , yk−1 ).

Numerical values of y1 , y2 , · · · , yk , · · · , yN −1 and yN can be computed from the two formulae as follows. If we let k = 1, the first formula give y1 ' y0 + (x1 − x0 ) y00 . Since y0 and y00 are known, we may calculate y1 . For k = 1, we may also calculate y10 from the second formula. If we let k = 2, the first formula gives y2 ' y1 + (x2 − x1 ) y10 which can be calculated as y1 and y10 are known from the earlier calculation for k = 1. Similarly, after calculating y20 using the second formula with k = 2, we let k = 3 to work out y3 and y30 , and so on. As in the case of the Euler’s method, we can write a computer program or use an electronic spreadsheet to execute the calculations described above. Example Use the numerical method described above to solve the ODE y 00 = −y − 2y 0 + 12 x subject to y(0) = 0 and y0 (0) = 1, over the interval 0 ≤ x ≤ 1, in order to compute y(1) approximately. Divide the interval 0 ≤ x ≤ 1 up into 184

N subintervals of equal size. Compute for N = 5 and N = 10. Firstly, note the ODE may be rewritten as y 00 +2y 0 +y = 12 x. In Chapter 3, we have learnt how to solve this ODE exactly. If we solve it exactly subject to y(0) = 0 and y 0 (0) = 1, we obtain 1 3 y (x) = −1 + x + e−x + xe−x . 2 2 Hence y(1) = −1 + 12 + e−1 + 32 e−1 = 0.4197 (accurate to 4 significant figures). Here we will solve the problem numerically using the method described above and compare the approximate numerical value of y(1) with the exact one. We divide up the interval 0 ≤ x ≤ 1 into N smaller subintervals of equal size h = (1−0)/N = 1/N. Thus, x0 = 0, x1 = h, x2 = 2h, · · · , xk = kh, · · · , xN−1 = (N − 1)h and xN = N h = 1. In this case, xk − xk−1 = h for all k = 1, 2, · · · , N. The required approximate value for y(1) is given by yN . The formulae for the calculations are 1 0 0 + h[−yk−1 − 2yk−1 + (k − 1)h] yk0 ' yk−1 2 0 yk ' yk−1 + hyk−1 . For the calculations, we prepare a spreadsheet using seven of its columns, that is, columns A, B, C, D, E, F and G defined as follows. Column A:= input N Column B:= value of h = 1/N Column C:= current value of the integer k Column D:= value of yk−1 for k given in column C 0 for k given in column C Column E:= value of yk−1

Column F:= value of yk for k given in column C 185

Column G:= value of yk0 for k given in column C We set up the spreadsheet following the steps below. 1. In the cell A1, we enter a value for N. 2. In B1, we insert the formula =1/A1 to compute h. 3. In C1, we enter the initial value of 1 for k . 4. In D1, we enter 0 for y(0) = 0. 5. In E1, we enter 1 for y0 (0) = 1. 6. In F1, we insert the formula =D1+B1*E1. 7. In G1, we insert the formula =E1 + B1*sum( − D1, − 2.0*E1,0.5*(C1 − 1.0)*B1). 8. In A2, we insert the formula =A1. 9. In B2, we insert the formula =B1. 10. In C2, we increase the value of k in C1 by 1 by inserting the formula =C1+1. 11. In D2, we insert the formula =F1. 12. In E2, we insert =G1. 13. Copy and paste the formula in F1 into F2. If this is properly done, the formula in F2 should be =D2 + B2*E2. 14. Copy and paste the formula in G1 into G2. If this is properly done, the formula in G2 should be =E2 + B2*sum( − D2, − 2.0*E2,0.5*(C2 − 1.0)*B2). 186

The spreadsheet is now ready for carrying out the required calculations. Table 6.4

1 2 3 4 5 Guide

A 5 5 5 5 5 N

B 0.2 0.2 0.2 0.2 0.2 h

C 1 2 3 4 5 k

D 0 0.2 0.32 0.388 0.424 yk−1

E 1 0.6 0.34 0.18 0.0904 0 yk−1

F 0.2 0.32 0.388 0.424 0.44208 yk

G 0.6 0.34 0.18 0.0904 0.04944 yk0

Table 6.4 shows the results for N = 5. Calculations for N = 5 yield y5 = 0.4421 (in cell F5) as an approximate numerical value for y(1). If we compare this with the exact value of 0.4197, the error in the approximate value is about 5%. This is a reasonably good result, if we take into consideration that we use only N = 5. If we repeat the calculations for N = 10, we obtain y10 = 0.4298 as an approximate numerical value for y(1). The percentage error is reduced to about 2%.

6.4 6.4.1

Oscillation of a pendulum Nonlinear ODE

Consider a swinging pendulum which consists of a body of mass m attached to a very thin taut string of length L as shown in Figure 6.2. The position of the body is given by the angle θ (in radian) as defined in Figure 6.2. It varies with time t ≥ 0. Hence θ is a function of t. If we regard θ as positive when the body is on the right of the dotted vertical line in Figure 6.2, then θ is negative when the body is on the other side. If the only forces acting on the body are due to gravity and the tension in the string, then θ(t) satisfies the ODE θ00 (t) + k 2 sin(θ) = 0 187

p where k = g/L and g is the acceleration due to gravity (g ' 9.81 meter per second per second). The ODE above comes from Newton’s law of motion. We will not go into the details of its derivation here.

Figure 6.2 The ODE above is not a linear one. It is a second order nonlinear ODE which is diﬃcult (if not impossible) to solve exactly.

6.4.2

ODE for ‘very small’ oscillation

In elementary physics courses, it is usual to assume that θ has a ‘very small’ magnitude so that we may approximate2 sin(θ) by θ. With sin(θ) ' θ, the nonlinear ODE may be replaced approximately by θ00 (t) + k2 θ(t) = 0. The simpler ODE above is only approximately true. It is a good approximation only if the pendulum has a ‘very small’ oscillation. It is a homogeneous 2nd order linear ODE with 2 The Taylor series for sin(x) about x = 0 is sin(x) = x−x3 /6+x5 /120− · · · . If x is very, very small, sin(x) can be approximated by x if we throw away all other terms in the series.

188

constant coeﬃcients and can be solved as explained in Chapter 3. Without going into details, the general solution of the ODE for ‘very small oscillation’ is θ(t) = C sin(kt) + D cos(kt) where C and D are arbitrary constants. We may determine C and D if we know the state of motion of the body at a particular time. For example, if the position of the pendulum at time t = 0 is known, that is, θ(0) = θ0 , where θ0 is a given constant, then D = θ0 . If it is further known that the body is at rest at t = 0, that is, θ0 (0) = 0, then C = 0. For this particular case, the swinging of the pendulum is given by θ(t) = θ0 cos(kt). Once again, bear in mind that this solution is only approximately true for ‘very small’ oscillation.

6.4.3

Numerical solution for ‘larger’ oscillation

If we wish to investigate the swinging of the pendulum without assuming that the angle θ has a very small magnitude, we have to solve the ‘complete’ nonlinear ODE given by θ00 (t) + k2 sin(θ) = 0. Rewrite the ODE as θ00 (t) = −k 2 sin(θ). We can use the approximate method in Section 6.3 to solve the nonlinear ODE. Before we proceed any further, let us give a specific value for the parameter k. We choose k = 1 (that is, g/L = 1 per second per second) and solve the ODE subject to θ(0) = 1 (radian) and θ0 (0) = 0, over the time interval 0 ≤ t ≤ 1, that is, during the first one second. We divide up the interval 0 ≤ t ≤ 1 into N smaller subintervals 0 ≤ t ≤ h, h ≤ t ≤ 2h, · · · , (k − 1)h ≤ 189

t ≤ kh, · · · , (N − 2)h ≤ t ≤ (N − 1)h and (N − 1)h ≤ t ≤ 1, where h = 1/N. We define θk = θ(kh) and θ0k = θ0 (kh). For the numerical solution of the ODE, we use the formulae on page 184 to obtain θ0k ' θ0k−1 − h sin(θk−1 ) θk ' θk−1 + hθ0k−1

together with θ0 = 1 and θ00 = 0. We carry out the calculations on an electronic spreadsheet as described in Section 6.3. For N = 100, we obtain approximate values of θ at t = 0.2, 0.4, 0.6, 0.8 and 1.0. When the calculations are repeated for N = 200, we observe that the approximate numerical values of θ agree to at least 2 significant figures with the ones for N = 100. In Table 6.5, the approximate values of θ for N = 200 are compared with those values from the solution θ(t) = cos(t) which is approximately true for ‘very small’ oscillation (linear solution). It is obvious that the approximate values of θ obtained from the nonlinear ODE are quite close to the corresponding ones from the linear ODE, only when time t is small. As t increases, the diﬀerence between the nonlinear and linear solutions becomes significant. At time t = 1, the diﬀerence between the two is about 12% the linear value. The linear solution is not a good approximation for the oscillation of the pendulum, as we give the pendulum quite a large angular displacement (of 1 radian or approximately 57◦ ) before allowing it to swing from rest. At least for 0 ≤ t ≤ 1, the linear solution appears to predict that the pendulum is swinging faster than it should. Is this according to your expectation? Table 6.5 Time t 0.20 0.40 0.60 0.80 1.00

Nonlinear θ 0.9836 0.9340 0.8522 0.7404 0.6016 190

Linear θ 0.9801 0.9211 0.8253 0.6967 0.5403

6.5

Numerical prudence

In essence, what a numerical method for an ODE does is to replace the task of solving the ODE by a set of simpler formulae which can be computed to give approximate values of the solution at selected points in the solution interval. Prudence (for care, caution and good judgment) is needed when applying a numerical method for solving an ODE. If possible, we should analyse mathematically the errors in our approximations and also round-oﬀ errors due to computer limitations. Such an error analysis is beyond the scope of our studies here. Even without an error analysis, we can still check the validity of a numerical method in several diﬀerent ways. We can check that the numerical method works well for problems which have known exact solutions. By comparing the numerical results with the exact solutions, we may have some idea of the accuracy of the method. When using a numerical method to solve ODEs which do not have exact solutions, we can examine whether the numerical solution converges (as it should) or not when we refine our calculations. If no convergence is observed, the numerical method may not be working properly. We may also compare our numerical results with those obtained using other (diﬀerent) numerical methods. We may examine the numerical results obtained to see if they are intuitively and physically acceptable. For example, in using a numerical method to compute the mass of a piece of melting ice, we know that something is amiss if the numerical results show that the mass is increasing with time!

6.6

Exercise X

1. Use the Euler’s method to solve y 0 = 2x2 y2 numerically subject to the condition y(0) = 1/2, over the interval 0 ≤ x ≤ 1. (a) Divide the interval 0 ≤ x ≤ 1 into 10 equal subintervals and obtain numerical values of y at x = 0.2, 0.4, 0.6, 0.8 and 1.0. (b) Repeat the calculation by 191

dividing the interval 0 ≤ x ≤ 1 into 20 equal subintervals. (c) Find the exact solution of the problem and compare the numerical values you obtained with the exact solution. 2. If the motion of the body in the swinging pendulum in Section 6.4 experiences resistance from its surrounding medium, it may be modeled using the ODE g α θ00 (t) = − sin(θ(t)) − θ0 (t), L m where α > 0 is the medium resistant coeﬃcient. For g/L = 1 per second per second and α/m = 1 per second, use the numerical method in Section 6.3 to solve the ODE subject to θ(0) = 1 and θ0 (0) = 0 over the interval 0 ≤ t ≤ 1. Divide the interval up into 100 equal subintervals. Tabulate the numerical values of θ at t = 0.20, 0.40, 0.60, 0.80 and 1.00. Repeat the exercise by dividing 0 ≤ t ≤ 1 up into 200 subintervals. Check for convergence when the calculations are refined. Compare the results you obtained with those in Table 6.5. Are they physically acceptable? 3. Assuming that the angle θ(t) can be expressed in terms of its Taylor series about t = 0, one can use the first few terms in the series to construct an approximate solution for the swinging pendulum. Specifically, for small time t, one can make the approximation θ(t) ' θ(0) + θ0 (0)t +

θ00 (0) 2 θ000 (0) 3 θ0000 (0) 4 t + t + t . 2 6 24

The values of θ(0) and θ0 (0) are known from the given conditions at t = 0 and those of θ00 (0), θ000 (0) and θ0000 (0) can be found from the ODE which governs the motion of the swinging pendulum in Problem 2 above. (Diﬀerentiate the ODE once and twice to obtain expressions for θ000 (t) and θ0000 (t).) Use the above to obtain a ‘small time’ approximate solution for the swinging pendulum in Problem 2 and compute the values of θ approximately at at 192

t = 0.20, 0.40, 0.60, 0.80 and 1.00. Are the values in good agreement with those obtained in Problem 2? 4. Extend the Euler’s method to devise a numerical procedure for solving the 3rd order ODE y000 = F (x, y, y 0 , y 00 ) subject to y(a) = c, y 0 (a) = d and y00 (a) = s, where a, b, c, d and s are given numbers. 5. Second order Runge-Kutta method. As mentioned on page 179, in the Euler’s method for solving the 1st order ODE y 0 (x) = F (x, y) subject to y(a) = c, the solution curve is approximated using a set of straight lines over the subintervals a = x0 ≤ x ≤ x1 , x1 ≤ x ≤ x2 , · · · , xk−1 ≤ x ≤ xk , · · · , xN−2 ≤ x ≤ xN−1 and xN −1 ≤ x ≤ xN , and the straight line over the k-th subinterval xk−1 ≤ x ≤ xk has a gradient given by F (xk−1 , yk−1 ). (Note that y0 = c = y(x0 ), y1 = y(x1 ), y2 = y(x2 ) and so on.) In the second order Runge-Kutta method, the gradient of the straight line over xk−1 ≤ x ≤ xk is modified to take the value of F (x, y(x)) at x = 12 (xk−1 + xk ) (the midpoint of the k-th subinterval). From the earlier approximation (in the Euler’s method), that is, y(x) ' yk−1 + (x − xk−1 )F (xk−1 , yk−1 ) for xk−1 ≤ x ≤ xk ,

we obtain 1 1 y( (xk−1 + xk )) ' yk−1 + (xk − xk−1 )F (xk−1 , yk−1 ). 2 2 Thus, if we define 1 (xk−1 + xk ) 2 1 = yk−1 + (xk − xk−1 )F (xk−1 , yk−1 ) 2

xk = yk

then the second order Runge-Kutta method for finding numerically y1 , y2 , · · · , yN −1 and yN is given by the for193

mula yk ' yk−1 + (xk − xk−1 )F (xk , yk ) for k = 1, 2, · · · , N.

Repeat Problem 1 above using the second order RungeKutta method. Do you obtain more accurate numerical results? 6. Extend the second order Runge-Kutta method in Problem 5 to solve the 2nd order ODE y00 = F (x, y, y0 ) subject to y(a) = c and y0 (a) = d. Apply the extended method to solve the problem in the example on page 184, that is, solve the ODE y00 = −y −2y 0 + 12 x subject to y(0) = 0 and y 0 (0) = 1, over the interval 0 ≤ x ≤ 1, in order to compute y(1) approximately. Do you obtain more accurate numerical results?

6.7

Solutions to Exercise X

1. Proceed following closely the example on page 179. In setting up the spreadsheet for the calculations, one difference is in the formula to insert in column E. The formula to insert in column E is in accordance with yk ' 2 2 = yk−1 + 2 h 3 (k − 1)2 · yk−1 for yk−1 + 2 h x2k−1 · yk−1 k = 1, 2, · · · , N, that is, in the cell E1, we insert the formula =D1+2.0*B1*B1*B1*(C1−1.0)*(C1−1.0)*D1*D1. Another diﬀerence is in the condition y(0) = 1/2. This requires us to insert 0.50 (instead of 1) in the cell D1. The given ODE is 1st order separable. It can be easily solved subject to y(0) = 1/2, as explained in Chapter 2, to give the exact solution y(x) = −3(2x3 − 6)−1 . The numerical values of y at x = 0.2, 0.4, 0.6, 0.8 and 1.0, for N = 10 and 20, are compared with the exact solution below. 194

x 0.20 0.40 0.60 0.80 1.00

N = 10 0.5005 0.5070 0.5286 0.5782 0.6834

N = 20 0.5009 0.5089 0.5334 0.5896 0.7128

Exact 0. 501 3 0. 510 9 0. 538 8 0. 602 9 0.7500

The numerical values appear to converge to the exact ones as N is increased from 10 to 20. 2. To set up the spreadsheet for calculations, follow closely the example on page 184. Formulae to use are given by θ0k ' θ0k−1 − h(sin(θk−1 ) + θ0k−1 ) and θk ' θk−1 + hθ0k−1 together with θ0 = 1 and θ00 = 0. The numerical values of θ at t = 0.20, 0.40, 0.60, 0.80 and 1.00, for N = 100 and N = 200, are given in the table below. In the third column of the table, the numerical values of θ (taken from Table 6.5) for the nonlinear motion in the absence of resistance from the surrounding medium are also given. t 0.20 0.40 0.60 0.80 1.00

N = 100 0.9850 0.9423 0.8780 0.7974 0.7053

N = 200 0.9846 0.9418 0.8774 0.7968 0.7049

From Table 6.5 0.9836 0.9340 0.8522 0.7404 0.6016

The numerical values for N = 100 and M = 200 agree to at least 2 significant figures. At a given time t, the value of θ is larger than the corresponding value from Table 6.5. Thus, the motion of the pendulum here is slower than that recorded in Table 6.5. This is expected as we have here resistance to the motion from the surrounding medium. 3. The ODE is θ00 (t) = − sin θ − θ0 (t). From θ(0) = 1 and θ0 (0) = 0, we find that θ00 (0) = − sin(1) ' −0. 841 47. Differentiating, we find that θ000 (t) = −θ0 (t) cos θ − θ00 (t) and 195

θ0000 (t) = (θ0 (t))2 sin θ − θ00 (t) cos θ − θ000 (t), hence θ000 (0) ' 0. 841 47 and θ0000 (0) ' −0. 386 82. It follows that 0. 841 47 2 0. 841 47 3 0. 386 82 4 t + t − t , 2 6 24 θ(0.2) ' 0. 984 2, θ(0.4) ' 0. 941 2, θ(0.60) ' 0. 876 7, θ(t) ' 1 −

θ(0.80) ' 0. 795 9, θ(1.00) ' 0. 703 4.

The approximate values of θ at t = 0.20, 0.40, 0.60, 0.80 and 1.00, for N = 100 and N = 200 are quite close to those obtained in Problem 2. 4. For the 3rd order ODE y000 = F (x, y, y0 , y 00 ), we make the approximations 0 yk ' yk−1 + (xk − xk−1 ) yk−1 0 00 yk0 ' yk−1 + (xk − xk−1 ) yk−1

00 0 00 yk00 ' yk−1 + (xk − xk−1 ) F (xk−1 , yk−1 , yk−1 , yk−1 ).

In the above, as before, we divide the solution interval a ≤ x ≤ b into N subintervals a = x0 ≤ x ≤ x1 , x1 ≤ x ≤ x2 , · · · , xk−1 ≤ x ≤ xk , · · · , xN−2 ≤ x ≤ xN −1 and xN−1 ≤ x ≤ xN , and define yk = y(xk ), yk0 = y 0 (xk ), yk00 = y 00 (xk ) and yk000 = y 000 (xk ). In the above formulae, if we let k = 1, we can calculate y1 , y10 and y20 using the given conditions at x = a, that is, y0 = c, y00 = d and y000 = s. Subsequently, we can let k = 2, 3, · · · in order to compute y2 , y3 and so on. 5. The solution domain 0 ≤ x ≤ 1 is divided into N equal subintervals as in Problem 1 above. Here xk = kh, where h = 1/N. For the problem here, the formulae for the second order Runge-Kutta method are given by 2 y k = yk−1 + h3 (k − 1)2 yk−1 1 yk ' yk−1 + h3 (2k − 1)2 y2k . 2

We will use columns A to F of the spreadsheet (Table 6.1) to carry out the calculations. We define them as follows. 196

Column A:= input N Column B:= value of h = 1/N Column C:= current value of the integer k Column D:= value of yk−1 for k given in column C Column E:= value of y k for k given in column C Coulmn F:= value of yk for k given in column C We set up the spreadsheet as outlined in the steps below. (1) In A1, enter the positive integer N . (2) In C1, set the initial value of k to 1. (3) In D1, enter the value 0.50 for y(0) = 1/2. (4) In B1, insert the formula =1.0/A1. (5) In E1, insert the formula =D1+B1*B1*B1*(C1−1.0)*(C1−1.0)*D1*D1. (6) In F1, insert the formula =D1+0.5*B1*B1*B1*(2*C1−1.0)*(2*C1−1.0)*E1*E1.

(7) In A2, insert the formula =A1. (8) In B2, insert the formula =B1. (9) In C2, insert the formula =C1+1. (10) In D2, insert the formula =F1. (11) Copy and paste the formula in E1 into E2. (12) Copy and paste the formula in F1 into F2.

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The numerical values of y at x = 0.2, 0.4, 0.6, 0.8 and 1.0, for N = 10 and 20, are compared with the exact solution below. x 0.20 0.40 0.60 0.80 1.00

N = 10 0.5012 0.5107 0.5383 0.6015 0.7446

N = 20 0.5013 0.5108 0.5387 0.6025 0.7485

Exact 0. 501 3 0. 510 9 0. 538 8 0. 602 9 0.7500

Convergence is obvious as N increases from 10 to 20. The numerical values here are definitely more accurate than those in Problem 1 above. 6. Recall that if we extend the Euler’s method to solve the 2nd order ODE y 00 = F (x, y, y 0 ), the formulae to use are 0 yk ' yk−1 + (xk − xk−1 ) yk−1

0 0 yk0 ' yk−1 + (xk − xk−1 ) F (xk−1 , yk−1 , yk−1 ).

The extension of the second order Runge-Kutta method to the 2nd order ODE are given by the following formulae: 1 (xk−1 + xk ) 2 1 0 = yk−1 + (xk − xk−1 )yk−1 2 1 0 0 = yk−1 + (xk − xk−1 )F (xk−1 , yk−1 , yk−1 ) 2 ' yk−1 + (xk − xk−1 ) y 0k

xk = yk y0k yk

0 yk0 ' yk−1 + (xk − xk−1 ) F (xk , y k , y0k ).

For the ODE y00 = −y − 2y 0 + 12 x, the interval 0 ≤ x ≤ 1 and the conditions y(0) = 0 and y 0 (0) = 1, if we divide the interval into N equal subintervals (as before), we obtain

198

xk = kh (h = 1/N ), y0 = 0 and y00 = 1 and the formulae xk = yk = y0k = yk ' yk0 '

1 (2k − 1)h 2 1 0 yk−1 + hyk−1 2 1 1 0 0 yk−1 + h(−yk−1 − 2yk−1 + (k − 1)h) 2 2 yk−1 + hy 0k 1 0 yk−1 + h(−y k − 2y 0k + (2k − 1)h). 4

We prepare the spreadsheet for calculations as follows. Column A:= input N Column B:= value of h = 1/N Column C:= current value of the integer k Column D:= value of yk−1 for k given in column C 0 for k given in column C Column E:= value of yk−1

Column F:= value of y k for k given in column C Column G:= value of y 0k for k given in column C Column H:= value of yk for k given in column C Column I:= value of yk0 for k given in column C (1) In A1, enter a value for N. (2) In B1, insert the formula =1/A1. (3) In C1, enter 1 for k . (4) In D1, enter 0 for y(0) = 0. (5) In E1, enter 1 for y 0 (0) = 1. (6) In F1, insert the formula =D1+0.5*B1*E1. 199

(7) In G1, insert the formula =E1 + 0.5*B1*sum( − D1, − 2.0*E1,0.5*(C1 − 1.0)*B1). (8) In H1, insert the formula =D1+B1*G1. (9) In I1, insert the formula =E1 + B1*sum( − F1, − 2.0*G1,0.25*(2*C1 − 1.0)*B1). (10) In A2, insert the formula =A1. (11) In B2, insert the formula =B1. (12) In C2, insert the formula =C1+1. (13) In D2, insert the formula =H1. (14) In E2, insert =I1. (15) Copy and paste the formula in F1 into F2. (16) Copy and paste the formula in G1 into G2. (17) Copy and paste the formula in H1 into H2. (18) Copy and paste the formula in I1 into I2. For N = 5, we obtain y(1) ' 0.4133 (about 1. 5% error) (compared to 5% error by the extended Euler’s method). For N = 10, y(1) ' 0.4183 (about 0. 3% error) (compared to 2% error by the extended Euler’s method). Yes, more accurate results are obtained here. For N = 50, we find that y(1) ' 0.4196 (the exact value of y(1) to 4 significant figures is 0.4197).

200