Introductory Notes in Ordinary Differential Equations for Physical

Lecture Notes in Mathematics

Arkansas Tech UniversityDepartment of Mathematics

Introductory Notes in OrdinaryDifferential Equations for Physical

Sciences and Engineering

Marcel B. Financ©All Rights Reserved

May 4, 2017

Preface

Differential equations arise from the study of problems in virtually every areaof physical sciences as well as in engineering. Many mathematical models ofreal world phenomena lead to a differential equation problem for which weseek its solution. In this course, we will explore analytical, numerical andgraphical techniques for finding the solutions.To gain a solid grasp of differential equations, it is essential for the reader tosolve ALL the exercises listed at the end of each section. Answers to mostexercises are available at the end of the book.A solution guide is available by request: Email [email protected]

Marcel B. FinanRussellville, ArkansasMay 2011

i

ii PREFACE

Contents

Preface i

1 Basic Terminology 3

2 Existence and Uniqueness of Solutions to First Order LinearIVP 13

3 Analytical Solution: The Method of Integrating Factor 19

4 Existence and Uniqueness of Solutions to First Order Nonlin-ear IVP 29

5 Separable Differential Equations 35

6 Exact Differential Equations 43

7 Substitution Techniques: Bernoulli and Riccati Equations 51

8 Graphical Solution: Direction Field of y′ = f(t, y) 59

9 Numerical Solutions to ODEs: Euler’s Method and its Vari-ants 69

10 Second Order Linear Differential Equations: Existence andUniqueness Results 75

11 The General Solution of 2nd Order Linear Homogeneous Equa-tions 81

1

2 CONTENTS

12 Second Order Linear Homogeneous Equations with ConstantCoefficients: Distinct Characterisitc Roots 91

13 Characteristic Equations with Repeated Roots 97

14 Characteristic Equations with Complex Roots 103

15 Series Solutions of Differential Equations 111

16 The Structure of the General Solution of Linear Nonhomo-geneous Equations 117

17 The Method of Variation of Parameters 123

18 The Laplace Transform: Basic Definitions and Results 129

19 Further Studies of Laplace Transform 141

20 The Laplace Transform and the Method of Partial Fractions151

21 Laplace Transforms of Periodic Functions 159

22 Convolution Integrals 167

Cheat Sheets 173

Answer Key 181

Index 223

1 Basic Terminology

In many models, we will have equations involving the derivatives of a depen-dent variable y with respect to one or more independent variables and areinterested in discovering this function y. Such equations are referred to asdifferential equations (abbreviated DE). They arise in many applicationssuch as population growth, decay of radioactive substance, the motion of afalling object, electrical network, Newton’s law of cooling and many moremodels.

A First Source of Differential Equations: Vertical Motion of anObjectWe consider the model of uniform motion. Suppose that an object initiallyat height y0 is moving straight up or down with initial velocity v0. Let y(t)denote the distance of the object from the ground , v(t) the object’s velocity,and a(t) the object’s acceleration at time t. We assume y to be positive inthe upward direction.If air resistance is neglected, then by Newton’s second law, which statesthat the net force is equal to the product of mass and acceleration, we havema(t) = −mg. The negative sign on the right-hand of the equation is dueto the fact that acceleration due to gravity is pointing downward. Using thefact that a(t) = y′′(t) and eliminating the mass, we obtain the equation

y′′ = −g.

To find the velocity v(t) we integrate for a first time and obtain

v(t) = y′(t) = −gt+ C1.

Since the initial velocity is v0, we obtain C1 = v(0) = v0 so that

v(t) = −gt+ v0.

3

4 1 BASIC TERMINOLOGY

Integrating for the second time we find the position function

y(t) = −1

2gt2 + v0t+ C2.

Since y0 is the initial height, we find C2 = y0 and so

y(t) = −1

2gt2 + v0t+ y0.

Example 1.1An object is dropped from the top of a cliff that is 144 feet above groundlevel.(a) When will the object reach ground level?(b) What is the velocity with which the object strikes the ground?

Solution.(a) The motion of the object translates to the differential equation y′′ = −32with solution y(t) = −16t2 + 144. The object reaches ground level when

y(t) = 0 or 16t2 = 144. Solving for t we find t =√

14416

= 3 sec. The object

will reach the ground 3 seconds after it is dropped from the cliff.(b) The object strikes the ground with velocity v(3) = −32(3) = −96 ft/sec

Basic Terms of Differential EquationsWe next discuss some basic notions of differential equations. There are twotypes of differential equations: ordinary and partial differential equations.By an ordinary differential equation (abbreviated ODE) we mean anequation that involves an unknown function (the dependent variable) ofa single variable, its independent variable, and one or more of its deriva-tives. The highest order derivative that appears in the equation is knownas the order of the equation. Thus, an nth order ordinary differentialequation is an equation of the form

y(n) = f(t, y, y′, · · · , y(n−1))

or alternativelyG(t, y, y′, y′′, · · · , yn) = 0.

A first-order ordinary differential equation, for example, takes the formf(t, y(t), y′(t)) = 0, and may alternatively be written as

y′(t) = g(t, y(t))

5

for all t in the interval of existence of y.Similarly, a second-order ordinary differential equation takes the formf(t, y(t), y′(t), y′′(t)) = 0 or y′′ = h(t, y, y′).

Example 1.2Determine the order of each equation.(a) y′ + 2ty = e−t

2.

(b) d2ydt2− 5dy

dt+ 6y(t) = 0.

(c) y′′ + 3ty′ + 2y = sin (5t).

Solution.(a) This is a first order differential equation because the highest derivative isthe first derivative.(b) and (c) are second order differential equations since the highest derivativein each equation is the second order derivative

When a dependent function is a function of two or more independent variablesthen the derivatives are known as partial derivatives. An equation thatinvolves a function of more than two independent variables and its partialderivatives is called a partial differential equation (abbreviated PDE).For example, the wave equation is a partial differential equation of theform

∂2u

∂x2− 1

c2∂2u

∂t2= 0.

In this course, when we use the term differential equation, we’ll mean anordinary differential equation.A solution of a differential equation is a function that satisfies the equation:When you substitute this function and its derivatives into the differentialequation, you get a true mathematical statement.

Example 1.3Show that the function y = 100+e−t is a solution to the differential equation

y′ = 100− y.

Solution.Indeed, finding the first order derivative of y we have y′ = −e−t. Also, 100−y = 100− (100 + e−t) = −e−t. Thus, y′ = 100− y so that y = 100 + e−t is asolution to the given DE


Example 1.4 (A piecewise-defined solution)Consider the differential equation ty′ − 4y = 0 on the interval (−∞,∞).Verify that the piecewise-defined function

y =

{−t4, t < 0t4, t ≥ 0

is a solution.

Solution.For t < 0, we have ty′ − 4y = t(−t4)′ − 4(−t4) = −4t4 + 4t4 = 0. For t ≥ 0,we have ty′ − 4y = t(t4)′ − 4t4 = 4t4 − 4t4 = 0. Thus, the given function is asolution

Solving a differential equation means finding all possible solutions of theequation.

Example 1.5Solve the differential equation:

y′′ = −2t.

Solution.Integrating twice, all the solutions have the form

y(t) = −t3

3+ C1t+ C2

Note that the function of the previous example defines all the solutions tothe differential equation. Such a function will be referred to as the generalsolution. The constants C1 and C2 are called the parameters. Specificvalues of C1 and C2 determine what is called a particular solution. To finda particular solution additional conditions on the values of the function orits derivatives must be given. Such conditions are called initial conditions.A differential equation together with a set of initial conditions is called aninitial value problem (abbreviated IVP).

Example 1.6Consider the differential equation y′′(t)− 1 = 0.(a) Find the general solution of this equation.(b) Find the solution that satisfies the initial conditions y(1) = 1 and y′(1) =4.

7

Solution.(a) Integrating twice we find the general solution

y(t) =t2

2+ C1t+ C2.

(b) Since y′(t) = t + C1 and y′(1) = 4, we find 4 = 1 + C1 so that C1 = 3.Hence, y(t) = t2

2+ 3t + C2. Now, since y(1) = 1, we have 1 = 1

2+ 3 + C2.

Solving for C2 we find C2 = −52. Hence, the solution to the IVP{y′′(t)− 1 = 0

y′(1) = 4, y(1) = 1

is

y(t) =t2

2+ 3t− 5

2

The graph of a particular solution is called a solution curve. The functiony(t) = Ce−3t + 2t + 1 is the general solution to the differential equationy′ + 3y = 6t + 5. A family of solution curves is shown in Figure 1.1. Noticefor C 6= 0 the solution curves have an oblique asymptote with equationy(t) = 2t+ 1.

Figure 1.1

Sometimes a differential equation possesses a solution that cannot be ob-tained by assigning values to the parameters in a family of solutions. Such asolution is called a singular solution.


Example 1.7Using the method of separation of variables (Chapter 5), the non-zero solu-

tions to the differential equation y′ = ty12 are given by y(t) = ( t

2

4+C)2. Find

the singular solution.

Solution.The function y(t) ≡ 0 is a solution to the differential equation. This is asingular solution since it cannot be obtained from the family for any choiceof the parameter C. The general solution consists of all the solutions of theform y(t) = ( t

2

4+ C)2 together with the zero solution

9

Practice Problems

Problem 1.1A car starts from rest and accelerates in a straight line at 1.6 m/sec2 for 10seconds.(a) What is its final speed?(b) How far has it travelled in this time?

Problem 1.2An object is thrown upward at time t = 0. After t seconds, its height is

y(t) = −4.9t2 + 7t+ 1.5

meters above the ground.(a) From what height was the object thrown?(b) What is the initial velocity of the object?(c) What is the acceleration due to gravity?

Problem 1.3An object thrown in the air on a planet in a distant galaxy is at heighty(t) = −25t2 + 72t + 40 feet at time t seconds after it is thrown. What isthe acceleration due to gravity on that planet? With what velocity was theobject thrown? From what height?

Problem 1.4An object is dropped from a 400-foot tower. When does it hit the groundand how fast is it going at the time of the impact?

Problem 1.5A ball that is dropped from a window hits the ground in five seconds. Howhigh is the window? (Give your answer in feet.)

Problem 1.6A ball is thrown straight up from ground level and reaches its greatest heightafter 5 seconds. Find the initial velocity of the ball and the value of itsmaximum height above ground level.

Problem 1.7At time t = 0 an object having mass m is released from rest at a height y0above the ground. Let g represent the constant gravitational acceleration.Derive an expression for the impact time (the time at which the object strikesthe ground). What is the velocity with which the object strikes the ground?


Problem 1.8At time t = 0, an object of mass m is released from rest at a height of252 ft above the floor of an experimental chamber in which gravitationalacceleration has been slightly modified. Assume (instead of the usual valueof 32 ft/sec2), that the acceleration has the form 32−ε sin

(πt4

)ft/sec2, where ε

is a positive constant. In addition, assume that the object strikes the groundexactly 4 sec after release. Can this information be used to determine theconstant ε? If so, determine ε.

Problem 1.9Find the order of the following differential equations.(a) ty′′ + y = t3.(b) y′ + y2 = 2.(c) sin (y′′′) + 3t2y = 6t.

Problem 1.10What is the order of the differential equation?(a) y′(t)− 1 = 0.(b) y′′(t)− 1 = 0.(c) y′′(t)− 2ty(t) = 0.

(d) y′′(t)(y′(t))12 − t

y(t)= 0.

Problem 1.11In the equation

∂u

∂x− ∂u

∂y= x− 2y,

identify the independent variable(s) and the dependent variable.

Problem 1.12Classify the following equations as either ordinary or partial.(a) (y′′′)4 + t2

(y′)2+4= 0.

(b) ∂u∂x

+ y ∂u∂y

= y−xy+x

.

(c) y′′ − 4y = 0.

Problem 1.13Solve the equation y′′′(t)− 2 = 0 by computing successive antiderivatives.

11

Problem 1.14Solve the initial-value problem

dy

dt= 3y, y(0) = 50.

What is the domain of the solution?

Problem 1.15For what real value(s) of λ is y = cosλt a solution of the equation y′′+9y = 0?

Problem 1.16For what value(s) of m is y = emt a solution of the equation y′′+3y′+2y = 0?

Problem 1.17Show that y(t) = et is a solution to the differential equation

y′′ −(

2 +2

t

)y′ +

(1 +

2

t

)y = 0.

Problem 1.18Show that any function of the form y(t) = C1 cosωt + C2 sinωt satisfies thedifferential equation

d2y

dt2+ ω2y = 0.

Problem 1.19Suppose y(t) = 2e−4t is the solution to the initial value problem y′ + ky =0, y(0) = y0. Find the values of k and y0.

Problem 1.20Consider t > 0. For what value(s) of the constant n, if any, is y(t) = tn asolution to the differential equation

t2y′′ − 2ty′ + 2y = 0?

Problem 1.21(a) Show that y(t) = C1e

2t +C2e−2t is a solution of the differential equation

y′′ − 4y = 0, where C1 and C2 are arbitrary constants.(b) Find the solution satisfying y(0) = 2 and y′(0) = 0.(c) Find the solution satisfying y(0) = 2 and limt→∞ y(t) = 0.


Problem 1.22Suppose that the graph below is the particular solution to the initial valueproblem y′(t) = m + 1, y(1) = y0. Determine the constants m and y0 andthen find the formula for y(t).

Problem 1.23Suppose that the graph below is the particular solution to the initial valueproblem y′(t) = mt, y(t0) = −1. Determine the constants m and t0 and thenfind the formula for y(t).

Problem 1.24Show that y(t) = e2t is not a solution to the differential equation y′′+4y = 0.

Problem 1.25Consider the initial-value problem

y′ + 3y = 6t+ 5, y(0) = 3.

(a) Show that y = Ce−3t + 2t + 1 is a solution to the above differentialequation.(b) Find the value of C.

2 Existence and Uniqueness ofSolutions to First Order LinearIVP

Solutions to differential equations can be given in one of the following forms:• by an explicit formula: For example, the function y =

√t3 + 1 is an explicit

solution to the initial value problem 2yy′ = 3t2, y(1) =√

2;

• by an implicit equation: The solution y to the equation y′ = −1+yety

1+tetyis

defined implicitly by the equation t+ y + ety = C;• by a power series representation. For example the general solution to theequation (t2 − 4)y′′ + 3ty′ + y = 0 is given by

y(t) = C1

(1 +

1

8t2 +

3

128t4 +

5

1024t6 + · · ·

)+C2

(t+

1

6t3 +

1

30t5 +

1

140t7 + · · ·

);

• numerically (i.e., Euler’s methods);• graphically (i.e., direction fields).

Before worrying about how to solve an initial value problem, either ana-lytically, graphically, or numerically, one should first resolve the basic issuesof existence and uniqueness. First, does a solution exist? If, not, it makes nosense trying to find one. Second, is the solution uniquely determined by theinitial conditions? Otherwise, the differential equation probably has littlerelevance for physical applications since we cannot use it as a predictive tool.In this section we discuss the conditions needed to guarantee the existence ofa unique solution to first order linear initial value problems. We start withthe definition of a first order linear differential equation.Any differential equation that can be written in the form

y′ + p(t)y = g(t) (2.1)

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142 EXISTENCE ANDUNIQUENESS OF SOLUTIONS TO FIRST ORDER LINEAR IVP

where p(t) and g(t) are functions with common domain a < t < b, is called afirst order linear differential equation. The term linear is used becauseL(y) = y′ + p(t)y is linear in y. That is, L(αy1 + βy2) = αL(y1) + βL(y2). ADE that is not linear is called non-linear.In mathematics and physics, linear generally means “simple” and non-linearmeans “complicated”. The theory for solving linear equations is very welldeveloped because linear equations are simple enough to be solvable. Non-linear equations can usually not be solved exactly and are the subject ofmuch on-going research.Now, we say that Equation (2.1) is homogeneous if g(t) ≡ 0 for all a <t < b. If there is a a < t < b such that g(t) 6= 0 then Equation (2.1) is callednon-homogeneous.

Example 2.1Classify each of the following first order differential equations as linear ornon-linear. If the equation is linear, decide whether it is homogeneous ornon-homogeneous.(a) dy

dt+ y

10= ty.

(b) t2 − 3y2 + 2ty dydt

= 0.

(c) tdydt

= t2 − 2y.

(d) dydt

= t−yt+y

.

Solution.(a) Notice that the given equation can be written as dy

dt+ ( 1

10− t)y = 0 which

is a homogeneous first order linear DE where p(t) = 110− t and g(t) = 0.

(b) This is non-linear because of the term y2.(c) This is a non-homogeneous first order linear DE since the right-hand sideis not identically zero on any interval. Here, we have p(t) = 2

tand g(t) = t.

(d) This is non-linear because of the y in the denominator

First order linear differential equations possess important linearity or su-perposition properties.

Theorem 2.1(a) If y1(t) and y2(t) are any two solutions of the homogeneous equationy′+p(t)y = 0 then for any constants c1 and c2 the linear combination c1y1(t)+c2y2(t) is also a solution of the homogeneous equation.(b) If y1(t) is a solution to the homogeneous equation y′ + p(t)y = 0 and

15

y2(t) is a solution to the non-homogeneous equation y′ + p(t)y = g(t) thenCy1(t) + y2(t) is also a solution to the non-homogeneous equation, where Cis an arbitrary constant.

Proof.(a) Since y1(t) and y2(t) are solutions to the homogeneous equation, we have

(c1y1 + c2y2)′ + p(t)(c1y1 + c2y2) = c1(y

′1 + p(t)y1) + c2(y

′2 + p(t)y2) = 0 + 0 = 0.

(b) We have

(Cy1 + y2)′ + p(t)(Cy1 + y2) = C(y′1 + p(t)y1) + y′2 + p(t)y2 = 0 + g(t) = g(t)

Remark 2.1Part (a) of the previous theorem is not true in general for non-homogeneous equa-tions. For example, consider the equation y′ = 1. Then y1(1) = t and y2(t) = t+ 1are both solutions to the DE. However, y1(t)+y2(t) = 2t+1 is not a solution since(y1 + y2)

′ = 2 6= 1

Next, we look at the conditions that guarantee the existence of a unique solutionto the IVP

y′ + p(t)y = g(t), y(t0) = y0. (2.2)

Theorem 2.2If p(t) and g(t) are continuous functions in the open interval I = (a, b) and t0 apoint inside I then the IVP (2.2) has a unique solution y(t) defined on I.

Example 2.2If p(t) is continuous on (a, b) and a < t0 < b then what is the unique solution tothe IVP

y′ + p(t)y = 0, y(t0) = 0?

Solution.Since the trivial solution y(t) ≡ 0 satisfies the initial value problem, by Theorem2.2, y(t) ≡ 0 is the unique solution

Remark 2.2Theorem 2.2 asserts that if the hypotheses are satisfied then a unique solutionexists on an interval containing t0. However, the solution might actually exist ona larger interval than what the theorem asserts. To be more precise, consider theIVP ty′+2y = 0, y(1) = 0. If we apply Theorem 2.2, then a unique solution existssay on the interval 0 < t < ∞ since this is the interval containing 1 and wherep(t) = 1

t is defined and continuous. Actually, the solution is y(t) ≡ 0. But one caneasily see that y(t) ≡ 0 is a solution for all −∞ < t < ∞. So our theorem assertsthe existence of a solution locally rather than globally


Practice Problems

Problem 2.1Find p(t) and y0 so that the function y(t) = 3et

2is the solution to the IVP

y′ + p(t)y = 0, y(0) = y0.

Problem 2.2For each of the initial conditions, determine the largest interval a < t < b on whichTheorem 2.2 guarantees the existence of a unique solution.

y′ +1

t2 + 1y = sin t

(a) y(0) = π (b) y(π) = 0.

Problem 2.3For each of the initial conditions, determine the largest interval a < t < b on whichTheorem 2.2 guarantees the existence of a unique solution

y′ +t

t2 − 4y =

et

t− 3

(a) y(5) = 2 (b) y(−32) = 1 (c) y(−6) = 2.

Problem 2.4(a) For what values of the constant C and the exponent r is y = Ctr the solutionof the IVP

2ty′ − 6y = 0, y(−2) = 8?

(b) Determine the largest interval of the form a < t < b on which Theorem 2.2guarantees the existence of a unique solution.(c) What is the actual interval of existence for the solution found in part (a)?

Problem 2.5What information does Theorem 2.2 give about the initial value problem ty′ =y + t3 cos t, y(1) = 1?y(−1) = 1?

Problem 2.6Consider the following differential equation

(t− 4)y′ + 3y =1

t2 + 5t.

Without solving, find the interval over which a unique solution is guaranteed foreach of the following initial conditions:(a) y(−3) = 4 (b) y(1.5) = −2 (c) y(−6) = 0 (d) y(4.1) = 3.

17

Problem 2.7Without solving the initial value problem, (t − 1)y′ + (ln t)y = 2

t−3 , y(t0) = y0,state whether or not a unique solution is guaranteed to exist for the y(t0) = y0listed below. If a unique solution is guaranteed, find the largest interval for whichthe solution satisfies the differential equation and the initial condition.(a) y(2) = 4 (b) y(0) = 0 (c) y(4) = 2.

Problem 2.8(a) State precisely the theorem (hypothesis and conclusion) for existence anduniqueness of a linear first order initial value problem.(b) Consider the equation y′ + t2y = et

3with initial conditions y(t0) = y0. Briefly

discuss if this has a solution, if it is unique and why.

Problem 2.9Is the differential equation linear or non-linear? If the equation is linear, decidewhether it is homogeneous or non-homogeneous.(a) y′ = ty2

(b) y′ = t2y(c) (cos t)y′ + ety = sin t

(d) y′

y + t3 = sin t, y > 0.

Problem 2.10Consider the initial value problem

y′ + p(t)y = g(t), y(3) = 1.

Suppose that p(t) and/or g(t) have discontinuities at t = −2, t = 0, and t = 5 butare continuous for all other values of t. What is the largest interval (a, b) on whichTheorem 2.2 is applied.

Problem 2.11Determine α and y0 so that the graph of the solution to the initial-value problem

y′ + αy = 0, y(0) = y0

passes through the points (1, 4) and (3, 1).

Problem 2.12Match the following equations with the correct description. Every equation matchesexactly one description.(a) y′ = 3y − 5t.

(b) ∂yt = ∂2y

∂t2+ ∂2y

∂x2.


(c) y′ − y2 = sin t.(d) y′ + 3y = 0.

(i) A partial differential equation.(ii) A homogeneous first order linear differential equation.(iii) A non-linear first order differential equation.(iv) A non-homogenous first order linear differential equation.

Problem 2.13Consider the differential equation y′ = −t2y + sin y. What is the order of thisequation? Is it linear or non-linear?

Problem 2.14Show that y′ = t+y

t is a linear first order non-homogeneous equation.

3 Analytical Solution: TheMethod of Integrating Factor

If a unique solution to the initial value problem (2.2) exists then this solutionis found by using the initial condition in the general solution to the differentialequation to determine the value of the parameter. The purpose of this section isto find analytically the general solution to the first order linear non-homogeneousequation

y′ + p(t)y = g(t) (3.1)

where p(t) and g(t) are continuous on the open interval a < t < b.

The Method of Integrating FactorWe discuss a technique for solving equation (3.1). Since p(t) is continuous, it hasan antiderivative namely

∫p(t)dt. Let µ(t) = e

∫p(t)dt. Multiply Equation (3.1) by

µ(t) and notice that the left hand side of the resulting equation is the derivativeof a product. Indeed,

d

dt(µ(t)y) = µ′(t)y(t) + µ(t)y′(t) = µ(t)(y′(t) + p(t)y(t)) = µ(t)g(t).

Integrate both sides of the last equation with respect to t to obtain

µ(t)y =

∫µ(t)g(t)dt+ C.

Now, divide through by µ(t), we find

y(t) =1

µ(t)

∫µ(t)g(t)dt+

C

µ(t)

or

y(t) = e−∫p(t)dt

∫e∫p(t)dtg(t)dt+ Ce−

∫p(t)dt.

19

203 ANALYTICAL SOLUTION: THEMETHODOF INTEGRATING FACTOR

One can write the above function in the form y(t) = Cy1(t) + y2(t) where y1(t) =e−

∫p(t)dt and y2(t) = e−

∫p(t)dt

∫e∫p(t)dtg(t)dt. Notice that y1 is a solution for the

homogeneous equation

y′ + p(t)y = 0

and y2 is a solution to the non-homogeneous equation as shown in the next ex-ample. Thus, the general solution to Equation (3.1) is the sum of a particularsolution of the non-homogeneous equation and the general solution of the homoge-neous equation. This solution structure will appear again when discussing secondorder linear equations.

Remark 3.1Notice that multiplying Equation (3.1) by µ(t) was a key factor in the integrationstep discussed above. That’s why µ(t) is known as the integrating factor.

Example 3.1Show that yp = e−

∫p(t)dt

∫e∫p(t)dtg(t)dt satisfies Equation (3.1).

Solution.We have

y′p + p(t)yp =− p(t)e−∫p(t)dt

∫e∫p(t)dtg(t)dt+ e−

∫p(t)dt · e

∫p(t)dtg(t)

+p(t)e−∫p(t)dt

∫e∫p(t)dtg(t)dt

=g(t)

Example 3.2Solve the initial-value problem

y′ − y

t= 4t, y(1) = 5.

Solution.By Theorem 2.2, the solution exists and is unique on the interval (0,∞) since 1belongs to that interval.We have p(t) = −1

t so that µ(t) = 1t . Multiplying the given equation by the

integrating factor and using the product rule we notice that(1

ty

)′= 4.

21

Integrating with respect to t and then solving for y we find that the general solutionis given by

y(t) = t

∫4dt+ Ct = 4t2 + Ct.

The condition y(1) = 5 implies C = 1 and hence the unique solution to the IVPis y(t) = 4t2 + t, 0 < t <∞

Example 3.3Find the general solution to the equation

y′ +2

ty = ln t, t > 0.

Solution.The integrating factor is µ(t) = e

∫2tdt = t2. Multiplying the given equation by t2

to obtain

(t2y)′ = t2 ln t.

Integrating with respect to t we find

t2y =

∫t2 ln tdt+ C.

The integral on the right-hand side is evaluated using integration by parts withu = ln t, dv = t2dt, du = dt

t , v = t3

3 obtaining

t2y =t3

3ln t− t3

9+ C.

Thus,

y =t

3ln t− t

9+C

t2

Remark 3.2Instead of using indefinite integrals in the above discussion one can use definiteintegrals. For example, replace

∫p(t)dt by

∫ tt0p(s)ds for some fixed t0. Using

definite integral is proven to be useful when p(t) does not have an elementaryfunction as an antiderivative. For example, when p(t) = sin t

t or p(t) = sin (t2). Weillustrate this idea in the next example.

Example 3.4Solve y′− 1

t y = sin t, y(1) = 3. Express your answer in terms of the sine integral,

Si(t) =∫ t0

sin ss ds.


Solution.Since p(t) = −1

t we find µ(t) = 1t . Thus,(

1

ty

)′=

sin t

t=

(∫ t

0

sin s

sds

)′1

ty(t) =Si(t) + C

y(t) =tSi(t) + Ct.

Since y(1) = 3, we find C = 3− Si(1). Hence, y(t) = tSi(t) + (3− Si(1))t

Case when either p(t) or g(t) has a jump discontinuityConsider the IVP

y′ + p(t)y = g(t), y(a) = y0, a ≤ t ≤ b (3.2)

where p(t) and g(t) are continuous in a ≤ t ≤ b except at t = c where either p(t)or g(t) has a jump discontinuity at a < c < b. Although y′ is not continuous at c,we seek a solution y(t) that is continuous at t = c.To solve this problem, we first solve the initial value problem on the intervala ≤ t < c where both p(t) and g(t) are continuous. Let y1(t) be the uniquesolution. Since we are seeking a continuous solution to (3.2), we expect y1(t) tohave a one-sided limit at c, i.e.,

limt→c−

y1(t) = y1(c−).

Next, we find the unique solution y2(t) to the IVP

y′ + p(t)y = g(t), y(c) = y1(c−)

where c ≤ t ≤ b. The unique solution to the original IVP is then given by

y(t) =

{y1(t), if a ≤ t < cy2(t) if c ≤ t ≤ b.

We illustrate this process in the next example.

Example 3.5Find the solution to the IVP

y′ +1

ty = g(t), y(1) = 1

where

g(t) =

{3t, if 1 ≤ t ≤ 20 if 2 < t ≤ 3.

The graph of g(t) is given in Figure 3.1(a).

23

Solution.First, we solve the IVP

y′ +1

ty = 3t, y(1) = 1, 1 ≤ t ≤ 2.

The integrating factor is µ(t) = t and the general solution is y1(t) = t2 + Ct . Since

y(1) = 1, we have C = 0. Hence, y1(t) = t2 and y1(2) = 4.Next, we solve the IVP

y′ +1

ty = 0, y(2) = 4, 2 < t ≤ 3.

The integrating factor is µ(t) = t and the general solution is y2(t) = Ct . Since

y2(2) = 4 we find C = 8. Thus,

y(t) =

{t2, if 1 ≤ t ≤ 28t if 2 < t ≤ 3.

The graph of y(t) is given in Figure 3.1(b). As you can see from the graph, y(t) iscontinuous on [1, 3] but not differentiable at t = 2

Figure 3.1


Practice Problems

Problem 3.1Solve the IVP

y′ = −2ty, y(1) = 1.

Problem 3.2Solve the IVP

y′ + ety = 0, y(0) = 2.

Problem 3.3Consider the first order linear non-homogeneous IVP

y′ + p(t)y = αp(t), y(t0) = y0.

(a) Show that the IVP can be reduced to a first order linear homogeneous IVP bythe change of variable z(t) = y(t)− α.(b) Solve this initial value problem for z(t) and use the solution to determine y(t).

Problem 3.4Apply the results of the previous problem to solve the IVP

y′ + 2ty = 6t, y(0) = 4.

Problem 3.5The unique solution to the IVP

ty′ − αy = 0, y(1) = y0

goes through the points (2, 1) and (4, 4). Find the values of α and y0.

Problem 3.6The table below lists values of t and ln [y(t)] where y(t) is the unique solution tothe IVP

y′ + tny = 0, y(0) = y0.

t 1 2 3 4

ln [y(t)] −0.25 −4.00 −20.25 −64.00

(a) Determine the values of n and y0.(b) What is y(−1)?

25

Problem 3.7Consider the differential equation y′ + p(t)y = 0. Find p(t) so that y = c

t is thegeneral solution.

Problem 3.8Consider the differential equation y′ + p(t)y = 0. Find p(t) so that y = ct3 is thegeneral solution.

Problem 3.9Solve the initial-value problem: y′ − 3

t y = 0, y(2) = 8.

Problem 3.10Solve the differential equation y′ − 2ty = 0.

Problem 3.11Find the function f(t) that crosses the point (0, 4) and whose slope satisfies f ′(t) =2f(t).

Problem 3.12Find the general solution to the differential equation y′′−2y′ = 0. Hint: Let y′ = z.

Problem 3.13Solve the IVP: y′ + 2ty = t, y(0) = 0.

Problem 3.14Find the general solution: y′ + 3y = t+ e−2t.

Problem 3.15Find the general solution: y′ + 1

t y = 3 cos t, t > 0.

Problem 3.16Find the general solution: y′ + 2y = cos (3t).

Problem 3.17Given that the solution to the IVP ty′+4y = αt2, y(1) = −1

3 exists on the interval−∞ < t <∞. What is the value of the constant α?

Problem 3.18Suppose that y(t) = Ce−2t+t+1 is the general solution to the equation y′+p(t)y =g(t). Determine the functions p(t) and g(t).


Problem 3.19Suppose that y(t) = −2e−t + et + sin t is the unique solution to the IVP y′ + y =g(t), y(0) = y0. Determine the constant y0 and the function g(t).

Problem 3.20Find the value (if any) of the unique solution to the IVP y′ + (1 + cos t)y =1 + cos t, y(0) = 3 in the long run, i.e., limt→∞ y(t)?

Problem 3.21Find the solution to the IVP

y′ + p(t)y = 2, y(0) = 1

where

p(t) =

{0, if 0 ≤ t ≤ 11t if 1 < t ≤ 2.


y′ + (sin t)y = g(t), y(0) = 3

where

g(t) =

{sin t, if 0 ≤ t ≤ π− sin t if π < t ≤ 2π.


y′ + y = g(t), t > 0, y(0) = 3

where

g(t) =

{1, if 0 ≤ t ≤ 10 if t > 1.


y′ + p(t)y = 0, y(0) = 3

where

p(t) =

2t− 1, if 0 ≤ t ≤ 1

0, if 1 < t ≤ 3−1t if 3 < t ≤ 4.

Sketch an accurate graph of the solution, and discuss the long-run behavior of thesolution. Is the solution differentiable on the interval (0,∞)? Explain your answer.

27

Problem 3.25Solve the initial-value problem y′ + y = ety2, y(0) = 1 using the substitutionu(t) = 1

y(t) .


4 Existence and Uniqueness ofSolutions to First OrderNonlinear IVP

When a mathematical model is constructed for physical systems, two reasonabledemands are made. First, solutions should exist if the model is to be useful atall. Second, to work effectively in predicting the future behavior of the physicalsystem, the model should produce only one solution for a particular set of initialconditions. Existence and uniqueness theorems help to meet these demands.In this section we discuss the conditions that guarantee the existence of a uniquesolution to the initial value problem

y′ = f(t, y), y(t0) = y0 (4.1)

where f(t, y) needs not be linear in y. Before we proceed to the major result ofthis section, we remind the reader of the definition of a partial derivative.

Partial DerivativesIf f(t, y) is a function of two variables t and y then the partial derivative ∂f

∂t off(t, y) is the derivative of f(t, y) with respect to t, while pretending y is a con-stant. The partial derivative ∂f

∂y is the derivative of f(t, y) with respect to y, whilepretending t is constant. The precise definitions are

∂f∂t (t, y) = limh→0

f(t+h,y)−f(t,y)h and ∂f

∂y (t, y) = limh→0f(t,y+h)−f(t,y)

h .

Example 4.1Find ∂f

∂t and ∂f∂y if f(t, y) = t4y3 + t5.

Solution.We have

∂f∂t (t, y) = 4t3y3 + 5t4 and ∂f

∂y (t, y) = 3t4y2

29

304 EXISTENCE ANDUNIQUENESS OF SOLUTIONS TO FIRST ORDER NONLINEAR IVP

The major existence and uniqueness theorem is stated next.

Theorem 4.1Suppose that f(t, y) and ∂f

∂y (t, y) are continuous on the open rectangle

R = {(t, y) : a < t < b, c < y < d}.

Then for any (t0, y0) in R the IVP

y′ = f(t, y), y(t0) = y0

has a unique solution defined on an interval of the form [t0 − h, t0 + h] ⊂ [a, b] forsome positive h.

Remark 4.1Although existence can be proved with no hypotheses on f beyond continuity,some assumption such as the continuity of ∂f

∂y is necessary for uniqueness. Forexample, the IVP

y′ = y13 , y(0) = 0

has as solutions the functions y(t) ≡ 0 and

y(t) =

{0 if t ≤ 0

(23 t)32 if t > 0.

Note that ∂f∂y (t, y) = 1

3y− 2

3 is not continuous at y = 0

Remark 4.2If the conditions of Theorem 4.1 are not satisfied then this does not mean that theproblem does not have a unique solution.

Example 4.2Consider the differential equation

y′ =y

13

t(y − 2).

Does the existence theorem guarantee the existence of a unique solution to thefollowing IVPs: (a) y(3) = 4 (b) y(0) = 7 (c) y(0) = 2 (d) y(1) = 2?

31

Solution.

The function f(t, y) = y13

t(y−2) is continuous for t 6= 0 and y 6= 2. The function

∂f

∂y(t, y) =

−2− 2y

3t(y − 2)2y23

is continuous for t 6= 0 and y 6= 0, 2. Thus, f and ∂f∂y are continuous for t 6= 0

and y 6= 0, 2. Theorem 4.1 guarantees the existence of a unique solution for theinitial value problem in (a) whereas there is no guarantee that there is a uniquesolution− or any solution− for the remaining IVPs

Example 4.3On what rectangle do we expect a unique solution to the IVP

y′ =y2

1− t2, y(0) = 0?

Solution.We have

f(t, y) =y2

1− t2and

∂f

∂y(t, y) =

2y

1− t2.

These are both continuous functions as long as we avoid the lines t = ±1. Theorem4.1 tells us that we can expect one and only one solution of

y′ =y2

1− t2, y(0) = 0

in the stripR = {(t, y) : −1 < t < 1,−∞ < y <∞}

Example 4.4Consider the IVP

y′ =1

2(−t+

√t2 + 4y), y(2) = −1.

(a) Show that y(t) = 1− t and y(t) = − t2

4 are two solutions to the above IVP.(b) Does this contradict Theorem 4.1?

Solution.(a) You can verify that the two functions are solutions by substitution.(b) Since f(t, y) = 1

2(−t+√t2 + 4y) and fy(t, y) = 1√

t2+4y, these two functions are

not continuous at (2,−1). Thus, we can not apply Theorem 4.1 for this problem


Example 4.5Consider the initial value problem: t2y′ − y2 = 0, y(1) = 1.(a) Determine the largest open rectangle in the ty−plane, containing the point(t0, y0) = (1, 1), in which the hypotheses of Theorem 4.1 are satisfied.(b) A solution of the initial value problem is y(t) = t. This solution exists on−∞ < t <∞. Does this fact contradicts Theorem 4.1? Explain your answer.

Solution.(a) We have f(t, y) = y2

t2, fy(t, y) = 2y

t2. So

R = {(t, y) : 0 < t <∞, −∞ < y <∞}.

(b) No. Theorem 4.1 is a local existence theorem and not a global one

Example 4.6Is it possible to find a function f(t, y) that is continuous and has continuous partialderivatives in the entire ty−plane such that the functions y1(t) = cos t and y2(t) =1− sin t are both solutions to the initial value problem y′ = f(t, y), y

(π2

)= 0?

Solution.Since f is continuous and has continuous partial derivatives in the entire ty-plane,the equation y′ = f(t, y) satisfies the conditions of Theorem 4.1. Notice thaty1(

π2 ) = y2(

π2 ) = 0, so the curves y1(t) = cos t and y2(t) = 1− sin t have a common

point (π2 , 0), so if they were both solutions of our equation, by the uniquenesstheorem they would have to agree on any rectangle containing (π2 , 0). Since theydo not, they cannot both be solutions of the equation y′ = f(t, y)

33

Practice Problems

For the given initial value problem in Problems 4.1 - 4.5,(a) Rewrite the differential equation, if necessary, to obtain the form

y′ = f(t, y), y(t0) = y0.

Identify the function f(t, y).(b) Compute ∂f

∂y . Determine where in the ty−plane both f(t, y) and ∂f∂y are con-

tinuous.(c) Determine the largest open rectangle in the ty−plane that contains the point(t0, y0) and in which the hypotheses of Theorem 4.1 are satisfied.

Problem 4.1

3y′ + 2t cos y = 1, y(π

2) = −1.

Problem 4.2

3ty′ + 2 cos y = 1, y(π

2) = −1.

Problem 4.3

2t+ (1 + y3)y′ = 0, y(1) = 1.

Problem 4.4

(y2 − 9)y′ + e−y = t2, y(2) = 2.

Problem 4.5

cos yy′ = 2 + tan t, y(0) = 0.

Problem 4.6Give an example of an initial value problem for which the open rectangle

R = {(t, y) : 0 < t < 4,−1 < y < 2}

represents the largest region in the ty−plane where the hypotheses of Theorem 4.1are satisfied.


Problem 4.7Can we apply Theorem 4.1 to the following problem ? Explain what (if anything)we can conclude, and why (or why not):

y′ =y√t, y(0) = 2.

Problem 4.8Consider the differential equation y′ = t−y

t+y . For which of the following initial valueconditions does Theorem 4.1 apply?(a) y(0) = 0 (b) y(1) = −1 (c) y(−1) = −1.

Problem 4.9Does the initial value problem y′ = y

t + 2, y(0) = 1 satisfy the conditions ofTheorem 4.1?

Problem 4.10Does the initial value problem y′ = y sin y+ t, y(0) = −1 satisfy the conditions ofTheorem 4.1?

5 Separable DifferentialEquations

In this section, we discuss an analytical method for solving a type of nonlinearfirst order differential equations, the separable differential equations.A first order differential equation is separable if it can be written with one variableonly on the left and the other variable only on the right:

f(y)y′ = g(t).

To solve this equation, we proceed as follows. Let F (t) be an antiderivative of f(t)and G(t) be an antiderivative of g(t). Then by the Chain Rule

d

dtF (y) =

dF

dy

dy

dt= f(y)y′.

Thus,

f(y)y′ − g(t) =d

dtF (y)− d

dtG(t) =

d

dt[F (y)−G(t)] = 0.

It follows thatF (y)−G(t) = C

which is equivalent to ∫f(y)y′dt =

∫g(t)dt+ C.

As you can see, the result is generally an implicit equation involving a function ofy and a function of t. It may or may not be possible to solve this to get y explicitlyas a function of t. For an initial value problem, substitute the values of t and y byt0 and y0 to get the value of C.

Remark 5.1If F is a differentiable function of y and y is a differentiable function of t and bothF and y are given then the chain rule allows us to find dF

dt given by

dF

dt=dF

dy· dydt.

35

36 5 SEPARABLE DIFFERENTIAL EQUATIONS

For separable equations, we are given f(y)y′ = dFdt and we are asked to find F (y).

This process is referred to as “reversing the chain rule.”

Example 5.1Solve the initial value problem y′ = 6ty2, y(1) = 1

25 .

Solution.Since f(t, y) = 6ty2 and fy(t, y) = 12ty are continuous in the rectangle

R = {(t, y) : −∞ < t <∞, −∞ < y <∞}

and R contains the point(1, 1

25

), by Theorem 4.1, the IVP has a unique solution

on some interval containing t = 1.Separating the variables and integrating both sides we obtain∫

y′

y2dt =

∫6tdt

or

−∫ (

1

y

)′dt =

∫6tdt.

Thus,

− 1

y(t)= 3t2 + C.

Since y(1) = 125 , we find C = −28. The unique solution to the IVP is then given

explicitly by

y(t) =1

28− 3t2.

The next question is the question of the interval of existence of this solution. Recallthat there are two conditions that define an interval of validity. First, it must be acontinuous interval with no breaks or holes in it. Second it must contain the valueof the independent variable in the initial condition, t = 1 in this case.There are three possible intervals where y(t) is continuous:

−∞ < t < −√

28

3, −

√28

3< t <

√28

3, t >

√28

3.

Only one of these will contain the value of t from the initial condition and so wecan see that

−√

28

3< t <

√28

3

37

must be the interval of existence for this solution. Figure 5.1 shows the graph ofthe solution

Figure 5.1

Example 5.2Solve the IVP yy′ = 4 sin (2t), y(0) = 1.

Solution.This is a separable differential equation. Integrating both sides we find∫ (

y2

2

)′dt = 4

∫sin (2t)dt.

Thus,y2 = −4 cos (2t) + C.

Since y(0) = 1, we find C = 5. Now, solving explicitly for y(t) we find

y(t) = ±√−4 cos t+ 5.

Since y(0) = 1, we find y(t) =√−4 cos t+ 5. The interval of existence of the

solution is the interval −∞ < t <∞

Example 5.3Solve the initial value problem

y′ =√

1− y2, y(0) = 0.

Solution.Separating the variables and then integrating we find∫

y′√1− y2

dt =

∫dt


orarcsin y = t+ C.

Since y(0) = 0, we find C = 0 and consequently y(t) = sin t where −π2 ≤ t ≤

π2 .

Now, notice that y1(t) = 1 and y1(t) = −1 are solutions to the differential equa-tions. Moreover, y(π2 ) = 1 and y(−π

2 ) = −1 so that the graph of y = arcsin t isconnected to the solution y1 at the point t = π

2 and to the solution y2 at the pointt = −π

2 . Thus, the solution to the given IVP is

y(t) =

−1, −∞ < t < −π

2sin t, −π

2 ≤ t ≤π2

1 π2 < t <∞.

The graph of this function is shown in Figure 5.2

Figure 5.2

Linear Versus Nonlinear Differential EquationsNext, we turn our attention to some general questions about differential equationsand explore in more detail some important ways in which nonlinear differentialequations differ from linear ones.

1. Existence and Uniqueness of SolutionsIf the functions p(t) and g(t) for a linear differential equation are continuous onan interval a < t < b containing the initial condition t0 then Theorem 2.2 assertsthe existence of a unique solution throughout the interval a < t < b. This intervalcan be identified by glancing at the functions p(t) and g(t).Turning now to nonlinear differential equations, the function f(t, y) = 6ty2 in Ex-ample 5.1 is continuous everywhere however the interval of existence of solution

to the initial value problem y′ = 6ty2 and y(1) = 125 is −

√283 < t <

√283 and one

cannot predict this by simply looking at the equation y′ = 6ty2.

2. General Solutions: There may not be a single formula that gives all solutions

39

For a first order linear differential equation the general solution involves a con-stant from which all particular solutions are obtained by specifying values for thisconstant. In contrast, this is not the case for nonlinear differential equations.If we consider the initial value problem y′ = 6ty2, y(0) = 0 then we see that thefamily of solutions y(t) = −1

3t2+Cdoes not include the zero solution. Indeed, there

is no choice of C that yields the unique solution y ≡ 0 guaranteed by Theorem4.1. Such a solution is called a singular solution.

3. Implicit SolutionsRecall again that, for an initial value problem of the form y′+p(t)y = g(t), y(t0) =y0, the unique solution is always given explicitly. In contrast, the situation for non-linear differential equations is much less satisfactory. For example, the solutionsin Examples 5.1 - 5.3, are defined explicitly. However, the unique solution to theinitial value problem (2y − sin y)y′ = sin t− t, y(0) = 0 is defined by the implicit

equation y2 + cos y+ cos t+ t2

2 = 2 (See Problem 5.11) where y can not be writtenexplicitly as a function of t.


Practice Problems

Problem 5.1Solve the (separable) differential equation

y′ = tet2−ln y2 .


y′ =t2y − 4y

t+ 2.


ty′ = 2(y − 4).


y′ = 2y(2− y).

Problem 5.5Solve the IVP:

y′ =4 sin (2t)

y, y(0) = 1.


yy′ = sin t, y(π

2) = −2.


y′ +1

y + 1= 0, y(1) = 0.


y′ − ty3 = 0, y(0) = 2.


y′ = 1 + y2, y(π

4) = −1.

41


y′ = t− ty2, y(0) =1

2.


(2y − sin y)y′ = sin t− t, y(0) = 0.

Problem 5.12For what values of the constants α, y0, and integer n is the function y(t) = (4+t)−

12

a solution of the initial value problem

y′ + αyn = 0, y(0) = y0?

Problem 5.13State an initial value problem, with initial condition imposed at t0 = 2, havingimplicit solution y3 + t2 + sin y = 4.


y′ = 2y2, y(0) = y0.

For what value(s) of y0 will the solution have a vertical asymptote at t = 4, wherethe t−interval of existence is −∞ < t < 4?

Problem 5.15Assume that y sin y− 3t+ 3 = 0 is an implicit solution of the initial value problemy′ = f(y), y(1) = 0. What is f(y)? What is an implicit solution to the initialvalue problem y′ = t2f(y), y(1) = 0?

Problem 5.16Find all the solutions to the differential equation y′ = 2ty

1+t .

Problem 5.17Solve the initial-value problem y′ = cos2 y cos2 t, y(0) = π

4 .

Problem 5.18Solve the initial-value problem y′ = et+y, y(0) = 0 and determine the interval onwhich the solution y(t) is defined.


Problem 5.19(a) Solve the initial-value problem

y′ =t2

e−y− ey

t2.

State the name of the method you are using.(b) Find the solution which satisfies the condition y(1) = 1.

6 Exact Differential Equations

We shall now present another technique for solving first order, non-linear, ordinarydifferential equations. This technique is a generalization of the one we used forseparable equations.We have seen that the solution procedure of separable equations consists of revers-ing the chain rule. This same procedure works for exact equations but this timethe chain rule is for functions of two variables.

The Extended Chain RuleYou recall the chain rule for functions of one variable: If u is differentiable att and f is differentiable at u(t) then the composite function y = f(u(t)) is alsodifferentiable at t with derivative given by

dy

dt=dy

du· dudt.

Example 6.1Find the derivative of the function y = e

√t.

Solution.Let u(t) =

√t and f(t) = et. Then du

dt = 12√t

and dydu = eu. Hence,

dy

dt= eu

1

2√t

=e√t

2√t

The above chain rule can be extended to functions of two variables. Suppose thatu and v are differentiable at t and f is a differentiable function of u and v. Thenthe function z(t) = f(u(t), v(t)) is differentiable at t with derivative

dz

dt=∂f

∂u

du

dt+∂f

∂v

dv

dt.

Example 6.2Let z = f(u, v) = u2 + 2u− uv + v2 where u(t) = t2 + 1 and v(t) = t3 − t2. Finddzdt

∣∣t=2

in two different ways.

43

44 6 EXACT DIFFERENTIAL EQUATIONS

Solution.First notice that u(2) = 5 and v(2) = 4. By using the extended chain rule we have

dz

dt=∂f

∂u

du

dt+∂f

∂v

dv

dt=(2u+ 2− v)(2t) + (2v − u)(3t2 − 2t).

Thus,dz

dt

∣∣∣∣t=2

= (10 + 2− 4)(4) + (8− 5)(8) = 56.

A different way for finding the derivative is to write z as only a function of tobtaining

z(t) = t6 − 3t5 + 3t4 − t3 + 5t2 + 3.

Finding the derivative of z(t)

z′(t) = 6t5 − 15t4 + 12t3 − 3t2 + 10t.

Finally, z′(2) = 56

Exact Differential EquationsThe basic idea underlying separable equations is to reverse the chain rule for func-tions of one variable. The basic idea underlying exact equations is to reverse theextended chain rule. To this end, consider the differential equation

M(t, y) +N(t, y)dy

dt= 0. (6.1)

Let H(t, y) be a function satisfying the two conditions

∂H

∂t(t, y) = M(t, y) and

∂H

∂y(t, y) = N(t, y). (6.2)

Then Equation (6.1) can be written as

∂H

∂t+∂H

∂y

dy

dt= 0. (6.3)

By the extended chain rule, Equation (6.3) is the same as

d

dtH(t, y) = 0.

Therefore, we obtain an implicitly defined solution given by

H(t, y) = C.

45

An equation like (6.1) is called exact if there is a function H(t, y) satisfying theconditions in (6.2).

Testing a Differentiable Equation for ExactnessThe next question is the question of telling whether or not Equation (6.1) is exact.This is answered by the following theorem.

Theorem 6.1Suppose that the functions M(t, y) and N(t, y) in (6.1) are continuous and havecontinuous first partial derivatives ∂M

∂y and ∂N∂t in an open rectangle

R = {(t, y) : a < t < b, c < y < d}.

Then (6.1) is exact in R if and only if

∂M

∂y=∂N

∂t

for all (t, y) in R.

Remark 6.1Every separable differential equation is exact. Indeed, since −g(t) + f(y)y′ = 0we have ∂g

∂y = 0 and ∂f∂t = 0. However, not every exact equation is separable.

For example, the differential equation (2t + y) + (2y + t)y′ = 0 is exact since∂M∂y = 1 = ∂N

∂t = 1. This equation is clearly not separable

Example 6.3Determine whether or not the equation is exact.(a) ty2 + t+ t2yy′ = 0.(b) y2 + 1 + tyy′ = 0.(c) cos y + (y2 + t sin y)y′ = 0.(d) cos y + (y2 − t sin y)y′ = 0.

Solution.(a) Since ∂

∂y (ty2 + t) = 2ty and ∂∂t(t

2y) = 2ty, the given equation is exact.

(b) Since ∂∂y (y2 + 1) = 2y and ∂

∂t(ty) = y, the given equation is not exact.

(c) Since ∂∂y (cos y) = − sin y and ∂

∂t(y2 + t sin y) = sin y, the given equation is not

exact.(d) Since ∂

∂y (cos y) = − sin y and ∂∂t(y

2 − t sin y) = − sin y, the equation is exact


Example 6.4Consider the initial value problem

t+ y + (t+ 2y)y′ = 0, y(0) = 1.

Show that the differential equation is exact and solve the IVP.

Solution.We have M(t, y) = t+ y and N(t, y) = t+ 2y. Since

∂M(t, y)

∂y=∂N(t, y)

∂t= 1

we have by Theorem 6.1 that the differential equation is exact. Thus,

H(t, y) =

∫(t+ y)dt = ty +

t2

2+ c1(y).

Hence,

t+ 2y =∂H(t, y)

∂y= t+ c′1(y).

It follows that

c1(y) =

∫(2y)dy = y2 + C.

Hence,

ty + y2 +t2

2= C.

Since y(0) = 1, we find C = 1. Thus, y satisfies the implicit equation

2ty + 2y2 + t2 = 2

47

Practice Problems

Problem 6.1Find ∂f

∂t and ∂f∂y if f(t, y) = y ln y − e−ty.

Problem 6.2Find ∂f

∂t and ∂f∂y if f(t, y) = ln (ty) + t2+1

y−5 .

Problem 6.3Let f(u, v) = 2u− 3uv where u(t) = 2 cos t and v(t) = 2 sin t. Find df

dt .

In Problems 6.4 - 6.8, determine whether the given differential equation is exact.If the equation is exact, find an implicit solution and (where possible) an explicitsolution.

Problem 6.4

yy′ + 3t2 − 2 = 0, y(−1) = −2.

Problem 6.5

y′ = (3t2 + 1)(y2 + 1), y(0) = 1.

Problem 6.6

(6t+ y3)y′ + 3t2y = 0, y(1) = 2.

Problem 6.7

(et+y + 2y)y′ + (et+y + 3t2) = 0, y(0) = 0.

Problem 6.8

(sin (t+ y) + y cos (t+ y) + t+ y)y′ + (y cos (t+ y) + y + t) = 0, y(1) = −1.

Problem 6.9For what values of the constants m,n, and α (if any) is the following differentialequation exact?

tmy2y′ + αt3yn = 0.


Problem 6.10Assume that N(t, y)y′+t2+y2 sin t = 0 is an exact differential equation. Determinethe general form of N(t, y).

Problem 6.11Assume that t3y + et + y2 = 5 is an implicit solution to the differential equation

N(t, y)y′ +M(t, y) = 0, y(0) = y0.

Determine possible functions M(t, y), N(t, y), and the possible value(s) for y0.

Problem 6.12Assume that y = −t−

√4− t2 is an explicit solution of the following initial value

problem(y + at)y′ + (ay + bt) = 0, y(0) = y0.

Determine values for the constants a, b and y0.

Problem 6.13Let k be a positive constant. Use the exactness criterion to determine whether ornot the population equation dP

dt = kP is exact. Do NOT try to solve the equationor carry out any further calculation.

Problem 6.14Consider the differential equation (2t + 3) + (2y − 2)y′ = 0. Determine whetherthis equation is exact or not. If it is, solve it.

Problem 6.15Consider the differential equation (ye2ty + t) + bte2tyy′ = 0. Determine for whichvalue of b this equation is exact, and then solve it with this value of b.

Problem 6.16Consider the differential equation y+ (2t− yey)y′ = 0. Check that this equation isnot exact. Now multiply the equation by y. Check that the new equation is exact,and solve it.

Problem 6.17(a) Consider the differential equation

y′ + p(t)y = g(t)

with p(t) 6= 0. Show that this equation is not exact.(b) Let µ(t) = e

∫p(t)dt. Show that the equation

µ(t)(y′ + p(t)y) = µ(t)g(t)

is exact and solve it.

49

Problem 6.18Use the method of the previous problem to solve the linear, first-order equationy′ − y

t = 1, with initial condition y(1) = 7. First, check that this equation is notexact. Next, find µ(t). Multiply the equation by µ(t) and check that the newequation is exact. Solve it, using the method of exact equations.

Problem 6.19Put the following differential equation in the “Exact Differential Equation” formand find the general solution

y′ =y3 − 2ty

t2 − 3ty2.

Problem 6.20The following differential equations are exact. Solve them by that method.(a) (4t3y + 4t+ 4)y′ = 8− 4y − 6t2y2, y(−1) = 1.(b) (6− 4y + 16t) + (10y − 4t+ 2)y′ = 0, y(1) = 2.


7 Substitution Techniques:Bernoulli and Riccati Equations

A well-known non-linear equation that reduces to a linear one with an appropriatesubstitution is the Bernoulli equation given by

y′ + p(t)y = g(t)yn (7.1)

where n is a real number different from 0 and 1. Notice that for n = 0 or n = 1the equation becomes linear.To solve (7.1), first notice that y(t) ≡ 0 is the trivial solution. If y(t) 6≡ 0 then(7.1) can be written as

y′

yn+ p(t)y1−n = g(t). (7.2)

Let z = y1−n. Then by the chain rule of differentiation we have z′ = (1− n)y−ny′

and therefore y′

yn = 11−nz

′. Thus, (7.1) reduces to

1

1− nz′ + p(t)z = g(t)

which is a first order linear differential equation that can be solved by the technique

of integrating factor. Once z is found then the desired solution is y(t) = z1

1−n .

Example 7.1Solve the Bernoulli equation

y′ − 1

ty = ty2, t > 0.

Solution.Divide by y2 and then let z = y−1 to obtain

z′ +1

tz = −t.

51

527 SUBSTITUTION TECHNIQUES: BERNOULLI AND RICCATI EQUATIONS

The integrating factor is µ(t) = t and the general solution is

z(t) =1

t

∫−t2dt+ Ct−1 = − t

2

3+ Ct−1.

Thus, the solutions to the differential equation are y = 1z = 1

− t23+Ct−1

and y(t) =

0

Example 7.2Solve the Bernoulli equation y′ + 3y = e3ty2.

Solution.Dividing by y2 to obtain

y2y′ + 3y−1 = e3t.

Let z = y−1. Then,z′ − 3z = −e3t.

Solving this equation using the integrating factor method with µ(t) = e−3t we find

z(t) = e3t∫e−3t(−e3t)dt+ Ce3t = −te3t + Ce3t = e3t(C − t).

The solutions are y(t) = e−3t(C − t)−1 and y(t) = 0

Example 7.3Solve: y′ + 2

t y = −t2y2 cos t.

Solution.Dividing through by y2 to obtain y−2y′ + 2

t y−1 = −t2 cos t. Letting z = y−1 to

obtain

z′ − 2

tz = t2 cos t.

Solving this equation by the method of integrating factor with µ(t) = 1t2

we find

z(t) = t2∫

cos tdt+ Ct2 = t2 sin t+ Ct2.

The solutions are y(t) = (t2 sin t+ Ct2)−1 and y(t) = 0

53

Riccati EquationA first order differential equation is called a Riccati equation if it can be writtenin the form:

y′ = a(t)y2 + b(t)y + c(t) (7.3)

where a, b and c are functions of t. Clearly, this is a first order nonlinear and non-separable differential equation.The solution of a Riccati equation requires knowledge of a particular solution to(7.3). To solve (7.3), we first find a particular solution y1 to (7.3). Then we usethe substitution 1

z = y − y1. Thus, y = 1z + y1. Now, (7.3) reduces to

− z′

z2+ y′1 =a(t)(

1

z2+ 2

y1z

+ y21) + b(t)(1

z+ y1) + c(t)

− z′

z2+ a(t)y21 + b(t)y1 + c(t) =

a(t)

z2+

2a(t)y1z

+ a(t)y21 +b(t)

z+ b(t)y1 + c(t)

z′ =− a(t)− [2a(t)y1 + b(t)]z.

Thus, a Riccati equation can be reduced to a linear equation

z′ + [2a(t)y1 + b(t)]z = −a(t) (7.4)

that can be solved by the method of integrating factor. Once z(t) is found thenthe solution to the original equation is y(t) = 1

z(t) + y1(t). As the next exampleillustrates, in many cases a solution of a Riccati equation cannot be expressed interms of elementary functions.

Example 7.4Solve : y′ = 2− 2ty + y2 given that y1(t) = 2t.

Solution.We have a(t) = 1, b(t) = −2t, and c(t) = 2. Substituting in (7.4) to obtain

z′ + 2tz = −1.

The integrating factor is µ(t) = et2

so that(et

2z)′

= −et2 .


Now, the integral∫ tt0es

2ds cannot be expressed in terms of elementary functions.

Thus, we write

et2z(t) = −

∫ t

t0

es2ds+ et

20z(t0).

Solving for z, we find

z(t) = e−t2(et

20z(t0)−

∫ t

t0

es2ds).

Finally,

y(t) =1

e−t2(et20z(t0)−

∫ tt0es2ds)

+ 2t

Example 7.5Solve the IVP: y′ = 2 + 2y + y2, y(0) = 0 using the method of separation ofvariables.

Solution.Notice first that 2 + 2y + y2 = 1 + (1 + y)2. Separating the variables we find

y′

1 + (1 + y)2= 1.

Integrating both sides with respect to t to obtain

arctan (1 + y) = t+ C.

But y(0) = 0 so that C = π4 . Thus,

arctan (1 + y) = t+π

4.

For the inverse arc tangent we must have −π2 < t+ π

4 <π2 or −3π

4 < t < π4 . Hence,

y(t) = tan (t+π

4)− 1, − 3π

4< t <

π

4

Example 7.6Solve the differential equation y′ = 5− t2 + 2ty − y2 given that y1(t) = t− 2 is aparticular solution.

55

Solution.Let 1

z = y − t+ 2. Then the given equation reduces to

z′ + 4z = 1.

Solving this equation by the method of integrating factor with µ(t) = e4t to obtain

z(t) = e−4t∫e4tdt+ Ce−4t =

1

4+ Ce−4t.

Thus,

y(t) = (1

4+ Ce−4t)−1 + t− 2


Practice Problems

Problem 7.1Solve the Bernoulli equation

y′ =t2 + 3y2

2ty, t > 0.

Problem 7.2Find the general solution of y′ + ty = te−t

2y−3.

Problem 7.3Solve the IVP ty′ + y = t2y2, y(0.5) = 0.5.

Problem 7.4Solve the IVP y′ − 1

t y = −y2, y(1) = 1.

Problem 7.5Solve the IVP y′ = y(1− y), y(0) = 1

2 .

Problem 7.6Solve y′ + y = ty4.

Problem 7.7Solve the differential equation y′ = 1 + t2 − y2 given that y1(t) = t is a particularsolution.

Problem 7.8Perform a change of variable that changes the Bernoulli equation y′ + y + y2 = 0into a linear equation in the new variable. Do NOT try to solve the equation orproceed further than with any calculations.

Problem 7.9Consider the equation

y′ = εy − σy3, ε > 0, σ > 0.

(a) Use the Bernoulli transformation to change this nonlinear equation into a linearequation.(b) Solve the resulting linear equation in part (a) and use the solution to find thesolution of the given differential equation above.

57

Problem 7.10Consider the differential equation

y′ = f(yt

).

(a) Show that the substitution z = yt leads to a separable differential equation in

z.(b) Use the above method to solve the initial-value problem

y′ =t+ y

t− y, y(1) = 0.

Problem 7.11Solve: y′ + y

3 = ety4.

Problem 7.12Solve: ty′ + y = ty3.

Problem 7.13Solve: ty′ + y = t2y2 ln t.

Problem 7.14Verify that y1(t) = 2 is a particular solution to the Riccati equation

y′ = −2− y + y2,

and then find the general solution.

Problem 7.15Verify that y1(t) = 2

t is a particular solution to the Riccati equation

y′ = − 4

t2− 1

ty + y2,

and then find the general solution.


8 Graphical Solution: DirectionField of y′ = f (t, y)

Since explicit solutions of differential equations are often unobtainable, we exploremethods of finding properties of solutions from the differential equation itself; theprincipal tool is the geometry of direction field.A direction field (also known as slope field) consists of an array of short linesegments in the ty-plane having the property that the line plotted at a point(t, y) has slope f(t, y). Direction fields are basically used to visualize the family ofsolutions of a given differential equation without the need of solving the equation.Direction fields give qualitative information about solutions of ODEs.In this section we use direction fields for solving initial value problems of the form

dy

dt= f(t, y), y(t0) = y0.

In the special case where f(t, y) = f(y), i.e. the independent variable t does notappear on the right side, the first order DE dy

dt = f(y) is called autonomous.Producing slope fields by hand is a daunting task. For that reason, slope fields areusually created by means of an electronic generator. One such a generator can befound at http://slopefield.nathangrigg.net/

Example 8.1Find the direction field of the differential equation

dy

dt= 2t.

What is the form of the general solution? Graph the particular solution goingthrough (0,−1).

Solution.Figure 8.1 shows the slope field and the graph of the particular solution to the given

59

60 8 GRAPHICAL SOLUTION: DIRECTION FIELD OF Y ′ = F (T, Y )

DE passing through the point (0,−1). The solution curves look like parabolas.Thus, the general solution is given by the equation y = t2 + C

Figure 8.1

Example 8.2Using direction field, guess the form of the solution curves of the differential equa-tion

dy

dt= − t

y.

Solution.The direction field is given in Figure 8.2. The solution curves look like circlescentered at the origin. Thus, the general solution is given implicitly by the equationt2 + y2 = C where C is a positive constant

Figure 8.2

61

Remark 8.1We point out here that even though one can draw solution curves, some do nothave simple formulas. For instance, the equation dy

dt = −1+yety

1+tety does not haveexplicit solutions.

Equilibrium Solutions and Stability for Autonomous EquationsA physical system is often said to be in equilibrium if it doesn’t change in time. Weadopt this idea and say that a solution to a differential equation is an equilibriumsolution if it is a constant function. Thus, in a direction field of an autonomousequation, equilibrium solutions are solution curves represented by horizontal lines.It follows that the equations of such solutions have the form y(t) ≡ c where c is aconstant. The following result tells us where to look for equilibrium solutions.

Theorem 8.1The function y(t) ≡ c, where c is a constant, is an equilibrium solution to y′ = f(y)if and only if c is a root of f(y) = 0.

Example 8.3Find the equilibrium solutions to the DE

dy

dt= 2y(1− y).

Solution.The roots of f(y) = 2y(1− y) = 0 are y = 0 and y = 1. According to the previoustheorem, the equilibrium solutions are y(t) ≡ 0 and y(t) ≡ 1. The direction fieldof the DE is shown in Figure 8.3

Figure 8.3


Remark 8.2Equilibrium solutions can be defined for nonautonomous differential equations. Forexample, the function y(t) ≡ 1 is an equilibrium solution to the DE y′ = (1− y)t2.

The direction field of a given differential equation indicates that as t increaseswithout bound, every solution either moves towards or moves away from an equi-librium solution.If all nearby solutions move towards a certain equilibrium solution, then that equi-librium solution is called asymptotically stable, stable, or attracting. Thesolution y = 1 in Figure 8.3 is attracting. An equilibrium solution is called un-stable or repelling when all nearby solutions move away from it. The solutiony = 0 in Figure 8.3 is repelling.If solutions on one side of an equilibrium solution move towards the equilibriumsolution and on the other side of the equilibrium solution move away from it thenwe call the equilibrium solution semi-stable.An equilibrium solution does not necessarily have to be either attracting or re-pelling. The next example illustrates this situation.

Example 8.4Sketch the field direction of the differential equation y′ = 4y(1 − y)2. Show thaty = 1 is neither stable or unstable.

Solution.The direction field is shown in Figure 8.4. Note that the equilibrium solutiony(t) ≡ 1 is neither stable or unstable. Nearby solutions that start below it areattracted upward towards it but nearby solutions that start above it are repelledupward and away from it

Figure 8.4

63

A very suitable qualitative representation of a differential equation for the studyof stability is the so-called phase line. A phase line consists of solid dots andarrows. The solid dots represent the equilibrium points and the arrows indicatethe directions that solutions move as t increases. Figure 8.5(a) shows an exampleof a phase line. We see that the equilibrium b is stable, whereas the equilibria aand c are unstable.

Example 8.5Consider the autonomous differential equation dy

dt = f(y) where the graph of f(y)is given in Figure 8.5(b).(a) Sketch the phase line.(b) Sketch the direction field of this differential equation.(c) Sketch the graph of the solution to the IVP y′ = f(y), y(0) = 3

2 . Findlimt→∞ y(t).(d) Sketch the graph of the solution to the IVP y′ = f(y), y(0) = 1

2 . Findlimt→∞ y(t).(e) Sketch the graph of the solution to the IVP y′ = f(y), y(0) = −3

2 . Findlimt→∞ y(t).

Figure 8.5

Solution.(a) Note that for y < −1, f(y) < 0 so that the solution y(t) is decreasing. For−1 < y(t) < 0, we have f(y) > 0 so that y(t) is increasing. For 0 < y(t) < 1, wehave that f(y) < 0 so that y(t) is decreasing. Finally, for y(t) > 1 we have thatf(y) > 0 so that y(t) is increasing. Hence, the phase line is given by Figure 8.5(c).(b) The direction field is given in Figure 8.6.(c) We notice from Figure 8.6 that limt→∞ y(t) =∞.(d) We notice from Figure 8.6 that limt→∞ y(t) = 0.(e) We notice from Figure 8.6 that limt→∞ y(t) = −∞


Figure 8.6

65

Practice Problems

Problem 8.1Sketch the direction field for the differential equation in the window −3 ≤ t ≤3,−3 ≤ y ≤ 3.(a) y′ = y (b) y′ = t− y.

Problem 8.2Match each direction field with the equation that the slope field could repre-sent. Each direction field is drawn in the portion of the ty-plane defined by−6 ≤ t ≤ 6,−4 ≤ y ≤ 4.(a) y′ = −t (b) y′ = sin t (c) y′ = 1− y (d) y′ = y(2− y).

Problem 8.3State whether or not the equation is autonomous.(a) y′ = −t (b) y′ = sin t (c) y′ = 1− y (d) y′ = y(2− y).

Problem 8.4Find all the equilibrium solutions of each of the autonomous differential equationsbelow(a) y′ = (y − 1)(y − 2).(b) y′ = (y − 1)(y − 2)2.(c) y′ = (y − 1)(y − 2)(y − 3).

Problem 8.5Find an autonomous differential equation with an equilibrium solution at y = 1and satisfying y′ < 0 for −∞ < y < 1 and 1 < y <∞.


Problem 8.6Find an autonomous differential equation with no equilibrium solutions and satis-fying y′ > 0.

Problem 8.7Find an autonomous differential equation with equilibrium solutions y = n

2 , wheren is an integer.

Problem 8.8Find an autonomous differential equation with equilibrium solutions y = 0 andy = 2 and satisfying the properties y′ > 0 for 0 < y < 2; y′ < 0 for y < 0 or y > 2.

Problem 8.9Classify whether the equilibrium solutions are stable, unstable, or neither.(a) y′ = 1− y2.(b) y′ = (y + 1)2.

Problem 8.10Consider the direction field below. Classify the equilibrium points, as asymptoti-cally stable, semi-stable, or unstable.

Problem 8.11Sketch the direction field of the equation y′ = y3. Sketch the solution satisfyingthe condition y(1) = −1. What is the domain of this solution?

Problem 8.12Find the equilibrium solutions and determine their stability

y′ = y2(y2 − 1).

67

Problem 8.13Find the equilibrium solutions of the equation

y′ = y2 − 4y

then decide whether they are stable or unstable. What is the long-run behavior ify(0) = 5?y(0) = 4?y(0) = 3?

Problem 8.14What is limt→∞ y(t) for the initial-value problem

y′ = sin (y(t)), y(0) =π

2?


9 Numerical Solutions toODEs: Euler’s Method and itsVariants

Whenever a mathematical problem is encountered in science or engineering, whichcannot readily or rapidly be solved by a traditional mathematical method, thena numerical method is usually sought and carried out. In this section, we studyEuler’s method and its variants for approximating the solution to the initial valueproblem

y′ = f(t, y), y(t0) = y0, a < t < b. (9.1)

Denote the (unique) solution by y(t). Then y(t) satisfies y′(t) = f(t, y(t)). A nu-merical method for solving the initial value problem provides an algorithm forcalculating approximate values y0, y1, y2, · · · , yn, · · · of the solution y(t) at a set ofpoints t0 < t1 < t2 < · · · < tn < · · · . Usually this is a step-by-step process.

Euler’s MethodEuler’s method is a very simple numerical method for solving the first-order initialvalue problem (9.1). Suppose we want to find approximate values y0, y1, y2, · · · , yn, · · ·of the solution y(t) at equally spaced points t0 < t1 < t2 < · · · < tn < · · · withstep size h. We begin by integrating Equation (9.1) between tn and tn+1 obtaining∫ tn+1

tn

y′(s)ds =

∫ tn+1

tn

f(s, y(s))ds.

By the fundamental theorem of calculus we can write∫ tn+1

tn

y′(s)ds = y(tn+1)− y(tn).

69

709 NUMERICAL SOLUTIONS TOODES: EULER’S METHODAND ITS VARIANTS

Thus,

y(tn+1) = y(tn) +

∫ tn+1

tn

f(s, y(s))ds. (9.2)

Computationally, this last equation can not be used since we do not know f(s, y(s))for tn ≤ s ≤ tn+1. If we estimate the integral by a left Riemann sum we find∫ tn+1

tn

f(s, y(s))ds ≈ hf(tn, y(tn)).

Hence,y(tn+1) ≈ y(tn) + hf(tn, y(tn)).

or

yn+1 = yn + hf(tn, yn) (9.3)

where yn ≈ y(tn). Equation (9.3) is known as Euler’s method. Therefore, we canview Euler’s method as a left Riemann sum approximation of the integral equation(9.2).

Example 9.1Suppose that y(0) = 1 and dy

dt = y. Estimate y(0.5) in 5 steps using Euler’s method.

Solution.The step size is h = 0.5−0

5 = 0.1. The following chart lists the steps needed:

k tk yk f(tk, yk)h

0 0 1 0.11 0.1 1.1 0.112 0.2 1.21 0.1213 0.3 1.331 0.13314 0.4 1.4641 0.146415 0.5 1.61051

Thus, y(0.5) ≈ 1.61051. Note that the exact value is y(0.5) = e0.5 = 1.6487213

The Improved Euler’s Method: Heun’s MethodEuler’s method is derived from estimating the integral equation by a left Riemannsum. A better estimate to the integral is obtained by using the trapezoid ruleobtaining ∫ tn+1

tn

f(s, y(s))ds ≈ h

2[f(tn, y(tn)) + f(tn+1, y(tn+1))].

71

This leads to the estimate

yn+1 = yn +h

2[f(tn, yn) + f(tn+1, yn+1)].

Note that yn+1 is defined implicitly so solving for yn+1 requires solving a nonlinearequation say by the Newton’s method for root-finding. Despite the inconvenienceof working with an implicit method, the trapezoid method has two advantagesover the ordinary Euler method: First, it is more accurate and second it is stable(this means that it is not sensitive to small changes in the initial values).Heun’s method is based on the evaluation of an estimate of f(tn+1, yn+1) by re-placing the value of yn+1 with an estimate derived from the original Euler method,with the resulting iteration scheme:

yn+1 = yn + hf(tn, yn), yn+1 = yn +h

2[f(tn, yn) + f(tn+1, yn+1)]

or

yn+1 = yn +h

2[f(tn, yn) + f(tn+1, yn + hf(tn, yn))].

This equation is known as Heun’s method or the improved Euler’s method.


dt = y. Estimate y(0.5) in 5 steps using the improvedEuler’s method.


5 = 0.1 The following chart lists the steps needed:

y0 =1

y1 =y0 +h

2[f(t0, y0) + f(t1, y0 + hf(t0, y0))] = 1.105

y2 =y1 +h

2[f(t1, y1) + f(t2, y1 + hf(t1, y1))] = 1.221025

y3 =y2 +h

2[f(t2, y2) + f(t3, y2 + hf(t2, y2))] = 1.349232625

y4 =y3 +h

2[f(t3, y3) + f(t4, y3 + hf(t3, y3))] = 1.490902050625

y5 =y4 +h

2[f(t4, y4) + f(t5, y4 + hf(t4, y4))] = 1.647446765940625

Thus, y(0.5) ≈ 1.647446765940625. Note that the exact value is y(0.5) = e0.5 =1.6487213


The Modified Euler’s MethodAnother numerical integration scheme is the midpoint rule. In this case, we have∫ tn+1

tn

f(s, y(s))ds ≈ hf(tn +

h

2, y

(tn +

h

2

)).

This implies that

y(tn+1) ≈ y(tn) + hf

(tn +

h

2, y

(tn +

h

2

)).

Using Euler’s method we can write

y

(tn +

h

2

)≈ y(tn) +

h

2f(tn, y(tn)).

Hence,

yn+1 = yn + hf

(tn +

h

2, yn +

h

2f(tn, yn)

).

This last equation is known as the modified Euler’s method.


dt = y. Estimate y(0.5) in 5 steps using the modifiedEuler’s method.


5 = 0.1. The following chart lists the steps needed:

y0 =1

y1 =y0 + hf(t0 +h

2, y0 +

h

2f(t0, y0)) = 1.105

y2 =y1 + hf(t1 +h

2, y1 +

h

2f(t1, y1)) = 1.221025

y3 =y2 + hf(t2 +h

2, y2 +

h

2f(t2, y2)) = 1.349232625

y4 =y3 + hf(t3 +h

2, y3 +

h

2f(t3, y3)) = 1.490902050625

y5 =y4 + hf(t4 +h

2, y4 +

h

2f(t4, y4)) = 1.647446765940625

Thus, y(0.5) ≈ 1.647446765940625. Note that the exact value is y(0.5) = e0.5 =1.6487213

73

Practice Problems

In Problems 9.1 - 9.3, answer the following questions:(a) Solve the initial value problem analytically, using an appropriate solution tech-nique.(b) For the given initial value problem express yn+1 in terms of yn using Heun’smethod.(c) For the given initial value problem express yn+1 in terms of yn using the Mod-ified Euler’s method.(d) Use a step size h = 0.1. Compute the first three approximations y1, y2, y3 usingthe method in part (b).(e) Use a step size h = 0.1. Compute the first three approximations y1, y2, y3 usingthe method in part (c).(f) For comparison, calculate and list the exact solution values y(t1), y(t2), y(t3).

Problem 9.1

y′ = 2t− 1, y(1) = 0.

Problem 9.2

y′ = −y, y(0) = 1.

Problem 9.3

y2y′ + t = 0, y(0) = 1.


y′ = 1 + y2, y(0) = −1.

(a) Find the exact solution of the given initial value problem.(b) Use step size h = 0.05. Compute 4 steps of Euler’s method, Heun’s method,and modified Euler’s method. Compare the numerical values obtained at t = 1 bycalculating the error |y(1)− y4|.


y′ + 2y = 4, y(0) = 3.


(a) Find the exact solution of the given initial value problem.(b) Use step size h = 0.05. Compute 4 steps of Euler’s method, Heun’s method,and modified Euler’s method. Compare the numerical values obtained at t = 1 bycalculating the error |y(1)− y4|.

In Problems 9.6 - 9.8, the given iteration is the result of applying N steps of Euler’smethod, Heun’s method, or the modified Euler’s method to an initial value problemof the form

y′ = f(t, y), y(t0) = y0, t0 ≤ t ≤ t0 + T.

Identify the numerical method and determine t0, N, T, and f(t, y).

Problem 9.6

yn+1 = yn + h(yn + t2ny3n), y0 = 1

tn = 2 + nh, h = 0.02, n = 0, 1, 2, · · · , 49.

Problem 9.7

yn+1 = yn + h

(tn +

h

2

)sin2

(yn +

h

2tn sin2 yn

), y0 = 1

tn = nh, h = 0.01, n = 0, 1, 2, · · · , 199.

Problem 9.8

yn+1 = yn +h

2[tny

2n + 2 + (tn + h)(yn + h(tny

2n + 1))2], y0 = 2

tn = 1 + nh, h = 0.05, n = 0, 1, 2, · · · , 99.

10 Second Order LinearDifferential Equations:Existence and UniquenessResults

To this point we have considered only first order differential equations. However,many of the most interesting differential equations involve second derivatives. In-deed, since acceleration is the second derivative of position, Newton’s second lawof motion, F = ma, is a second order differential equation. In this and the comingsections we turn our attention to linear second-order differential equations.By a second-order linear differential equation we mean an equation of theform

y′′ + p(t)y′ + q(t)y = g(t).

If g(t) ≡ 0 we say that the equation is homogeneous. Otherwise the equationis non-homogeneous. Initial-value problems for second-order linear differentialequations require two initial-conditions. In this section we will consider the ques-tion of existence and uniqueness of solutions to the initial value problem

y′′ + p(t)y′ + q(t)y = g(t), y(t0) = y0, y′(t0) = y′0. (10.1)

Existence and uniqueness results similar to first-order equations exist for second-order equations as well. The following theorem tells us the conditions for theexistence and uniqueness of solution to (10.1). Note how this theorem is analogousto the corresponding theorem for first order linear ODE’s with initial conditions.

Theorem 10.1If p(t), q(t), and g(t) are continuous functions over an interval a < t < b containingt0 then the initial value problem (10.1) has a unique solution in the interval a <t < b.

75

7610 SECONDORDER LINEAR DIFFERENTIAL EQUATIONS: EXISTENCE ANDUNIQUENESS RESULTS

Example 10.1Find the largest interval where

(t2 − 1)y′′ + 3ty′ + (cos t)y = et, y(0) = 4, y′(0) = 5

is guaranteed to have a unique solution.

Solution.We first put it into the standard form

y′′ +3t

t2 − 1y′ +

cos t

t2 − 1y =

et

t2 − 1, y(0) = 4, y′(0) = 5.

We see that p, q, and g are all continuous except at t = −1 and t = 1. Theorem10.1 tells us that there is a unique solution on −1 < t < 1 since this intervalcontains 0

Remark 10.1The above theorem does not give the largest t−interval of existence as shown inthe next example. Thus, Theorem 10.1 is a local existence theorem.

Remark 10.2If the conditions of the above theorem are not guaranteed, this does not mean thethe problem does not have a solution.

Example 10.2Consider the initial value problem t2y′′ − ty′ + y = 0, y(1) = 1, y′(1) = 1.(a) What is the largest interval on which Theorem 10.1 guarantees the existenceof a unique solution?(b) Show by direct substitution that the function y(t) = t is the unique solutionto this initial value problem. What is the interval on which this solution actuallyexists?(c) Does this example contradict the assertion of Theorem 10.1? Explain.

Solution.(a) Writing the equation in standard form to obtain

y′′ − 1

ty′ +

1

t2y = 0.

From this, we see that the functions p(t) = −1t and q(t) = 1

t2are continuous for all

t 6= 0. Since t0 = 1, the largest t−interval according to Theorem 10.1 is 0 < t <∞.(b) If y(t) = t then y′(t) = 1 and y′′(t) = 0 so that y′′ − 1

t y′ + 1

t2y = 0− 1

t + 1t =

0, y(1) = y′(1) = 1. So y(t) = t is a solution so that by Theoren 10.1 it is the onlysolution. The t−interval for this solution is −∞ < t <∞.(c) No, because the theorem is a local existence theorem and not a global one

77

Example 10.3Consider the graphs shown. Each graph displays a portion of the solution of oneof the four initial value problems given. Match each graph with the appropriateinitial value problem.(a) y′′ + y = 2− sin t, y(0) = 1, y′(0) = −1;(b) y′′ + y = −2t, y(0) = 1, y′(0) = −1;(c) y′′ − y = t2, y(0) = y′(0) = 1;(d) y′′ − y = −2 cos t, y(0) = y′(0) = 1.

Solution.(a) B since y′(0) = −1 < 0 and y′′(0) = 2− sin 0− y(0) = 1 > 0.(b) D since y′(0) = −1 < 0 and y′′(0) = −2(0)− y(0) = −1 < 0.(c) A since y′(0) = 1 > 0 and y′′(0) = 02 + y(0) = 1 > 0.(d) C since y′(0) = 1 > 0 and y′′(0) = −2 cos 0 + y(0) = −1 < 0

Example 10.4Give an example of an initial value problem of the form y′′ + p(t)y′ + q(t)y =0, y(t0) = y0, y

′(t0) = y′0 for which the given t−interval −1 < t < 5 is the largeston which Theorem 10.1 guarantees a unique solution.

Solution.One such an answer is


y′′ + 1t+1y

′ + y = 1t−5 , y(0) = 1, y′(0) = 2

Example 10.5Is there a solution y(t) to the initial value problem

y′′ + 2y′ +1

t− 3y = 0, y(1) = 1, y′(1) = 2

such that limt→0+ y(t) =∞? That is, y(t) has an infinite discontinuity at t = 0.

Solution.Since p(t) = 2, q(t) = 1

t−3 , and t0 = 1, according to Theorem 10.1 the largestinterval for which the solution y(t) is defined is −∞ < t < 3. Since 0 is in thatinterval and y(t) is continuous , the limit cannot hold

79

Practice ProblemsIn Problems 10.1 - 10.6, determine the largest t−interval on which Theorem 10.1guarantees the existence of a unique solution.

Problem 10.1

y′′ + y′ + 3ty = tan t, y(π) = 1, y′(π) = −1.

Problem 10.2

ety′′ + 1t2−1y = 4

t , y(−2) = 1, y′(−2) = 2.

Problem 10.3

ty′′ + sin 2tt2−9 y

′ + 2y = 0, y(1) = 0, y′(1) = 1.

Problem 10.4

ty′′ − (1 + t)y′ + y = t2e2t, y(−1) = 0, y′(−1) = 1.

Problem 10.5

(sin2 t)y′′ − (2 sin t cos t)y′ + (cos2 t+ 1)y = sin3 t, y(π4 ) = 0, y′(π4 ) =√

2.

Problem 10.6

t2y′′ + ty′ + y = sec (ln t), y(π3 ) = 0, y′(π3 ) = −1.

In Problems 10.7 - 10.8, give an example of an initial value problem of the formy′′ + p(t)y′ + q(t)y = 0, y(t0) = y0, y

′(t0) = y′0 for which the given t−interval isthe largest on which Theorem 10.1 guarantees a unique solution.

Problem 10.7

−∞ < t <∞.


Problem 10.8

3 < t <∞.

Problem 10.9Determine the largest interval in which the initial-value problem

(t− 3)y′′ + ty′ + (ln |t|)y = 0, y(1) = 0, y′(1) = 1

is certain to have a unique solution.

Problem 10.10Theorem 10.1 tells us that the initial-value problem

y′′ + t2y = 0, y(0) = y′(0) = 0

defines exactly one function y(t). Using only Theorem 10.1, show that this functionhas the additional property y(−t) = y(t). That is, y(t) is an even function.

Problem 10.11State the existence and uniqueness theorem for second order linear differentialequations with initial consitions.

Problem 10.12Consider the homogeneous second order linear differential equation

ty′′ − y′ + 4t3y = 0.

Show that y = cos (t2) is a solution to this differential equation.

Problem 10.13Find the largest interval where√

16− t2y′′ + ln (t+ 1)y′ + (cos t)y = 0, y(0) = 2, y′(0) = 0


Problem 10.14Find the largest interval where

(t+ 2)y′′ + ty′ + cot ty = t2 + 1, y(2) = 11, y′(2) = −2


Problem 10.15Define L(y) = y′′+p(t)y′+q(t)y. Show that L(αy1+βy2) = αL(y1)+βL(y2), wherey1 and y2 are twice differentiable functions. That is, L is a linear transformation.

11 The General Solution of 2nd

Order Linear HomogeneousEquations

In this section we discuss the structure of the general solution to the homogeneoussecond order linear differential equation

y′′ + p(t)y′ + q(t)y = 0 (11.1)

where p(t) and q(t) are continuous functions for a < t < b.The first key property of (11.1) is its linear property also known as the principleof superposition.

Theorem 11.1 (Principle of Superposition)If y1 and y2 are respective solutions of (11.1) then for any constants c1 and c2, thefunction y = c1y1 + c2y2 is also a solution to (11.1).

The function c1y1 + c2y2 is called a linear combination of the functions y1 andy2.

Example 11.1Write y = 3 cos

(2t+ π

4

)as a linear combination of y1 = cos 2t and y2 = sin 2t.

Solution.Using the identity

cos (α+ β) = cosα cosβ − sinα sinβ

81

8211 THE GENERAL SOLUTION OF 2ND ORDER LINEAR HOMOGENEOUS EQUATIONS

we arrive at

3 cos(

2t+π

4

)=3 cos 2t cos

π

4− 3 sin 2t sin

π

4

=[3 cos

π

4

]cos 2t+

[−3 sin

π

4

]sin 2t

=

(3√

2

2

)cos 2t+

(−3√

2

2

)sin 2t

Theorem 11.1 states that if y1 and y2 are two given solutions of (11.1) then onecan build many new solutions by taking a linear combination y = c1y1 + c2y2.However, this theorem does not say if every solution to (11.1) has to be writtenas a linear combination of y1 and y2. So our next interest is to find out if one canexpress every solution of (11.1) as a linear combination of two solutions of (11.1).If there are such solutions y1 and y2, we shall say that the set {y1, y2} forms afundamental set of solutions to (11.1).It follows that if we know a fundamental set of solutions {y1, y2} then we knowthe general solution to (11.1) which is given by

y(t) = c1y1(t) + c2y2(t).

Identifying Fundamental SetsGiven a particular homogeneous differential equation and two solutions of thatdifferential equation, is there a convenient way for checking whether or not thesetwo solutions form a fundemental set of solutions? The answer is in the affirmativeaccording to the following theorem.

Theorem 11.2Let y1(t) and y2(t) be two solutions to the homogeneous second order linear dif-ferential equation

y′′ + p(t)y′ + q(t)y = 0, a < t < b

where p(t) and q(t) are continuous in a < t < b. If there is a a < t0 < b such that

W (y1(t0), y2(t0)) =

∣∣∣∣ y1(t0) y2(t0)y′1(t0) y′2(t0)

∣∣∣∣ = y1(t0)y′2(t0)− y′1(t0)y2(t0) 6= 0

then {y1, y2} is a fundamental set of solutions. We call the function W the Wron-skian function.

Proof.We need to show that if y(t) is a solution to (11.1) then we can write y(t) as alinear combination of y1 and y2. That is

y(t) = c1y1(t) + c2y2(t).

83

So the problem reduces to finding the constants c1 and c2. These are found bysolving the following linear system of two equations in the unknowns c1 and c2:

c1y1(t0) + c2y2(t0) = y(t0)

c1y′1(t0) + c2y

′2(t0) = y′(t0).

By the method of elimination we find

c1 =y(t0)y

′2(t0)− y′(t0)y2(t0)

W (y1(t0), y2(t0))

and

c2 =y′(t0)y1(t0)− y(t0)y

′1(t0)

W (y1(t0), y2(t0)).

Note that c1 and c2 exist since W (y1(t0), y2(t0)) 6= 0


y′′ + 4y = 0. (11.2)

(a) Show that y1(t) = cos 2t and y2(t) = sin 2t are solutions to (11.2).(b) Show that {cos 2t, sin 2t} is a fundamental set of solutions.(c) Write the solution y(t) = 3 cos (2t+ π

4 ) as a linear combination of y1 and y2.

Solution.(a) A simple calculation shows

y′′1 + 4y1 = −4 cos 2t+ 4 cos 2t = 0

y′′2 + 4y2 = −4 sin 2t+ 4 sin 2t = 0.

(b) For any t we have

W (y1(t), y2(t)) =

∣∣∣∣ cos 2t sin 2t−2 sin 2t 2 cos 2t

∣∣∣∣ = 2 cos2 2t+ 2 sin2 2t = 2 6= 0.

Thus, {y1, y2} is a fundamental set of solutions.(c) Using the formulas for c1 and c2 with t0 = 0 we find

c1 =y(0)y′2(0)− y′(0)y2(0)

W (y1(0), y2(0))

=6 cos π4 cos 0 + 6 sin π

4 sin 0

2=

3√

2

2


and

c2 =y′(0)y1(0)− y(0)y′1(0)

W (y1(0), y2(0))

=−6 sin π

4 cos 0 + 6 cos π4 sin 0

2= −3

√2

2

Theorem 11.2 says that if one can find a < t0 < b such that W (y1(t0), y2(t0)) 6= 0then the set {y1, y2} is a fundamental set of solutions. The following theoremshows that the condition W (y1(t0), y2(t0)) 6= 0 implies that W (y1(t), y2(t)) 6= 0for all t in the interval (a, b). That is, the theorem tells us that we can choose ourtest point t0 on the basis of convenience−any test point t0 will do. That’s why wechose t0 = 0 in the previous example.

Theorem 11.3 (Abel’s)Let y1(t) and y2(t) be two solutions to the homogeneous second order linear dif-ferential equation

y′′ + p(t)y′ + q(t)y = 0, a < t < b

where p(t) and q(t) are continuous in a < t < b. If t0 is any point in (a, b) then

W (y1(t), y2(t)) = W (y1(t0), y2(t0))e−

∫ tt0p(s)ds

.

Thus, if W (y1(t0), y2(t0)) 6= 0 then W (y1(t), y2(t)) 6= 0 for all a < t < b.

Remark 11.1It is easy to check that W satisfies the differential equation W ′ + pW = 0. Thus,Abel’s formula is just the solution to the initial value problem

W ′ + pW = 0, W (y1(t0), y2(t0)) = α 6= 0.

In the case y1 and y2 form a fundamental set of solutions then W (y1(t), y2(t)) isnever zero in the interval a < t < b as shown in the following theorem.

Theorem 11.4Suppose that {y1, y2} is a fundamental set of solutions to (11.1). ThenW (y1(t), y2(t)) 6=0 for all a < t < b.

Combining Theorem 11.2 and Theorem 11.4 we obtain the following corollarycharacterizing a fundamental set of solutions.

Corollary 11.1Let y1(t) and y2(t) be two solutions of (11.1). Let W (y1(t), y2(t)) denote theWronskian of y1 and y2. Then {y1, y2} is a fundamental set of solution if and onlyif W (y1(t), y2(t)) 6= 0 for all a < t < b.

85

Example 11.3Consider the initial value problem

y′′ − 1

ty′ − 3

t2y = 0, y(1) = 4, y′(1) = 8, 0 < t <∞.

(a) Show that y1(t) = t3 and y2(t) = t−1 are solutions to the differential equation.(b) Show that {y1, y2} is a fundamental set of solutions to the differential equation.(c) Solve the given initial value problem.

Solution.(a) By substitution and simple calculation we find

y′′1 −1

ty′1 −

3

t2y1 = 6t− 1

t· 3t2 − 3

t2· t3 = 0

y′′2 −1

ty′2 −

3

t2y2 = 2t−3 − 1

t· (−t−2)− 3

t2· t−1 = 0.

(b) Finding the Wronskian at t0 = 1 we see

W (y1(1), y2(1)) =

∣∣∣∣ 1 13 −1

∣∣∣∣ = −4 6= 0.

Thus, {y1, y2} is a fundamental set of solution.(c) The general solution to the differential equation has the form y(t) = c1y1(t) +c2y2(t). The initial conditions yield the following linear system in the unknownsc1 and c2.

c1y1(1) + c2y2(1) =4

c1y′1(1) + c2y

′2(1) =8

or

c1 + c2 =4

3c1 − c2 =8.

Solving this system by the method of elimination we find c1 = 3 and c2 = 1. Thus,y(t) = 3t3 + 1

t , 0 < t <∞

Example 11.4Suppose that y1(t) = t is a solution to the differential equation

t2y′′ − (t+ 2)ty′ + (t+ 2)y = 0.

Find a second solution y2.


Solution.Rewriting the given equation in the form

y′′ − (2

t+ 1)y′ + (

2

t2+

1

t)y = 0.

Thus, p(t) = −(2t + 1). But W ′ + pW = 0 so that

W ′ − (2

t+ 1)W = 0.

Using the method of integrating factor we find

W (t) = Ct2et.

To find y2, note that(y2t

)′= W (t)

t2= et. Integrating both sides, we find

y2(t) = t

∫etdt = tet

where we chose the constant of integration to be 0

87

Practice Problems

In Problems 11.1-11.7, the t−interval of solution is −∞ < t < ∞ unless indi-cated otherwise.(a) Determine whether the given functions are solutions to the differential equa-tion.(b) If both functions are solutions, calculate the Wronskian. Does this calculationshow that the two functions form a fundamental set of solutions?(c) If the two functions have been shown in (b) to form a fundamental set, con-struct the general solution and determine the unique solution satisfying the initialvalue problem.

Problem 11.1

y′′ − 4y = 0, y1(t) = e2t, y2(t) = 2e−2t, y(0) = 1, y′(0) = −2.

Problem 11.2

y′′ + y = 0, y1(t) = sin t cos t, y2(t) = sin t, y(π2 ) = 1, y′(π2 ) = 1.

Problem 11.3

y′′ − 4y′ + 4y = 0, y1(t) = e2t, y2(t) = te2t, y(0) = 2, y′(0) = 0.

Problem 11.4

ty′′ + y′ = 0, y1(t) = ln t, y2(t) = ln 3t, y(3) = 0, y′(3) = 3, 0 < t <∞.

Problem 11.5

t2y′′ − ty′ − 3y = 0, y1(t) = t3, y2(t) = −t−1, y(−1) = 0, y′(−1) = −2, t < 0.

Problem 11.6

y′′ = 0, y1(t) = t+ 1, y2(t) = −t+ 2, y(1) = 4, y′(1) = −1.


Problem 11.7

4y′′ + 4y′ + y = 0, y1(t) = et2 , y2(t) = te

t2 , y(1) = 1, y′(1) = 0.

Problem 11.8The functions y1(t) = t and y2(t) = t ln t form a fundamental set of solutions tothe differential equation

t2y′′ − ty′ + y = 0, 0 < t <∞.

(a) Show that y(t) = 2t+ t ln 3t is a solution to the differential equation.(b) Find c1 and c2 such that y(t) = c1y1(t) + c2y2(t).

Problem 11.9The functions y1(t) = e3t and y2(t) = e−3t are known to be solutions of y′′+αy′+βy = 0, where α and β are constants. Determine α and β.

Problem 11.10The functions y1(t) = t and y2(t) = et are known to be solutions of y′′ + p(t)y′ +q(t)y = 0.(a) Determine the functions p(t) and q(t).(b) On what t−intervals are the functions p(t) and q(t) continuous?(c) Compute the Wronskian of these two functions. On what t−intervals is theWronskian nonzero?(d) Are the observations in (b) and (c) consistent with Abel’s Theorem?

Problem 11.11It is known that two solutions of y′′+ ty′+2y = 0 has a Wronskian W (y1(t), y2(t))that satisfies W (y1(1), y2(1)) = 4. What is W (y1(2), y2(2))?

Problem 11.12The pair of functions {y1, y2} is known to form a fundamental set of solutions ofy′′ + αy′ + βy = 0, where α and β are constants. One solution is y1(t) = e2t, andthe Wronskian formed by these two solutions is W (y1(t), y2(t)) = e−t. Determinethe constants α and β.

Problem 11.13The Wronskian of a pair of solutions of y′′+ p(t)y′+ 3y = 0 is W (t) = e−t

2. What

is the coefficient function p(t)?

89

Problem 11.14Prove that if y1 and y2 have maxima or minima at the same point in an intervalI, then they cannot be a fundamental set of solutions on that interval.

Problem 11.15Without solving the equation, find the Wronskian of two solutions of Bessel’sequation

t2y′′ + ty′ + (t2 − µ2)y = 0.

Problem 11.16If W (y1, y2) = t2et and y1(t) = t then find y2(t).

Problem 11.17The functions t2 and 1/t are solutions to a 2nd order, linear homogeneous ODEon t > 0. Verify whether or not the two solutions form a fundamental solution set.

Problem 11.18Show that t3 and t4 can’t both be solutions to a differential equation of the formy′′ + p(t)y′ + q(t)y = 0 where p and q are continuous functions defined on the realnumbers.

Problem 11.19Suppose that t2 + 1 is the Wronskian of two solutions to the differential equationy′′ + p(t)y′ + q(t)y = 0. Find p(t).


12 Second Order LinearHomogeneous Equations withConstant Coefficients: DistinctCharacterisitc Roots

In the previous section we discussed the structure of the general solution of a secondorder linear homogeneous differential equation. As we saw, the general solutionis a linear combination of two solutions that form a fundamental set of solutions.In this and the next two sections we discuss methods for finding the fundamentalset of solutions for second order homogeneous equations with constant coefficients,i.e., equations of the form

ay′′ + by′ + cy = 0 (12.1)

where a, b and c are constants with a 6= 0.Notice first that for b = 0 and c 6= 0 the function y′′ is a constant multiple of y.So it makes sense to look for a function with such property. One such function isy(t) = ert. Substituting this function into (12.1) leads to

ay′′ + by′ + cy = ar2ert + brert + cert = (ar2 + br + c)ert = 0.

Since ert > 0 for all t, the previous equation leads to

ar2 + br + c = 0. (12.2)

Thus, a function y(t) = ert is a solution to (12.1) when r satisfies equation (12.2).We call (12.2) the characteristic equation for (12.1) and the polynomial C(r) =ar2 + br + c is called the characteristic polynomial.

Example 12.1Solve: y′′ − 5y′ − 6y = 0.

91

9212 SECONDORDER LINEAR HOMOGENEOUS EQUATIONSWITH CONSTANT COEFFICIENTS: DISTINCT CHARACTERISITC ROOTS

Solution.The characteristic polynomial for this equation is C(r) = r2−5r−6 = (r−2)(r−3).Thus, the roots of the characteristic equation are r = 2 and r = 3. Since

W (t) =

∣∣∣∣ e2t e3t

2e2t 3e3t

∣∣∣∣ = e5t 6= 0

the functions y1(t) = e2t and y2(t) = e3t form a fundamental set of solutions.Hence, the general solution is given by y(t) = c1e

2t + c2e3t where c1 and c2 are

arbitrary constants

We conclude from the previous example that the two distinct real solutions tothe characteristic equation lead to the general solution. Does this result applyto any equation (12.1) whose characteristic equation has distinct solutions? Theanswer is in the affirmative. To see this, let r1 and r2 be the two distinct solutionsto (12.2). Then

W (t) =

∣∣∣∣ er1t er2t

r1er1t r2e

r2t

∣∣∣∣ = r2e(r1+r2)t − r1e(r1+r2)t = (r2 − r1)e(r1+r2)t 6= 0

since both r1 − r2 and e(r1+r2)t are not equal to 0. Hence, er1t and er2t form afundamental set of solutions. As a result, the general solution of (12.1) is given byy(t) = c1e

r1t + c2er2t where c1 and c2 are arbitrary constants.


y′′ − y′ − 6y = 0, y(0) = 1, y′(0) = 2.

Describe the behavior of the solution y(t) as t→ −∞ and t→∞.

Solution.The characteristic polynomial is C(r) = r2 − r − 6 = (r − 3)(r + 2) so that thecharacteristic equation r2 − r − 6 = 0 has the solutions r1 = 3 and r2 = −2. Thegeneral solution is then given by

y(t) = c1e3t + c2e

−2t.

Taking the derivative to obtain

y′(t) = 3c1e3t − 2c2e

−2t.

93

The conditions y(0) = 1 and y′(0) = 2 lead to the system

c1 + c2 = 13c1 − 2c2 = 2.

Solving this system by the method of elimination we find c1 = 45 and c2 = 1

5 .Hence, the unique solution to the initial value problem is

y(t) =1

5(4e3t + e−2t).

As t → −∞, e3t → 0 and e−2t → ∞. Thus, y(t) → ∞. Similarly, y(t) → ∞ ast→∞

Remark 12.1In this section, we have only considered the case when (12.2) has two distinct realsolutions. In Section 13, we discuss the case of (12.2) having repeated solutionsand in Section 14 we look at the complex solutions

Example 12.3Consider the initial value problem y′′ + αy′ + βy = 0, y(0) = 1, y′(0) = y′0,where α, β, and y′0 are constants. It is known that one solution of the differentialequation is y1(t) = e−3t and that the solution of the initial value problem satisfieslimt→∞ y(t) = 2. Determine the constants α, β, and y′0.

Solution.Since r = −3 is a solution to the characteristic equation, we have (−3)2 +α(−3) +β = 0 or −3α + β = −9. Also, since limt→∞ y(t) = 2, the second root for thecharacteristic equation must be r = 0. In this case, β = 0 and solving for α wefind α = 3. Hence, y(t) = c1e

−3t + c2. Since limt→−∞ y(t) = 2, we find c2 = 2.Since y(0) = 1, we find c1 + 2 = 1 so that c1 = −1. Thus, y(t) = −e−3t + 2 andy′(t) = 3e−3t. Therefore, y′0 = y′(0) = 3

Example 12.4Consider the initial value problem y′′ + αy′ + βy = 0, y(0) = 3, y′(0) = 5. Thedifferential equation has a fundamental set of solutions {y1, y2}. It is known thaty1(t) = e−t and that the Wronskian formed by the two members of the fundamentalset is W (t) = 4e2t.(a) Determine y2(t).(b) Determine the constants α and β.(c) Solve the initial value problem.


Solution.(a) The second solution is of the form y2(t) = ert. In this case,

W (t) =

∣∣∣∣ e−t ert

−e−t rert

∣∣∣∣ = (r + 1)e(r−1)t.

But W (t) = 4e2t and this leads to r = 3. Hence, y2(t) = e3t.(b) Since r = −1 and r = 3 are the roots for the characteristic equation, we have(r + 1)(r − 3) = 0 or r2 − 2r − 3 = 0. This implies that y′′ − 2y′ − 3y = 0 so thatα = −2 and β = −3(c) The initial conditions and y′(t) = −c1e−t+3c2e

3t lead to the system c1+c2 = 3and −c1 + 3c2 = 5. Solving this system we find c1 = 1 and c2 = 2. Thus,

y(t) = e−t + 2e3t

Example 12.5Solve the initial-value problem 4y′′ − y = 0, y(0) = 2, y′(0) = β. Then find β sothat the solution approaches zero as t→∞.

Solution.The characteristic equation 4r2 − 1 = 0 has roots r = −1

2 and r = 12 . Thus,

y(t) = c1e− t

2 + c2et2 .

The initial conditions and y′(t) = − c12 e− t

2 + c22 e

t2 lead to the system c1 + c2 = 2

and c1 − c2 = −2β. Solving this system we find c1 = 1− β and c2 = 1 + β. Thus,

y(t) = (1− β)e−t2 + (1 + β)e

t2 .

Since limt→∞ y(t) = 0 and limt→∞ et =∞, we must have 1+β = 0. Thus, β = −1

Example 12.6Show that if λ is a root of aλ3 + bλ2 + cλ + d = 0, then eλt is a solution ofay′′′ + by′′ + cy′ + dy = 0.

Solution.We have

ay′′′ + by′′ + cy′ + dy = aλ3eλt + bλ2eλt + cλeλt + deλt

= (aλ3 + bλ2 + cλ+ d)eλt = 0

95

Practice Problems

Problem 12.1Solve the initial value problem

y′′ + y′ − 2y = 0, y(0) = 3, y′(0) = −3.



y′′ − 4y′ + 3y = 0, y(0) = −1, y′(0) = 1.



y′′ − y = 0, y(0) = 1, y′(0) = −1.



y′′ + 5y′ + 6y = 0, y(0) = 1, y′(0) = −1.



y′′ − 4y = 0, y(3) = 0, y′(3) = 0.



2y′′ − 3y′ = 0, y(−2) = 3, y′(−2) = 0.




y′′ + 4y′ + 2y = 0, y(0) = 0, y′(0) = 4.



2y′′ − y = 0, y(0) = −2, y′(0) =√

2.


Problem 12.9Obtain the general solution to the differential equation y′′′ − 5y′′ + 6y′ = 0.

Problem 12.10A particle of mass m moves along the x−axis and is acted upon by a drag forceproportional to its velocity. The drag constant is denoted by k. If x(t) representsthe particle position at time t, Newton’s law of motion leads to the differentialequation mx′′(t) = −kx′(t).(a) Obtain the general solution to this second order linear differential equation.(b) Solve the initial value problem if x(0) = x0 and x′(0) = v0.(c) What is limt→∞ x(t)?

Problem 12.11Find a homogeneous second-order linear ordinary differential equation whose gen-eral solution is y(t) = c1e

2t + c2e−t.

Problem 12.12Find the general solution of the differential equation y′′ − 3y′ − 4y = 0.

Problem 12.13Find the general solution of the differential equation y′′ + 4y′ − 5y = 0.

Problem 12.14Find the general solution of the differential equation −3y′′ + 2y′ + y = 0.

Problem 12.15Solve the initial-value problem: y′′ + 3y′ − 4y = 0, y(0) = −1, y′(0) = 1.

Problem 12.16Solve the initial-value problem: 2y′′ + 5y′ − 3y = 0, y(0) = 2, y′(0) = 1.

13 Characteristic Equationswith Repeated Roots

In this section we consider the question of the characteristic equation having arepeated real solution. This occurs when b2 − 4ac = 0. The two equal roots aregiven by

r1 = r2 = − b

2a.

The computation based on the trial form y(t) = ert yields only one solution,namely

y1(t) = e−b2at.

Since a fundamental set of solutions consists of two functions having a nonzeroWronskian, there must be another solution having a different functional form.The second solution follows from the following theorem.

Theorem 13.1Suppose that y1(t) is a nontrivial solution to the differential equation

y′′ + p(t)y′ + q(t)y = 0. (13.1)

Then any solution y2(t) can be written in the form

y2(t) = C

(∫e−

∫p(t)dt

y21(t)dt

)y1(t) + C ′y1(t) (13.2)

where C and C ′ are arbitrary constants.

Proof.First, recall that the Wronskian W (t) of any two solutions to (13.1) satisfies thedifferential equation W ′+p(t)W = 0 so that W (t) = Ce−

∫p(t)dt. If y2 is a solution

to (13.1) then (y2y1

)′=W (t)

y21(t)= C

e−∫p(t)dt

y21(t).

97

98 13 CHARACTERISTIC EQUATIONS WITH REPEATED ROOTS

Integrating this last equation we find

y2(t) = C

(∫e−

∫p(t)dt

y21(t)dt

)y1(t) + C ′y1(t)

where C and C ′ are arbitrary constantsNotice that the term C ′y1(t) is simply a constant multiple of y1(t). Since thegeneral solution of the differential equation (13.1) contains y1(t) multiplied by anarbitrary constant, we lose no generality by setting C ′ = 0. We can likewise takeC = 1 since y2(t) will also be multiplied by an arbitrary constant in the generalsolution. With these simplification the second solution is

y2(t) =

(∫e−

∫p(t)dt

y21(t)dt

)y1(t).

Now, for the equation

ay′′ + by′ + cy = 0 (13.3)

we have p(t) = ba . If y1(t) = e−

b2at then

y2(t) =

(∫e−

bat

e−batdt

)e−

b2at = te−

b2at.

Hence, the general solution to (13.3) is given by

y(t) = c1e− b

2at + c2te

− b2at.

Example 13.1Solve the initial value problem: y′′ + 2y′ + y = 0, y(0) = 1, y′(0) = −1.

Solution.The characteristic equation r2 + 2r + 1 = 0 has a repeated root: r1 = r2 = −1.Thus, the general solution is given by

y(t) = c1e−t + c2te

−t.

The two conditions y(0) = 1 and y′(0) = −1 lead to c2 = 0 and c1 = 1. Hence, theunique solution is y(t) = e−t

99


t2y′′ + 2ty′ − 2y = 0, 0 < t <∞.

Find the general solution given that y1(t) = t is a solution to the differentialequation.

Solution.Since t > 0, we can rewrite the given equation in the form

y′′ +2

ty′ − 2

t2y = 0.

In this case, p(t) = 2t and

y2(t) =

(∫e−

∫2tdt

t2dt

)t =

(∫1

t4dt

)t = − 1

3t2.

Hence, the general solution is y(t) = c1t+ c2t−2

Example 13.3The graph of a solution y(t) of the differential equation 4y′′ + 4y′ + y = 0 passes

through the points (1, e−12 ) and (2, 0). Determine y(0) and y′(0).

Solution.The characteristic equation 4r2 + 4r + 1 = 0 has the roots r1 = r2 = −1

2 so thatthe general solution is

y(t) = c1e− t

2 + c2te− t

2 .

Since y(2) = 0, we find c1 + 2c2 = 0. Since y(1) = e−12 , we have c1 + c2 = 1.

Solving the system of two equations, we find c1 = 2 and c2 = −1. Hence,

y(t) = 2e−t2 − te−

t2 .

Now, y(0) = 2. Also, replacing t = 0 in y′(t) = −2e−t2 + t

2e− t

2 to obtain y′(0) =−2

Example 13.4Find a homogeneous second order linear differential equation whose general solu-tion is given by y(t) = c1e

−3t + c2te−3t.


Solution.The characteristic equation has the double roots r1 = r2 = −3 so that r2+6r+9 =0. Hence, the differential equation is

y′′ + 6y′ + 9y = 0

Example 13.5Show that if λ is a double root of at3 + bt2 + ct+ d = 0, then teλt is also a solutionof ay′′′ + by′′ + cy′ + dy = 0.

Solution.Since λ is a double root, we have aλ3 + bλ2 + cλ+ d = 0 and 3aλ2 + 2bλ+ c = 0.Let y(t) = teλt. Then y′ = eλt + λteλt, y′′ = 2λeλt + λ2teλt, y′′′ = 3λ2eλt + λ3teλt.Substituting into the equation we find

ay′′′ + by′′ + cy′ + dy =[(aλ3 + bλ2 + cλ+ d)t+ (3aλ2 + 2bλ+ c)]eλt

=0

101

Practice Problems

In Problems 13.1 - 13.5 answer the following questions.(a) Obtain the general solution of the differential equation.(b) Impose the initial conditions to obtain the unique solution of the initial valueproblem.(c) Describe the behavior of the solution as t→ −∞ and t→∞.

Problem 13.1

9y′′ − 6y′ + y = 0, y(3) = −2, y′(3) = −5

3.

Problem 13.2

25y′′ + 20y′ + 4y = 0, y(5) = 4e−2, y′(5) = −3

5e−2.

Problem 13.3

y′′ − 4y′ + 4y = 0, y(1) = −4, y′(1) = 0.

Problem 13.4

y′′ + 2√

2y′ + 2y = 0, y(0) = 1, y′(0) = 0.

Problem 13.5

3y′′ + 2√

3y′ + y = 0, y(0) = 2√

3, y′(0) = 3.

Problem 13.6Find the general solution of y′′ − 6y′ + 9y = 0.

Problem 13.7Find the general solution of 4y′′ − 4y′ + y = 0.

Problem 13.8Solve the initial-value problem: y′′ + y′ + y

4 = 0, y(0) = 2, y′(0) = 0.


14 Characteristic Equationswith Complex Roots

In this section, we solve the linear second order homogeneous differential equationwith constant coefficients

ay′′ + by′ + cy = 0, a 6= 0 (14.1)

when the roots of the characteristic equation

ar2 + br + c = 0 (14.2)

are complex numbers. This occurs when b2 − 4ac < 0. In this case, the complexroots of equation (14.2) are given by

r1,2 =−b± i

√4ac− b2

2a

where i =√−1. We will write

r1,2 = α± iβ

where α = − b2a and β =

√4ac−b22a . Like before, we would like to conclude that the

functions

e(α+iβ)x and e(α−iβ)x

are solutions to (14.1). These are complex solutions, we would like to have realsolutions to the original real differential equation. This requires the use of theso-called the complex exponential function which we introduce and discusssome of its properties.For any complex number z = α+ iβ we define the Euler’s function

ez = eα(cosβ + i sinβ).

103

104 14 CHARACTERISTIC EQUATIONS WITH COMPLEX ROOTS

The exponential function satisfies the usual laws of exponentials such as

ezew = ez+w.

To see this, we let z = α1 + iβ1 and w = α2 + iβ2. Then

ezew =eα1(cosβ1 + i sinβ1)eα2(cosβ2 + i sinβ2)

=eα1+α2 [(cosβ1 cosβ2 − sinβ1 sinβ2) + i(sinβ1 cosβ2 + cosβ1 sinβ2)]

=eα1+α2 [cos (β1 + β2) + i sin (β1 + β2)]

=ez+w.

From the above rule we can write

(ez)n = ez · ez · · · ez = ez+z+···+z = enz

where n is a positive integer.

It follows from the above discussion that the complex solutions to the differ-ential equation are linear combinations of eαt cosβt and eαt sinβt. Now lettingy1(t) = eαt cosβt and y2(t) = eαt sinβt we find

ay′′1 + by′1 + cy1 =a(α2eαt cosβt− β2eαt cosβt− 2αβeαt sinβt)

+b(αeαt cosβt− βeαt sinβt) + ceαt cosβt

=eαt cosβt(a(α2 − β2) + bα+ c)− eαt sinβt(2aαβ + bβ)

=eαt cosβt

(a

(b2

4a2− 4ac− b2

4a2

)+ b

(−b2a

)+ c

)−eαt sinβt

(2a

(− b

2a

√4ac− b2

2a

)+b√

4ac− b22a

)= 0.

Thus, y1(t) = eαt cosβt is a solution to equation (14.1). Similarly, we show thaty2(t) = eαt sinβt is a solution to equation (14.1). Moreover,

W (t) =

∣∣∣∣ eαt cosβt eαt sinβtαeαt cosβt− βeαt sinβt αeαt sinβt+ βeαt cosβt

∣∣∣∣ = βe2αt 6= 0.

Hence, {y1, y2} is a fundamental set of solutions to equation (14.1) so that thegeneral solution is given by

y(t) = eαt(c1 cosβt+ c2 sinβt)

where c1 and c2 are real numbers.

105

Example 14.1Solve: y′′ + 2y′ + 5y = 0.

Solution.The characteristic equation r2 + 2r+ 5 = 0 has complex roots r1,2 = −1± 2i. Thegeneral solution is

y(t) = e−t(c1 cos 2t+ c2 sin 2t)


y′′ − 10y′ + 29y = 0, y(0) = 1, y′(0) = 3.

Solution.The characteristic equation r2 − 10r+ 29 = 0 has the complex roots r1,2 = 5± 2i.Thus, the general solution is given by the expression

y(t) = e5t(c1 cos 2t+ c2 sin 2t).

Finding y′ we obtain

y′(t) = e5t[(5c1 + 2c2) cos 2t+ (5c2 − 2c1) sin 2t].

The initial conditions yield c1 = 1 and c2 = −1. Thus, the unique solution to theinitial value problem is

y(t) = e5t(cos 2t− sin 2t)

Next, we consider the question of representing the general solution y(t) = eαt(c1 cosβt+c2 sinβt) in the form y(t) = Keαt cos (βt− δ), where 0 ≤ δ < 2π. For this, we letP (c1, c2) be a coordinate point in the plane and let δ be the angle between the

t-axis and ray−−→OP . See Figure 14.1. Then

cos δ = c1K and sin δ = c2

K

where K =√c21 + c22. Then in terms of K and δ we can write

c1 cosβt+ c2 sinβt =K( c1K

cosβt+c2K

sinβt)

=K(cos δ cosβt+ sin δ sinβt) = K cos (βt− δ).

It follows thaty(t) = Keαt cos (βt− δ).


Figure 14.1

We call Keαt the amplitude of the oscillations. This means that the graph ofy(t) is bounded by the graphs of ±Keαt. The angle δ

β is the phase shift. Theterm “phase shift” reflects the fact that we obtain the graph of cos (βt− δ) =

cosβ(t− δ

β

)by shifting the graph of cosβt to the right by an amount t = δ

β .

Example 14.3Put the solution of the initial value problem

y′′ − 2y′ + 17y = 0, y(0) = −4, y′(0) = 8

in the form y(t) = Keαt cos (βt− δ).

Solution.The characteristic equation r2 − 2r + 17 = 0 has the complex roots r1,3 = 1± 4i.Thus, the general solution to the differential equation is

y(t) = et(c1 cos 4t+ c2 sin 4t).

Since y(0) = −4, we find c1 = −4.Also y′(t) = et(c1 cos 4t+c2 sin 4t)+et(4c2 cos 4t−4c1 sin 4t) and y′(0) = 8 imply 8 = −4 + 4c2 so that c2 = 3. Thus, the unique so-lution to the initial value problem is

y(t) = et(3 sin 4t− 4 cos 4t).

Now, K =√

9 + 16 =√

25 = 5. Thus, tan δ = −34 so that δ = arctan (−3

4). Hence,

y(t) = 5et cos

(4t+

(arctan

3

4

)).

107

The graph of y(t) together with the envelope containing it is shown in Figure 14.2

Figure 14.2


Practice Problems

Problem 14.1For any z = α + iβ we define the conjugate of z to be the complex numberz = α− iβ. Show that α = 1

2(z + z) and β = 12i(z − z).

Problem 14.2Write each of the complex numbers in the form α + iβ, where α and β are realnumbers.1. 2ei

π3 .

2. (2− i)ei3π2 .

3. (√

2eiπ6 )4.

Problem 14.3Write each function in the form eαt(A cosβt+ iB sinβt), where α, β,A, and B arereal numbers.1. 2ei

√2t.

2. −12e

2t+i(t+π).

3. (√

3e(1+i)t)3.

In Problems 14.4 - 14.8(a) Determine the roots of the characteristic equation.(b) Obtain the general solution as a linear combination of real-valued solutions.(c) Impose the initial conditions and solve the initial value problem.

Problem 14.4

y′′ + 2y′ + 2y = 0, y(0) = 3, y′(0) = −1.

Problem 14.5

2y′′ − 2y′ + y = 0, y(−π) = 1, y′(−π) = −1.

Problem 14.6

y′′ + 4y′ + 5y = 0, y(π

2) =

1

2, y′(

π

2) = −2.

Problem 14.7

y′′ + 4π2y = 0, y(1) = 2, y′(1) = 1.

109

Problem 14.8

9y′′ + π2y = 0, y(3) = 2, y′(3) = −π.

In Problems 14.9 - 14.10, the function y(t) is a solution of the initial value problemy′′+ay′+by = 0, y(t0) = y0, y

′(t0) = y′0, where the point t0 is specified. Determinethe constants a, b, y0, and y′0.

Problem 14.9

y(t) = 2 sin 2t+ cos 2t, t0 =π

4.

Problem 14.10

y(t) = et−π6 cos 2t− et−

π6 sin 2t, t0 =

π

6.

In Problems 14.11 - 14.13, rewrite the function y(t) in the form y(t) = Keαt cos (βt− δ),where 0 ≤ δ < 2π. Use this representation to sketch a graph of the given function,on a domain sufficiently large to display its main features.

Problem 14.11y(t) = sin t+ cos t.

Problem 14.12y(t) = et cos t+

√3et sin t.

Problem 14.13y(t) = e−2t cos 2t− e−2t sin 2t.

Problem 14.14Consider the differential equation y′′ + ay′ + 9y = 0, where a is a real number.Suppose that we know the Wronskian of a fundamental set of solutions of thisdifferential equation is constant: W (t) = 1 for all real numbers t. Find the generalsolution of this differential equation.

Problem 14.15Rewrite y(t) = 2 cos 7t− 11 sin 7t in phase-angle form. Give the exact function (soyour answer will involve the inverse tangent function.)


Problem 14.16Find a homogeneous linear ordinary differential equation whose general solutionis y(t) = c1e

2t cos (3t) + c2e2t sin (3t).

Problem 14.17Rewrite y(t) = 5e(5−2i)t − 3e(5+2i)t, without complex exponents, using sines andcosines. What ODE of the form ay′′ + by′ + cy = 0, has y as a solution?

Problem 14.18Consider the function y(t) = 3 cos 2t−4 sin 2t. Find a second order linear IVP thaty satisfies.

Problem 14.19An equation of the form

t2y′′ + αty′ + βy = 0, t > 0

where α and β are real constants is called an Euler equation. Show that thesubstitution x = ln t transforms Euler equation into an equation with constantcoefficients. Hint: dy

dt = dydx

dxdt .

Problem 14.20Use the result of the previous problem to solve the differential equation t2y′′ +ty′ + y = 0.

15 Series Solutions ofDifferential Equations

We know how to find the general solution of a second order linear differentialequation with constant coefficients. In this section, we discuss finding the generalsolution to the homogeneous equation

y′′ + p(t)y′ + q(t)y = 0 (15.1)

where p(t) and q(t) are not necessarily constants.Let t = t0. We are interested in answering the question: When is it possible torepresent the general solution of (15.1) in terms of a power series that converge insome open interval centered at t0?In order to answer this question, we need to introduce the following definitions:We say that a function f(t) is analytic at t = t0 if f(t) can be expressed as aTaylor series

f(t) =∞∑n=0

f (n)(t0)

n!(t− t0)n.

Examples of analytic functions are polynomial functions and rational functions.We say that t = t0 is an ordinary point when both p(t) and q(t) are analytic att0. If p(t) and/or q(t) is not analytic at t0, we say that t0 is a singular point.


(1− t2)y′′ + (tan t)y′ + t53 y = 0

in the open interval −2 < t < 2. Classify each point in this interval as an ordinarypoint or a singular point.

Solution.We first rewrite the equation in the form (15.1),

y′′ +tan t

1− t2y′ +

t53

1− t2y = 0.

111

112 15 SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS

Therefore,

p(t) = tan t1−t2 and q(t) = t

53

1−t2 .

The function p(t) is not analytic at t = ±1 (where the denominator vanishes) andat t = ±π

2 (where the tangent has a vertical asymptotes). The function q(t) isnot analytic at t = ±1 and at t = 0 (q(t) does not have a second derivative at 0.)Thus, in the interval −2 < t < 2, the five points t = 0,±1,±π

2 are singular pointsand all other points are ordinary points

The following theorem answers our earlier question about a series solution to theequation (15.1).

Theorem 15.1Let p(t) and q(t) be analytic at t0 and let R be the smaller of the two radii ofconvergence1 of their respective Taylor series representations. The general solutionto (15.1) can be expressed as

y(t) =∞∑n=0

an(t− t0)n = c1y1(t) + c2y2(t),

where c1 and c2 are arbitrary constants. The functions y1(t) and y2(t) form afundamental set of solutions, analytic in the interval |t− t0| < R.

Example 15.2The point t0 = 0 is an ordinary point of the differential equation

y′′ + (t3 + 1)y′ − ty = 0.

Determine the value of R as stated in the previous theorem.

Solution.Since p(t) and q(t) are polynomial functions, we have Rp = Rq = ∞. Hence,R =∞

The Method of SolutionWe assume that t0 is an ordinary point of the equation (15.1), so it has a seriessolution near t0 of the form y(t) =

∑∞n=0 an(t− t0)n that converges for |t− t0| < R.

1The radius of convergence of a function f(t), about t0, is the distance between t0 andthe nearest singularity of f(t).

113

We want to determine the coefficients a0, a1, a2, · · · in the series solution.Let’s differentiate the series termwise twice:

y(t) =a0 + a1(t− t0) + a2(t− t0)2 + · · ·y′(t) =a1 + 2a2(t− t0) + 3a3(t− t0)2 + · · ·y′′(t) =2a2 + 6a3(t− t0) + 12a4(t− t0)2 + · · ·

Substituting in the differential equation (15.1), we get

K0 +K1(t− t0) +K2(t− t0)2 + · · · = 0 (15.2)

where Kn is function of certain coefficients ai.In order that series (15.2) be valid for all t in the interval |t − t0| < R, we musthave

Kn = 0, n = 0, 1, 2, · · · .This leads to a set of conditions that the coefficients an must satisfy.

Example 15.3Find the series solution of the differential equation near the ordinary point 0.

(t2 − 4)y′′ + 3ty′ + y = 0.

Solution.The singular points of the given differential equation are −2 and 2. Thus, we willfind a series solution of the differential equation in the interval |t| < 2. The solutionis of the form

y(t) =

∞∑n=0

antn.

Differentiating this series twice we find

y′(t) =∑∞

n=1 nantn−1 and y′′(t) =

∑∞n=2 n(n− 1)ant

n−2.

Substituting these series in the given equation we find

t2∞∑n=2

n(n− 1)antn−2 − 4

∞∑n=2

n(n− 1)antn−2 + 3t

∞∑n=1

nantn−1 +

∞∑n=0

antn =0

∞∑n=2

n(n− 1)antn −

∞∑n=2

4n(n− 1)antn−2 +

∞∑n=1

3nantn +

∞∑n=0

antn =0

∞∑n=0

n(n− 1)antn −

∞∑n=0

4(n+ 2)(n+ 1)an+2tn +

∞∑n=0

3nantn +

∞∑n=0

antn =0

∞∑n=0

[n(n− 1)an − 4(n+ 1)(n+ 2)an+2 + 3nan + an] tn =0.


Equating all the coefficients to zero we find

(n+ 1)2an − 4(n+ 1)(n+ 2)an+2 = 0

and obtain the recurrence formula,

an+2 =(n+ 1)an4(n+ 2)

, n ≥ 0.

From this formula, we find

a2 =a08

a3 =a16

a4 =3a0128

a5 =a130

a6 =5a01024

a7 =a1140

......

Notice that each even coefficient is expressed in terms of a0 and each odd coefficientis expressed in terms of a1. The general solution is:

y(t) = a0

(1 +

1

8t2 +

3

128t4 +

5

1024t6 + · · ·

)+a1

(t+

1

6t3 +

1

30t5 +

1

140t7 + · · ·

)

115

Practice Problems

Problem 15.1Identify the singular and ordinary points of the differential equation

y′′ + ty′ + (t2 + 2)y = 0.


(t2 − 1)y′′ + ty′ +1

ty = 0.


(t2 − 4)y′′ + 3ty′ + y = 0.

Problem 15.4The point t0 = 0 is an ordinary point of the differential equation

(t2 − 4)y′′ + 3ty′ + y = 0.

Determine a value of R as stated in Theorem 15.1.

Problem 15.5Find the series solution near the ordinary point 0 to the differential equation

y′′ + ty′ + (t2 + 2)y = 0.


y′′ − ty′ − t2y = 0.


(t2 + 1)y′′ − y′ + y = 0.

Problem 15.8Write the following as a single power series:∑

n=0

nan(t− 2)n+1 +∑n=2

n2an(t− 2)n.



y′′ + ty′ + y = 0.


(1− t)2y′′ − 2y = 0.

16 The Structure of theGeneral Solution of LinearNonhomogeneous Equations

In this section, we consider the question of finding the general solution to thedifferential equation

y′′ + p(t)y′ + q(t)y = g(t) (16.1)

where p(t), q(t) and g(t) are continuous functions for a < t < b. The followingtheorem provides the structure of the general solution to equation (16.1).

Theorem 16.1Let {y1(t), y2(t)} be a fundamental set of solutions to the homogeneous equationy′′ + p(t)y′ + q(t)y = 0 and yp(t) be a particular solution of the nonhomogeneousequation

y′′ + p(t)y′ + q(t)y = g(t), a < t < b.

The general solution of the nonhomogeneous equation is given by

y(t) = yp(t) + c1y1(t) + c2y2(t)

for constants c1 and c2.

Proof.Let y(t) be any solution to equation (16.1). Because yp(t) is also a solution then

(y − yp)′′ + p(t)(y − yp)′ + q(t)(y − yp) =(y′′ + p(t)y′ + q(t)y)

−(y′′p + p(t)y′p + q(t)yp)

=g(t)− g(t) = 0.

117

11816 THE STRUCTUREOF THEGENERAL SOLUTIONOF LINEAR NONHOMOGENEOUS EQUATIONS

Therefore, y − yp is a solution to the homogeneous equation. But {y1, y2} isa fundamental set of solutions to the homogeneous equation so that there existunique constants c1 and c2 such that y(t)− yp(t) = c1y1(t) + c2y2(t). Hence,

y(t) = yp(t) + c1y1(t) + c2y2(t)

It follows from the above theorem that finding the general solution to nonhomo-geneous equations consists of three steps:1. Find the general solution of the associated homogeneous equation y′′+ p(t)y′+q(t)y = 0.2. Find a single solution of the original equation y′′ + p(t)y′ + q(t)y = g(t).3. Add together the solutions found in steps 1 and 2.

Example 16.1Use the fact that yp(t) = 3t − 1 to find the unique solution to the initial valueproblem

y′′ − 2y′ − 3y = −9t− 3, y(0) = 1, y′(0) = 3.

Solution.Because we are given yp then we need to find the general solution of the ho-mogeneous equation y′′ − 2y′ − 3y = 0. The associated characteristic equationr2 − 2r − 3 = 0 has roots r1 = −1 and r2 = 3. Hence, the general solution to thedifferential equation is

y(t) = c1e−t + c2e

3t + 3t− 1.

The derivative of this function is given by y′(t) = −c1e−t+3c2e3t+3. The condition

y(0) = 1 leads to c1 + c2 = 2. The condition y′(0) = 3 leads to −c1 + 3c2 = 0.Solving for c1 and c2 we find c1 = 3

2 and c2 = 12 . The unique solution is given by

y(t) =3

2e−t +

1

2e3t + 3t− 1

Note that y1(t) = 3t − 1 and y2(t) = e3t + 3t − 1 both are particular solutions tothe given differential equation. However, the sum u(t) = y1(t)+y2(t) = e3t+6t−2is not a solution since

u′′ − 2u′ − 3u = −18t− 6 6= −9t− 3.

This shows that the superposition of solutions is valid only for homogeneous equa-tions and not true in general for nonhomogeneous equations. However, we canhave a property of superposition of nonhomogeneous if one is adding two solutionsof two different nonhomogeneous equations. More precisely, we have

119

Theorem 16.2Let y1(t) be a solution of y′′ + p(t)y′ + q(t)y = g1(t) and y2(t) a solution ofy′′ + p(t)y′ + q(t)y = g2(t). Then for any constants c1 and c2 the function Y (t) =c1y1(t) + c2y2(t) is a solution of the equation

y′′ + p(t)y′ + q(t)y = c1g1(t) + c2g2(t).

Proof.We have

Y ′′ + p(t)Y ′ + q(t)Y =c1y′′1 + c2y

′′2 + p(t)c1y

′1 + p(t)c2y

′2 + q(t)c1y1 + q(t)c2y2

=c1(y′′1 + p(t)y′1 + q(t)y1) + c2(y

′′2 + p(t)y′2 + q(t)y2)

=c1g1(t) + c2g2(t)

Example 16.2The functions u1(t) and u2(t) are solutions to the following differential equations

u′′1 + p(t)u′1 + q(t)u1 =2e−t − t− 1

u′′2 + p(t)u′2 + q(t)u2 =3t

Use the functions u1 and u2 to construct a particular solution of the differentialequation

u′′ + p(t)u′ + q(t)u = 4e−t − 2.

Solution.The right-hand side of the given equation can be written as 4e−t − 2 = 2(2e−t −t− 1) + 2

3(3t) so that by the previous theorem, the function u(t) = 2u1(t) + 23u2(t)

is the required particular solution

Example 16.3Consider the IVP

y′′ − y′ − 2y = 4e−t, y(0) = 0, y′(0) = 0, yp(t) = −4

3te−t.

(a) Verify that the given function, yp(t), is a particular solution of the differentialequation.(b) Determine the general solution,yh, of the homogeneous equation.(c) Find the general solution to the differential equation and impose the initialconditions to obtain the unique solution of the initial value problem.


Solution.(a) y′p = −4

3e−t + 4

3 te−t, y′′p = 8

3e−t − 4

3 te−t.

y′′p − y′p − 2yp =8

3e−t − 4

3te−t +

4

3e−t − 4

3te−t +

8

3te−t

=4e−t

(b) The associated characteristic equation r2 − r − 2 = 0 has roots r1 = −1 andr2 = 2. Hence, the general solution to the homogeneous differential equation is

yh(t) = c1e−t + c2e

2t

(c) The general solution to the differential equation is y(t) = c1e−t+ c2e

2t− 43 te−t.

The derivative of this function is given by y′(t) = −c1e−t + 2c2e2t − 4

3e−t + 4

3 te−t.

The condition y(0) = 0 leads to c1 + c2 = 0. The condition y′(0) = 0 leads to−c1 + 2c2 = 4

3 . Solving for c1 and c2 we find c1 = −49 and c2 = 4

9 . The uniquesolution is given by

y(t) =4

9(e2t − e−t − 3te−t)

Example 16.4Determine the function g(t) if

y′′ − 2y′ − 3y = g(t), yp(t) = 3e5t.

Solution.We have y′p = 15e5t and y′′p = 75e5t. Thus,

g(t) =y′′p − 2y′p − 3yp = 75e5t − 30e5t − 9e5t = 36e5t

Example 16.5The general solution of the nonhomogeneous differential equation y′′+αy′+ βy =g(t) is given below, where c1 and c2 are arbitrary constants. Determine the con-stants α and β and the function g(t).

y(t) = c1et + c2e

2t + 2e−2t.

Solution.From the given general solution we see that the roots of the characteristic equationare r1 = 1 and r2 = 2. Thus, the characteristic equation is (r − 1)(r − 2) =r2 − 3r + 2 = 0. The associated differential equation is y′′ − 3y′ + 2y = 0. Hence,α = −3 and β = 2. Now, letting yp(t) = 2e−t, we have

g(t) =y′′p − 3y′p + 2yp

=8e−2t + 12e−2t + 4e−2t = 24e−2t

121

Practice Problems

In Problems 16.1- 16.6, answer the following three question.(a) Verify that the given function, yp(t), is a particular solution of the differentialequations.(b) Determine the general solution,yh, of the homogeneous equation.(c) Find the general solution to the differential equation and impose the initialconditions to obtain the unique solution of the initial value problem.

Problem 16.1

y′′ − 2y′ − 3y = e2t, y(0) = 1, y′(0) = 0, yp(t) = −1

3e2t.

Problem 16.2

y′′ − y′ − 2y = 10, y(−1) = 0, y′(−1) = 1, yp(t) = −5.

Problem 16.3

y′′ + y′ = 2e−t, y(0) = 2, y′(0) = 2, yp(t) = −2te−t.

Problem 16.4

y′′ + 4y = 10et−π, y(π) = 2, y′(π) = 0, yp(t) = 2et−π.

Problem 16.5

y′′ − 2y′ + 2y = 5 sin t, y(π

2) = 1, y′(

π

2) = 0, yp(t) = 2 cos t+ sin t.

Problem 16.6

y′′ − 2y′ + y = t2 + 4 + 2 sin t, y(0) = 1, y′(0) = 3, yp(t) = t2 + 4t+ 10 + cos t.

The functions u1, u2, and u3 are solutions to the following differential equations

u′′ + p(t)u′ + q(t)u =2et + 1

u′′ + p(t)u′ + q(t)u =2e−t − t− 1

u′′ + p(t)u′ + q(t)u =3t.

In Problems 16.7 - 16.8, use the functions u1, u2 and u3 to construct a particularsolution of the differential equation.


Problem 16.7

u′′ + p(t)u′ + q(t)u = et + 2t+1

2.

Problem 16.8

u′′ + p(t)u′ + q(t)u =et + e−t

2.

In Problems 15.9 - 15.11, determine the function g(t).

Problem 16.9

y′′ − 2y′ = g(t), yp(t) = 3t+√t, t > 0.

Problem 16.10

y′′ + y′ = g(t), yp(t) = ln (1 + t), t > −1.

Problem 16.11

y′′ + 2y′ + y = g(t), yp(t) = t− 2.

In Problems 16.12 - 16.13, the general solution of the nonhomogeneous differentialequation y′′ + αy′ + βy = g(t) is given, where c1 and c2 are arbitrary constants.Determine the constants α and β and the function g(t).

Problem 16.12

y(t) = c1et + c2te

t + t2et.

Problem 16.13

y(t) = c1 sin 2t+ c2 cos 2t− 1 + sin t.

Problem 16.14Given that the function et

5 satisfies the differential equation y′′ + 4y = et, write ageneral solution of the differential equation y′′ + 4y = et.

Problem 16.15Find the general solution to the differential equation

y(4) + 9y′′ = 24 + 108t2

given a particular solution yp(t) = cos 3t+ sin 3t+ t4.

17 The Method of Variation ofParameters

In this section, we discuss a method for finding a particular solution to a nonho-mogeneous differential equation

y′′ + p(t)y′ + q(t)y = g(t), a < t < b. (17.1)

This method has no prior conditions to be satisfied by either p(t), q(t), or g(t).To use this method, we first find the general solution to the homogeneous equation

y(t) = c1y1(t) + c2y2(t).

Then we replace the parameters c1 and c2 by two functions u1(t) and u2(t) to bedetermined. From this the method got its name. Thus, obtaining

yp(t) = u1(t)y1(t) + u2(t)y2(t).

Observe that if u1 and u2 are constant functions then the above y is just thehomogeneous solution to the differential equation.In order to determine the two functions one has to impose two constraints. Findingthe derivative of yp we obtain

y′p = (y′1u1 + y′2u2) + (y1u′1 + y2u

′2).

Finding the second derivative to obtain

y′′p = y′′1u1 + y′1u′1 + y′′2u2 + y′2u

′2 + (y1u

′1 + y2u

′2)′.

Since it is up to us to choose u1 and u2 we decide to do that in such a way to makeour computation simple. One way to achieving that is to impose the condition

y1u′1 + y2u

′2 = 0. (17.2)

123

124 17 THE METHOD OF VARIATION OF PARAMETERS

Under such a constraint y′p and y′′p are simplified to

y′p = y′1u1 + y′2u2

andy′′p = y′′1u1 + y′1u

′1 + y′′2u2 + y′2u

′2.

In particular, y′′p does not involve u′′1 and u′′2.Inserting yp, y

′p, and y′′p into equation (17.1) to obtain

[y′′1u1 + y′1u′1 + y′′2u2 + y′2u

′2] + p(t)(y′1u1 + y′2u2) + q(t)(u1y1 + u2y2) = g(t).

Rearranging terms,

[y′′1 + p(t)y′1 + q(t)y1]u1 + [y′′2 + p(t)y′2 + q(t)y2]u2 + [u′1y′1 + u′2y

′2] = g(t).

Since y1 and y2 are solutions to the homogeneous equation, the previous equa-tion yields our second constraint

u′1y′1 + u′2y

′2 = g(t). (17.3)

Combining equation (17.2) and (17.3) we find the system of two equations in theunknowns u′1 and u′2

y1u′1 + y2u

′2 =0

u′1y′1 + u′2y

′2 =g(t).

Since {y1, y2} is a fundamental set, the expression W (t) = y1y′2 − y′1y2 is nonzero

so that one can find unique u′1 and u′2. Using the method of elimination, thesefunctions are given by

u′1(t) = −y2(t)g(t)W (t) and u′2(t) = y1(t)g(t)

W (t) .

Computing antiderivatives to obtain

u1(t) =∫−y2(t)g(t)

W (t) dt and u2(t) =∫ y1(t)g(t)

W (t) dt.

Example 17.1Find the general solution of

y′′ − y′ − 2y = 2e−t

using the method of variation of parameters.

125

Solution.The characteristic equation r2 − r − 2 = 0 has roots r1 = −1 and r2 = 2. Thus,y1(t) = e−t, y2(t) = e2t and W (t) = 3et. Hence,

u1(t) = −∫e2t · 2e−t

3etdt = −2

3t

and

u2(t) =

∫e−t · 2e−t

3etdt = −2

9e−3t.

The particular solution is

yp(t) = −2

3te−t − 2

9e−t.

The general solution is then given by

y(t) = c1e−t + c2e

2t − 2

3te−t − 2

9e−t

Example 17.2Find the general solution to (2t− 1)y′′ − 4ty′ + 4y = (2t− 1)2e−t if y1(t) = t andy2(t) = e2t form a fundamental set of solutions to the equation.

Solution.First we rewrite the equation in standard form

y′′ − 4t

2t− 1y′ +

4

2t− 1y = (2t− 1)e−t.

Since W (t) = (2t− 1)e2t we find

u1(t) = −∫e2t · (2t− 1)e−t

(2t− 1)e2tdt = e−t

and

u2(t) =

∫t · (2t− 1)e−t

(2t− 1)e2tdt = −1

3te−3t − 1

9e−3t.

Thus,

yp(t) = te−t − 1

3te−t − 1

9e−t =

2

3te−t − 1

9e−t.

The general solution is

y(t) = c1t+ c2e2t +

2

3te−t − 1

9e−t


Example 17.3Find the general solution to the differential equation y′′ + y′ = ln t, t > 0.

Solution.The characterisitc equation r2 + r = 0 has roots r1 = 0 and r2 = −1 so thaty1(t) = 1, y2(t) = e−t, and W (t) = −e−t. Hence,

u1(t) =−∫e−t ln t

−e−tdt =

∫ln tdt = t ln t− t

u2(t) =

∫ln t

−e−tdt = −

∫et ln tdt = −et ln t+

∫et

tdt

Thus,

yp(t) = t ln t− t− ln t+ e−t∫et

tdt

and

y(t) = c1 + c2e−t + t ln t− t− ln t+ e−t

∫et

tdt

Example 17.4Find the general solution of

y′′ + y =1

2 + sin t.

Solution.Since the characteristic equation r2 + 1 = 0 has roots r = ±i, the general solutionof the corresponding homogeneous equation y′′ + y = 0 is given by

yh(t) = c1 cos t+ c2 sin t

Since W (t) = 1 we find

u1(t) =−∫

sin t

2 + sin tdt = −t+

∫2

2 + sin tdt

u2(t) =

∫cos t

2 + sin tdt = ln (2 + sin t)

Hence, the particular solution is

yp(t) = sin t ln (2 + sin t) + cos t(

∫2

2 + sin tdt− t)

and the general solution is

y(t) = c1 cos t+ c2 sin t+ yp(t)

127

Practice Problems

Problem 17.1Solve y′′ + y = sec t by variation of parameters.

Problem 17.2Solve y′′ − y = et by variation of parameters.

Problem 17.3Solve the following 2nd order equation using the variation of parameter method:

y′′ + 4y = t2 + 8 cos 2t.

Problem 17.4Find a particular solution by the variation of parameters to the equation

y′′ + 2y′ + y = e−t ln t.

Problem 17.5Solve the following initial value problem by using variation of parameters:

y′′ + 2y′ − 3y = tet, y(0) = − 1

64, y′(0) =

59

64.

Problem 17.6(a) Verify that {e

√t, e−

√t} is a fundamental set for the equation

4ty′′ + 2y′ − y = 0

on the interval (0,∞). You may assume that the given functions are solutions tothe equation.(b) Use the method of variation of parameters to find one solution to the equation

4ty′′ + 2y′ − y = 4√te√t.

Problem 17.7Use the method of variation of parameters to find the general solution to theequation

y′′ + y = sin t.


Problem 17.8Consider the differential equation

t2y′′ + 3ty′ − 3y = 0, t > 0.

(a) Determine r so that y = tr is a solution.(b) Use (a) to find a fundamental set of solutions.(c) Use the method of variation of parameters for finding a particular solution to

t2y′′ + 3ty′ − 3y =1

t3, t > 0.

Problem 17.9Use the method of variation of parameters to find the general solution to thedifferential equation

y′′ + y = sin2 t.

18 The Laplace Transform:Basic Definitions and Results

Laplace transform is yet another operational tool for solving constant coefficientslinear differential equations. The process of solution consists of three main steps:• The given “hard” problem is transformed into a “simple” equation.• This simple equation is solved by purely algebraic manipulations.• The solution of the simple equation is transformed back to obtain the solutionof the given problem.In this way the Laplace transformation reduces the problem of solving a differentialequation to an algebraic problem. The third step is made easier by tables, whoserole is similar to that of integral tables in integration.The above procedure can be summarized by Figure 18.1.

Figure 18.1

In this section we introduce the concept of Laplace transform and discuss some ofits properties.The Laplace transform is defined in the following way. Let f(t) be defined fort ≥ 0. Then the Laplace transform of f, which is denoted by L[f(t)] or by F (s),is defined by the following equation

L[f(t)] = F (s) = limT→∞

∫ T

0f(t)e−stdt =

∫ ∞0

f(t)e−stdt.

The integral which defines a Laplace transform is an improper integral. An im-proper integral may converge or diverge, depending on the integrand. Whenthe improper integral is convergent then we say that the function f(t) possesses

129

13018 THE LAPLACE TRANSFORM: BASIC DEFINITIONS AND RESULTS

a Laplace transform. So what types of functions possess Laplace transforms, thatis, what type of functions guarantees a convergent improper integral?

Example 18.1Find the Laplace transform, if it exists, of each of the following functions.

(a) f(t) = eat (b) f(t) = 1 (c) f(t) = t (d) f(t) = et2.

Solution.(a) Using the definition of Laplace transform we see that

L[eat] =

∫ ∞0

e−(s−a)tdt = limT→∞

∫ T

0e−(s−a)tdt.

But ∫ T

0e−(s−a)tdt =

{T if s = a

1−e−(s−a)T

s−a if s 6= a.

For the improper integral to converge we need s > a. In this case,

L[eat] = F (s) =1

s− a, s > a.

(b) In a similar way to what was done in part (a), we find

L[1] =

∫ ∞0

e−stdt = limT→∞

∫ T

0e−stdt =

1

s, s > 0.

(c) We have

L[t] =

∫ ∞0

te−stdt =

[− te

−st

s− e−st

s2

]∞0

=1

s2, s > 0.

(d) Again using the definition of Laplace transform we find

L[et2] =

∫ ∞0

et2−stdt.

If s ≤ 0 then t2 − st ≥ 0 so that et2−st ≥ 1 and this implies that

∫∞0 et

2−stdt ≥∫∞0 dt. Since the integral on the right is divergent, by the comparison theorem of

improper integrals the integral on the left is also divergent. Now, if s > 0 then∫∞0 et(t−s)dt ≥

∫∞s dt. By the same reasoning the integral on the left is divergent.

This shows that the function f(t) = et2

does not possess a Laplace transform

131

The above example raises the question of what class or classes of functions possessa Laplace transform. Looking closely at Example 18.1(a), we notice that for s > athe integral

∫∞0 e−(s−a)tdt is convergent and a critical component for this conver-

gence is the type of the function f(t). To be more specific, if f(t) is a continuousfunction such that

|f(t)| ≤Meat, t ≥ C (18.1)

where M ≥ 0 and a and C are constants, then this condition yields∫ ∞0

f(t)e−stdt ≤∫ C

0f(t)e−stdt+M

∫ ∞C

e−(s−a)tdt.

Since f(t) is continuous in 0 ≤ t ≤ C, by letting A = max{|f(t)| : 0 ≤ t ≤ C} wehave ∫ C

0f(t)e−stdt ≤ A

∫ C

0e−stdt = A

(1

s− e−sC

s

)<∞.

Also, by Example 18.1(a), the integral∫∞C e−(s−a)tdt is convergent for s > a.

By the comparison theorem of improper integrals the integral on the left is alsoconvergent. That is, f(t) possesses a Laplace transform.We call a function that satisfies condition (18.1) a function with an exponentialorder a. Graphically, this means that the graph of f(t) is contained in the regionbounded by the graphs of y = Meat and y = −Meat for t ≥ C. Note also thatthis type of functions controls the negative exponential in the transform integralso that to keep the integral from blowing up. If C = 0 then we say that thefunction is exponentially bounded. Note that f has exponential order a iflimt→∞ e

−atf(t) = 0.

Example 18.2(a) Show that f(t) = tn where n is a positive integer, has an exponential order.(b) Show that any bounded function f(t) for t ≥ 0 is exponentially bounded.(c) Show that f(t) = et

2is not of exponential order.

Solution.(a) Let a > 0. Using L’Hopital’s rule n times we can see that limt→∞ e

−attn = 0.Thus, there is a positive constant C such that e−attn ≤ 1 for all t ≥ C. That is,tn ≤ eat for all t ≥ C.(b) Since f(t) is bounded for t ≥ 0, there is a positive constant M such that|f(t)| ≤ M for all t ≥ 0. But this is the same as (18.1) with a = 0 and C = 0.Thus, f(t) is exponentially bounded.(c) Suppose not. That is, let M and C be non-negative constants such that


et2 ≤Meat for some constant a and for all t ≥ C. Hence, et

2−at ≤M for all t ≥ C.Letting t→∞ we find M ≥ ∞ which is impossible since M <∞. It follows thatf(t) = et

2is not of exponential order

Another question that comes to mind is whether it is possible to relax the conditionof continuity on the function f(t). Let’s look at the following situation.

Example 18.3Show that the square wave function whose graph is given in Figure 18.2 possessesa Laplace transform.

Figure 18.2

Note that the function is periodic of period 2.

Solution.Since f(t)e−st ≤ e−st,

∫∞0 f(t)e−stdt ≤

∫∞0 e−stdt. But the integral on the right is

convergent for s > 0 so that the integral on the left is convergent as well. That is,L[f(t)] exists for s > 0The function of the above example belongs to a class of functions that we definenext.A function is called piecewise continuous on an interval if it consists of a finitenumber of continuous pieces with possibly either removable or jump discontinu-ities (but no inifinite discontinuities). Below is a sketch of a piecewise continuousfunction.

Figure 18.3

An example of a function that is not piecewise continuous is the function f(t) = 1t−1

on the interval [0,∞) since at t = 1 the continuity is infinite.

133

Example 18.4Show that the following function is piecewise continuous and exponentially bounded.

f(t) =

{et sin (t−1) if t ≥ 1

12 if 0 ≤ t < 1.

Solution.Since et sin (t−1) = (esin (t−1))t ≤ (e1)t = et for all t ≥ 0, f(t) is piecewise continuousand exponentially bounded

The following theorem guarantees the existence of the Laplace transform for allfunctions that are piecewise continuous and have exponential order.

Theorem 18.1 (Existence)Suppose that f(t) is piecewise continuous on t ≥ 0 and has an exponential orderwith |f(t)| ≤Meat for t ≥ C. Then the Laplace transform

F (s) =

∫ ∞0

f(t)e−stdt

exists as long as s > a. Note that the two conditions above are sufficient, but notnecessary, for F (s) to exist.

Example 18.5Consider the floor function f(t) = btc, where for any integer n we have btc = n forall n ≤ t < n+ 1.(a) Is f(t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuous on0 ≤ t <∞, or neither?(b) Are there fixed numbers a and M such that |f(t)| ≤Meat for 0 ≤ t <∞?

Solution.(a) The floor function is a piecewise continuous function on 0 ≤ t <∞.(b) Since btc ≤ t < et for 0 ≤ t <∞ we find M = 1 and a = 1

Example 18.6Consider the function f(t) = t2e−t.(a) Is f(t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuous on0 ≤ t <∞, or neither?(b) Are there fixed numbers a and M such that |f(t)| ≤Meat for 0 ≤ t <∞?


Solution.(a) Since t2 and e−t are continuous everywhere, f(t) = t2e−t is continuous on0 ≤ t <∞.(b) By L’Hopital’s rule one has

limt→∞

t2

et= 0.

Since f(0) = 0, f(t) is bounded. Since f ′(t) = (2t − t2)e−t, f(t) has a maximumwhen t = 2. The value of this maximum is f(2) = 4e−2. Hence, M = 4e−2 anda = 0

In what follows, we will denote the class of all piecewise continuous functionswith exponential order by PE . The next theorem shows that any linear combi-nation of functions in PE is also in PE . The same is true for the product of twofunctions in PE .

Theorem 18.2Suppose that f(t) and g(t) are two elements of PE with

|f(t)| ≤M1ea1t, t ≥ C1 and |g(t)| ≤M2e

a1t, t ≥ C2.

(i) For any constants α and β the function αf(t) + βg(t) is also a member of PE .Moreover

L[αf(t) + βg(t)] = αL[f(t)] + βL[g(t)].

(ii) The function h(t) = f(t)g(t) is an element of PE .

Example 18.7Use the linearity property of Laplace transform to find L[5e−7t + t + 2e2t]. Findthe domain of F (s).

Solution.We have L[e−7t] = 1

s+7 , s > −7, L[t] = 1s2, s > 0, and L[e2t] = 1

s−2 , s > 2. Hence,

L[5e−7t + t+ 2e2t] = 5L[e−7t] + L[t] + 2L[e2t] =5

s+ 7+

1

s2+

2

s− 2, s > 2

We next discuss the problem of how to determine the function f(t) if F (s) is given.That is, how do we invert the transform. The following result on uniqueness pro-vides a possible answer. This result establishes a one-to-one correspondence be-tween the set PE and its Laplace transforms. Alternatively, the following theoremasserts that the Laplace transform of a member in PE is unique.

135

Theorem 18.3Let f(t) and g(t) be two elements in PE with Laplace transforms F (s) and G(s)such that F (s) = G(s) for some s > a. Then f(t) = g(t) for all t ≥ 0 where bothfunctions are continuous.

With the above theorem, we can now officially define the inverse Laplace transformas follows: For a piecewise continuous function f with exponential order whoseLaplace transform is F, we call f the inverse Laplace transform of F and writef = L−1[F (s)]. Symbolically

f(t) = L−1[F (s)]⇐⇒ F (s) = L[f(t)].

Example 18.8

Find L−1(

1s−1

), s > 1.

Solution.From Example 18.1(a), we have that L[eat] = 1

s−a , s > a. In particular, for a = 1

we find that L[et] = 1s−1 , s > 1. Hence, L−1

(1s−1

)= et, t ≥ 0

The above theorem states that if f(t) is continuous and has a Laplace transformF (s), then there is no other function that has the same Laplace transform. Tofind L−1[F (s)], we can inspect tables of Laplace transforms of known functions tofind a particular f(t) that yields the given F (s).When the function f(t) is not continuous, the uniqueness of the inverse Laplacetransform is not assured. The following example addresses the uniqueness issue.

Example 18.9Consider the two functions f(t) = h(t)h(3− t) and g(t) = h(t)− h(t− 3) where

h(t) =

{1 if t ≥ 00 otherwise.

(a) Are the two functions identical?(b) Show that L[f(t)] = L[g(t).

Solution.(a) We have

f(t) =

{1, 0 ≤ t ≤ 30, t > 3

and

g(t) =

{1, 0 ≤ t < 30, t ≥ 3.


Since f(3) = 1 and g(3) = 0, f and g are not identical.(b) We have

L[f(t)] = L[g(t)] =

∫ 3

0e−stdt =

1− e−3s

s, s > 0.

Thus, both functions f(t) and g(t) have the same Laplace transform even thoughthey are not identical. However, they are equal on the interval(s) where they areboth continuous

The inverse Laplace transform possesses a linear property as indicated in the fol-lowing result.

Theorem 18.4Given two Laplace transforms F (s) and G(s) then

L−1[aF (s) + bG(s)] = aL−1[F (s)] + bL−1[G(s)]

for any constants a and b.

Example 18.10

Find L−1(

2s+2 + 2

s−2

).

Solution.We have

L−1(

2

s+ 2+

2

s− 2

)= 2L−1

(1

s+ 2

)+ 2L−1

(1

s− 2

)= 2(e−2t + e2t), t ≥ 0

137

Practice Problems

Problem 18.1Determine whether the integral

∫∞0

11+t2

dt converges. If the integral converges,give its value.


∫∞0

t1+t2

dt converges. If the integral converges,give its value.


∫∞0 e−t cos (e−t)dt converges. If the integral con-

verges, give its value.

Problem 18.4Using the definition, find L[e3t], if it exists. If the Laplace transform exists thenfind the domain of F (s).

Problem 18.5Using the definition, find L[t− 5], if it exists. If the Laplace transform exists thenfind the domain of F (s).

Problem 18.6Using the definition, find L[e(t−1)

2], if it exists. If the Laplace transform exists

then find the domain of F (s).

Problem 18.7Using the definition, find L[(t − 2)2], if it exists. If the Laplace transform existsthen find the domain of F (s).

Problem 18.8Using the definition, find L[f(t)], if it exists. If the Laplace transform exists thenfind the domain of F (s).

f(t) =

{0, 0 ≤ t < 1

t− 1, t ≥ 1.

Problem 18.9Using the definition, find L[f(t)], if it exists. If the Laplace transform exists thenfind the domain of F (s).

f(t) =

0, 0 ≤ t < 1

t− 1, 1 ≤ t < 20, t ≥ 2.


Problem 18.10Let n be a positive integer. Using integration by parts establish the reductionformula ∫

tne−stdt = − tne−st

s+n

s

∫tn−1e−stdt, s > 0.

Problem 18.11For s > 0 and n a positive integer evaluate the limits

(a) limt→0 tne−st (b) limt→∞ t

ne−st.

Problem 18.12(a) Use the previous two problems to derive the reduction formula for the Laplacetransform of f(t) = tn,

L[tn] =n

sL[tn−1], s > 0.

(b) Calculate L[tk], for k = 1, 2, 3, 4, 5.(c) Formulate a conjecture as to the Laplace transform of f(t), tn with n a positiveinteger.

From a table of integrals,∫eαu sinβudu =eαu

α sinβu− β cosβu

α2 + β2∫eαu cosβudu =eαu

α cosβu+ β sinβu

α2 + β2

Problem 18.13Use the above integrals to find the Laplace transform of f(t) = cosωt, if it exists.If the Laplace transform exists, give the domain of F (s).

Problem 18.14Use the above integrals to find the Laplace transform of f(t) = sinωt, if it exists.If the Laplace transform exists, give the domain of F (s).

Problem 18.15Use the above integrals to find the Laplace transform of f(t) = cosω(t− 2), if itexists. If the Laplace transform exists, give the domain of F (s).

Problem 18.16Use the above integrals to find the Laplace transform of f(t) = e3t sin t, if it exists.If the Laplace transform exists, give the domain of F (s).

139

Problem 18.17Consider the function f(t) = tan t.(a) Is f(t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuous on0 ≤ t <∞, or neither?(b) Are there fixed numbers a and M such that |f(t)| ≤Meat for 0 ≤ t <∞?

Problem 18.18Consider the function f(t) = et

2

e2t+1.

(a) Is f(t) continuous on 0 ≤ t < ∞, discontinuous but piecewise continuous on0 ≤ t <∞, or neither?(b) Are there fixed numbers a and M such that |f(t)| ≤Meat for 0 ≤ t <∞?

Problem 18.19Find L−1

(3s−2

).


(− 2s2

+ 1s+1

).


19 Further Studies of LaplaceTransform

Properties of the Laplace transform enable us to find Laplace transforms withouthaving to compute them directly from the definition. In this section, we establishproperties of Laplace transform that will be useful for solving ODEs.

Laplace Transform of the Heaviside Step FunctionThe Heaviside step function is a piecewise continuous function defined by

h(t) =

{1, t ≥ 00, t < 0.

Figure 19.1 displays the graph of h(t).

Figure 19.1

Taking the Laplace transform of h(t) we find

L[h(t)] =

∫ ∞0

h(t)e−stdt =

∫ ∞0

e−stdt =

[−e−st

s

]∞0

=1

s, s > 0.

141

142 19 FURTHER STUDIES OF LAPLACE TRANSFORM

A Heaviside function at α ≥ 0 is the shifted function h(t − α) (α units to theright). For this function, the Laplace transform is

L[h(t− α)] =

∫ ∞0

h(t− α)e−stdt =

∫ ∞α

e−stdt =

[−e−st

s

]∞α

=e−sα

s, s > 0.

Laplace Tranform of eat

The Laplace transform for the function f(t) = eat is

L[eat] =

∫ ∞0

e−(s−a)tdt =

[−e−(s−a)t

s− a

]∞0

=1

s− a, s > a.

Laplace Tranforms of sin at and cos atUsing integration by parts twice we find

L[sin at] =

∫ ∞0

e−st sin atdt

=

[−e−st sin at

s− ae−st cos at

s2

]∞0

− a2

s2

∫ ∞0

e−st sin atdt

=− a

s2− a2

s2L[sin at](

s2 + a2

s2

)L[sin at] =

a

s2

L[sin at] =a

s2 + a2, s > 0.

A similar argument shows that

L[cos at] =s

s2 + a2, s > 0.

Laplace Transforms of cosh at and sinh atUsing the linear property of L we can write

L[cosh at] =1

2

(L[eat] + L[e−at]

)=

1

2

(1

s− a+

1

s+ a

), s > |a|

=s

s2 − a2, s > |a|.

A similar argument shows that

L[sinh at] =a

s2 − a2, s > |a|.

143

Laplace Transform of a PolynomialLet n be a positive integer. Using integration by parts we can write∫ ∞

0tne−stdt = −

[tne−st

s

]∞0

+n

s

∫ ∞0

tn−1e−stdt.

By repeated use of L’Hopital’s rule we find limt→∞ tne−st = limt→∞

n!snest = 0 for

s > 0. Thus,

L[tn] =n

sL[tn−1], s > 0.

Using induction on n = 0, 1, 2, · · · one can easily establish that

L[tn] =n!

sn+1, s > 0.

Using the above result together with the linearity property of L one can find theLaplace transform of any polynomial.The next two results are referred to as the first and second shift theorems. Aswith the linearity property, the shift theorems increase the number of functionsfor which we can easily find Laplace transforms.

Theorem 19.1 (First Shifting Theorem)If f(t) is a piecewise continuous function for t ≥ 0 and has exponential order with|f(t)| ≤Meat, t ≥ C, then for any real number α we have

L[eαtf(t)] = F (s− α), s > a+ α

where L[f(t)] = F (s).

Theorem 19.2 (Second Shifting Theorem)If f(t) is a piecewise continuous function for t ≥ 0 and has exponential order with|f(t)| ≤Meat, t ≥ C, then for any real number α ≥ 0 we have

L[f(t− α)h(t− α)] = e−αsF (s), s > a

where L[f(t)] = F (s) and h(t) is the Heaviside step function.

Example 19.1Find

(a) L[e2tt2] (b) L[e3t cos 2t] (c) L−1[ e−2s

s2].


Solution.(a) By Theorem 19.1, we have L[e2tt2] = F (s−2) where L[t2] = 2!

s3= F (s), s > 0.

Thus, L[e2tt2] = 2(s−2)3 , s > 2.

(b) As in part (a), we have L[e3t cos 2t] = F (s − 3) where L[cos 2t] = F (s). ButL[cos 2t] = s

s2+4, s > 0. Thus,

L[e3t cos 2t] =s− 3

(s− 3)2 + 4, s > 3.

(c) Since L[t] = 1s2, by Theorem 19.2, we have

e−2s

s2= L[(t− 2)h(t− 2)].

Therefore,

L−1[e−2s

s2

]= (t− 2)h(t− 2) =

{0, 0 ≤ t < 2

t− 2, t ≥ 2

The following result relates the Laplace transform of derivatives and integrals tothe Laplace transform of the function itself.

Theorem 19.3Suppose that f(t) is continuous for t ≥ 0 and f ′(t) is piecewise continuous andexponentially bounded. Then(a) f(t) is exponentially bounded.(b) L[f ′(t)] = sL[f(t)]− f(0) = sF (s)− f(0), s > max{a, 0}+ 1.(c) L[f ′′(t)] = s2L[f(t)]−sf(0)−f ′(0) = s2F (s)−sf(0)−f(0), s > max{a, 0}+1.

(d) L[∫ t

0 f(u)du]

= L[f(t)]s = F (s)

s , s > max{a, 0}+ 1.

The results in (b) and (c) of the previous theorem can be extended to higher orderderivatives.

Theorem 19.4Let f(t), f ′(t), · · · , f (n−1)(t) be continuous and f (n)(t) be piecewise continuous andexponentially bounded. Then

L[f (n)(t)] = snL[f(t)]−sn−1f(0)−sn−2f ′(0)−· · ·−f (n−1)(0), s > max{a, 0}+1.

We next illustrate the use of the previous theorem in solving initial value problems.

145

Example 19.2Use Laplace transform technique to solve the initial value problem

y′′ − 4y = e3t, y(0) = 0, y′(0) = 0.

Solution.Taking the Laplace transform of the DE and using linearity we find

L[y′′]− 4L[y] = L[e3t].

But L[y′′] = s2L[y]− sy(0)− y′(0) = s2L[y]. Letting L[y] = Y (s) we obtain

s2Y (s)− 4Y (s) =1

s− 3.

Solving for Y (s) we find

Y (s) =1

(s− 3)(s− 2)(s+ 2).

Using partial fraction decomposition

1

(s− 3)(s− 2)(s+ 2)=

A

s− 3+

B

s+ 2+

C

s− 2

we find A = 15 , B = 1

20 , and C = −14 . Thus,

y(t) =L−1[

1

(s− 3)(s− 2)(s+ 2)

]=

1

5L−1

[1

s− 3

]+

1

20L−1

[1

s+ 2

]− 1

4L−1

[1

s− 2

]=

1

5e3t +

1

20e−2t − 1

4e2t, t ≥ 0

Example 19.3Determine the constants α, β, y0, and y′0 so that Y (s) = s

(s+1)2is the Laplace

transform of the solution to the initial value problem

y′′ + αy′ + βy = 0, y(0) = y0, y′(0) = y′0.

Solution.Taking the Laplace transform of both sides we find

s2Y (s)− sy0 − y′0 + αsY (s)− αy0 + βY (s) = 0.


Y (s) =sy0 + (y′0 + αy0)

s2 + αs+ β=

s

s2 + 2s+ 1.

By identification we find α = 2, β = 1, y0 = 1, and y′0 = −2



y′′ − 4y′ + 9y = t, y(0) = 0, y′(0) = 1.

Solution.We apply Theorem 19.4 that gives the Laplace transform of a derivative. By thelinearity property of the Laplace transform we can write

L[y′′]− 4L[y′] + 9L[y] = L[t].

Now since

L[y′′] =s2L[y]− sy(0)− y′(0) = s2Y (s)− 1

L[y′] =sY (s)− y(0) = sY (s)

L[t] =1

s2

where L[y] = Y (s), we obtain

s2Y (s)− 1− 4sY (s) + 9Y (s) =1

s2.

Rearranging gives

(s2 − 4s+ 9)Y (s) =s2 + 1

s2.

Thus,

Y (s) =s2 + 1

s2(s2 − 4s+ 9)

and

y(t) = L−1[

s2 + 1

s2(s2 − 4s+ 9)

]In the next section we will discuss a method for finding the inverse Laplace trans-form of the above expression.

147

f(t) F(s)

h(t) =

{1, t ≥ 00, t < 0

1s , s > 0

tn, n = 1, 2, · · · n!sn+1 , s > 0

eαt 1s−α , s > α

sin (ωt) ωs2+ω2 , s > 0

cos (ωt) ss2+ω2 , s > 0

sinh (ωt) ωs2−ω2 , s > |ω|

cosh (ωt) ss2−ω2 , s > |ω|

eαtf(t), with |f(t)| ≤Meat F (s− α), s > α+ a

eαth(t) 1s−α , s > α

eαttn, n = 1, 2, · · · n!(s−α)n+1 , s > α

eαt sin (ωt) ω(s−α)2+ω2 , s > α

eαt cos (ωt) s−α(s−α)2+ω2 , s > α

f(t− α)h(t− α), α ≥ 0 e−αsF (s), s > awith |f(t)| ≤Meat


f(t) F(s) (continued)

h(t− α), α ≥ 0 e−αs

s , s > 0

tf(t) -F ′(s)

t2ω sinωt s

(s2+ω2)2, s > 0

12ω3 [sinωt− ωt cosωt] 1

(s2+ω2)2, s > 0

f ′(t), with f(t) continuous sF (s)− f(0)and |f ′(t)| ≤Meat s > max{a, 0}+ 1

f ′′(t), with f ′(t) continuous s2F (s)− sf(0)− f ′(0)and |f ′′(t)| ≤Meat s > max{a, 0}+ 1

f (n)(t), with f (n−1)(t) continuous snF (s)− sn−1f(0)− · · ·and |f (n)(t)| ≤Meat -sf (n−2)(0)− f (n−1)(0)

s > max{a, 0}+ 1∫ t0 f(u)du, with |f(t)| ≤Meat F (s)

s , s > max{a, 0}+ 1

Table L

149

Practice Problems

Problem 19.1Use Table L to find L[2et + 5].

Problem 19.2Use Table L to find L[e3t−3h(t− 1)].

Problem 19.3Use Table L to find L[sin2 ωt].

Problem 19.4Use Table L to find L[sin 3t cos 3t].

Problem 19.5Use Table L to find L[e2t cos 3t].

Problem 19.6Use Table L to find L[e4t(t2 + 3t+ 5)].

Problem 19.7Use Table L to find L−1[ 10

s2+25+ 4

s−3 ].

Problem 19.8Use Table L to find L−1[ 5

(s−3)4 ].

Problem 19.9Use Table L to find L−1[ e−2s

s−9 ].

Problem 19.10Use Table L to find L−1[ e

−3s(2s+7)s2+16

].

Problem 19.11Graph the function f(t) = h(t− 1) +h(t− 3) for t ≥ 0, where h(t) is the Heavisidestep function, and use Table L to find L[f(t)].

Problem 19.12Graph the function f(t) = t[h(t−1)−h(t−3)] for t ≥ 0, where h(t) is the Heavisidestep function, and use Table L to find L[f(t)].

Problem 19.13Graph the function f(t) = 3[h(t−1)−h(t−4)] for t ≥ 0, where h(t) is the Heavisidestep function, and use Table L to find L[f(t)].


Problem 19.14Graph the function f(t) = |2 − t|[h(t − 1) − h(t − 3)] for t ≥ 0, where h(t) is theHeaviside step function, and use Table L to find L[f(t)].

Problem 19.15Graph the function f(t) = h(2 − t) for t ≥ 0, where h(t) is the Heaviside stepfunction, and use Table L to find L[f(t)].

Problem 19.16Graph the function f(t) = h(t− 1) +h(4− t) for t ≥ 0, where h(t) is the Heavisidestep function, and use Table L to find L[f(t)].

Problem 19.17The graph of f(t) is given below. Represent f(t) as a combination of Heavisidestep functions, and use Table L to calculate the Laplace transform of f(t).

Problem 19.18The graph of f(t) is given below. Represent f(t) as a combination of Heavisidestep functions, and use Table L to calculate the Laplace transform of f(t).

Problem 19.19Use Laplace transform technique to solve the initial value problem

y′ + 4y = g(t), y(0) = 2

where

g(t) =

0, 0 ≤ t < 112, 1 ≤ t < 30, t ≥ 3.

20 The Laplace Transform andthe Method of Partial Fractions

In the last example of the previous section we encountered the equation

y(t) = L−1[

s2 + 1

s2(s2 − 4s+ 9)

].

We would like to find an explicit expression for y(t). This can be done using themethod of partial fractions which is the topic of this section. According to this

method, finding L−1(N(s)D(s)

), where N(s) and D(s) are polynomials, require de-

composing the rational function into a sum of simpler expressions whose inverseLaplace transform can be recognized from a table of Laplace transform pairs.The method of partial fractions decomposition consists of writing a rational func-tion, i.e., function of the form

R(s) =N(s)

D(s)

where N(s) and D(s) are polynomials, as a sum of simpler fractions called partialfractions. This can be done in the following way:

Step 1. Use long division to find two polynomials r(s) and q(s) such that

N(s)

D(s)= q(s) +

r(s)

D(s).

Note that if the degree of N(s) is smaller than that of D(s) then q(s) = 0 andr(s) = N(s).

Step 2. Write D(s) as a product of factors of the form (as+ b)n or (as2 + bs+ c)n

where as2 + bs+ c is irreducible, i.e. as2 + bs+ c = 0 has no real zeros.

151

15220 THE LAPLACE TRANSFORMANDTHEMETHODOF PARTIAL FRACTIONS

Step 3. Decompose r(s)D(s) into a sum of partial fractions in the following way:

(1) For each factor of the form (as+ b)k write

A1

as+ b+

A2

(as+ b)2+ · · ·+ Ak

(as+ b)k,

where the numbers A1, A2, · · · , Ak are to be determined.(2) For each factor of the form (as2 + bs+ c)k write

B1s+ C1

as2 + bs+ c+

B2s+ C2

(as2 + bs+ c)2+ · · ·+ Bks+ Ck

(as2 + bs+ c)k,

where the numbers B1, B2, · · · , Bk and C1, C2, · · · , Ck are to be determined.

Step 4. Multiply both sides of r(s)D(s) by D(s) and simplify. This leads to an expres-

sion of the form

r(s) = a polynomial whose coefficients are combinations of Ai,Bi, and Ci.

Finally, we find the constants, Ai, Bi, and Ci by equating the coefficients of likepowers of x on both sides of the last equation.

Example 20.1Decompose into partial fractions R(s) = s3+s2+2

s2−1 .

Solution.Step 1. s3+s2+2

s2−1 = s+ 1 + s+3s2−1 .

Step 2. s2 − 1 = (s− 1)(s+ 1).Step 3. s+3

(s+1)(s−1) = As+1 + B

s−1 .

Step 4. Multiply both sides of the last equation by (s− 1)(s+ 1) to obtain

s+ 3 = A(s− 1) +B(s+ 1).

Expand the right hand side, collect terms with the same power of s, and identifycoefficients of the polynomials obtained on both sides:

s+ 3 = (A+B)s+ (B −A).

Hence, A+B = 1 and B−A = 3. Adding these two equations gives B = 2. Thus,A = −1 and so

s3 + s2 + 2

s2 − 1= s+ 1− 1

s+ 1+

2

s− 1

Now, after decomposing the rational function into a sum of partial fractions all weneed to do is to find the Laplace transform of expressions of the form A

(s−α)n orBs+C

(as2+bs+c)n.

153

Example 20.2

Find L−1[

1s(s−3)

].

Solution.We write

1

s(s− 3)=A

s+

B

s− 3.

Multiply both sides by s(s− 3) and simplify to obtain

1 = A(s− 3) +Bs

or1 = (A+B)s− 3A.

Now equating the coefficients of like powers of s to obtain −3A = 1 and A+B = 0.Solving for A and B we find A = −1

3 and B = 13 . Thus,

L−1[

1

s(s− 3)

]=− 1

3L−1

[1

s

]+

1

3L−1

[1

s− 3

]=− 1

3h(t) +

1

3e3t, t ≥ 0

where h(t) is the Heaviside unit step function

Example 20.3

Find L−1[

3s+6s2+3s

].

Solution.We factor the denominator and split the integrand into partial fractions:

3s+ 6

s(s+ 3)=A

s+

B

s+ 3.

Multiplying both sides by s(s+ 3) to obtain

3s+ 6 = A(s+ 3) +Bs= (A+B)s+ 3A

Equating the coefficients of like powers of x to obtain 3A = 6 and A + B = 3.Thus, A = 2 and B = 1. Finally,

L−1[

3s+ 6

s2 + 3s

]= 2L−1

[1

s

]+ L−1

[1

s+ 3

]= 2h(t) + e−3t, t ≥ 0.


Example 20.4

Find L−1[

s2+1s(s+1)2

].

Solution.We factor the denominator and split the rational function into partial fractions:

s2 + 1

s(s+ 1)2=A

s+

B

s+ 1+

C

(s+ 1)2.

Multiplying both sides by s(s+ 1)2 and simplifying to obtain

s2 + 1 =A(s+ 1)2 +Bs(s+ 1) + Cs

=(A+B)s2 + (2A+B + C)s+A.

Equating coefficients of like powers of s we find A = 1, 2A + B + C = 0 andA+B = 1. Thus, B = 0 and C = −2. Now finding the inverse Laplace transformto obtain

L−1[s2 + 1

s(s+ 1)2

]= L−1

[1

s

]− 2L−1

[1

(s+ 1)2

]= h(t)− 2te−t, t ≥ 0.

Example 20.5Use Laplace transform to solve the initial value problem

y′′ + 3y′ + 2y = e−t, y(0) = y′(0) = 0.

Solution.By the linearity property of the Laplace transform we can write

L[y′′] + 3L[y′] + 2L[y] = L[e−t].

Now since

L[y′′] =s2L[y]− sy(0)− y′(0) = s2Y (s)

L[y′] =sY (s)− y(0) = sY (s)

L[e−t] =1

s+ 1

where L[y] = Y (s), we obtain

s2Y (s) + 3sY (s) + 2Y (s) =1

s+ 1.

155

Rearranging gives

(s2 + 3s+ 2)Y (s) =1

s+ 1.

Thus,

Y (s) =1

(s+ 1)(s2 + 3s+ 2).

and

y(t) = L−1[

1

(s+ 1)(s2 + 3s+ 2)

].

Using the method of partial fractions we can write

1

(s+ 1)(s2 + 3s+ 2)=

1

s+ 2− 1

s+ 1+

1

(s+ 1)2.

Thus,

y(t) = L−1[

1

s+ 2

]−L−1

[1

s+ 1

]+ L−1

[1

(s+ 1)2

]= e−2t − e−t + te−t, t ≥ 0


Practice Problems

In Problems 20.1 - 20.4, give the form of the partial fraction expansion for F (s).You need not evaluate the constants in the expansion. However, if the denomi-nator has an irreducible quadratic expression then use the completing the squareprocess to write it as the sum/difference of two squares.

Problem 20.1

F (s) =s3 + 3s+ 1

(s− 1)3(s− 2)2.

Problem 20.2

F (s) =s2 + 5s− 3

(s2 + 16)(s− 2).

Problem 20.3

F (s) =s3 − 1

(s2 + 1)2(s+ 4)2.

Problem 20.4

F (s) =s4 + 5s2 + 2s− 9

(s2 + 8s+ 17)(s− 2)2.


[1

(s+1)3

].


[2s−3

s2−3s+2

].


[4s2+s+1s3+s

].


[s2+6s+8s4+8s2+16

].

Problem 20.9Use Laplace transform to solve the initial value problem

y′ + 2y = 26 sin 3t, y(0) = 3.

157


y′ + 2y = 4t, y(0) = 3.


y′′ + 3y′ + 2y = 6e−t, y(0) = 1, y′(0) = 2.


y′′ + 4y = cos 2t, y(0) = 1, y′(0) = 1.


y′′ − 2y′ + y = e2t, y(0) = 0, y′(0) = 0.


y′′ + 9y = g(t), y(0) = 1, y′(0) = 3

where

g(t) =

{6, 0 ≤ t < π0, π ≤ t <∞

Problem 20.15Determine the constants α, β, y0, and y′0 so that Y (s) = 2s−1

s2+s+2is the Laplace

transform of the solution to the initial value problem

y′′ + αy′ + βy = 0, y(0) = y0, y′(0) = y′0.


21 Laplace Transforms ofPeriodic Functions

In many applications, the nonhomogeneous term in a linear differential equation isa periodic function. In this section, we derive a formula for the Laplace transformof such periodic functions.Recall that a function f(t) is said to be T−periodic if T is the smallest positiveinterger such that f(t+T ) = f(t) whenever t and t+T are in the domain of f(t).For example, the sine and cosine functions are 2π−periodic whereas the tangentand cotangent functions are π−periodic.If f(t) is T−periodic for t ≥ 0 then we define the function

fT (t) =

{f(t), 0 ≤ t ≤ T

0, t > T.

The Laplace transform of this function is then

L[fT (t)] =

∫ ∞0

fT (t)e−stdt =

∫ T

0f(t)e−stdt.

The Laplace transform of a T−periodic function is given next.

Theorem 21.1If f(t) is a T−periodic, piecewise continuous function for t ≥ 0 then

L[f(t)] =L[fT (t)]

1− e−sT, s > 0.

Example 21.1Determine the Laplace transform of the function

f(t) =

1, 0 ≤ t ≤ T

2f(t+ T ) = f(t), t ≥ 0.

0, T2 < t < T.

159

160 21 LAPLACE TRANSFORMS OF PERIODIC FUNCTIONS

Solution.The graph of f(t) is shown in Figure 21.1.

Figure 21.1

We have

L[fT (t)] =

∫ T2

0e−stdt

=1− e−

sT2

s.

By Theorem 21.1, we have

L[f(t)] =1−e−

sT2

s

1− e−sT=

1

s(1 + e−sT2 ), s > 0

Example 21.2Find the Laplace transform of the sawtooth curve shown in Figure 21.2

Figure 21.2

161

Solution.The given function is periodic of period b. For the first period the function isdefined by

fb(t) =a

bt[h(t)− h(t− b)].

So we have

L[fb(t)] =

∫ b

0e−st

a

btdt

=a

b

[− tse−st − e−st

s2

]b0

=a

b

(1

s2− bse−bs + e−bs

s2

).

Hence,

L[f(t)] =L[fb(t)]

1− e−bs=a

b

[1− e−bs − bse−bs

s2(1− e−bs)

]Example 21.3

Find L−1[

1s2− e−s

s(1−e−s)

].

Solution.Note first that

1

s2− e−s

s(1− e−s)=

1− e−s − se−s

s2(1− e−s).

According to the previous example with a = 1 and b = 1 we find that L−1[

1s2− e−s

s(1−e−s)

]is the sawtooth function shown in Figure 21.2

Linear Time Invariant Systems and the Transfer FunctionThe Laplace transform is a powerful technique for analyzing linear time-invariantsystems such as electrical circuits, harmonic oscillators, optical devices, and me-chanical systems, to name just a few. A mathematical model described by a lineardifferential equation with constant coefficients of the form

any(n) + an−1y

(n−1) + · · ·+ a1y′ + a0y = bmu

(m) + bm−1u(m−1) + · · ·+ b1u

′ + b0u

is called a linear time invariant system. The function y(t) denotes the systemoutput and the function u(t) denotes the system input. The system is called time-invariant because the parameters of the system are not changing over time and aninput now will give the same result as the same input later.


Applying the Laplace transform on the linear differential equation with null initialconditions we obtain

ansnY (s)+an−1s

n−1Y (s)+· · ·+a0Y (s) = bmsmU(s)+bm−1s

m−1U(s)+· · ·+b0U(s).

The function

Φ(s) =Y (s)

U(s)=bms

m + bm−1sm−1 + · · ·+ b1s+ b0

ansn + an−1sn−1 + · · ·+ a1s+ a0

is called the system transfer function. That is, the transfer function of a lineartime-invariant system is the ratio of the Laplace transform of its output to theLaplace transform of its input.

Example 21.4Consider the mathematical model described by the initial value problem

my′′ + γy′ + ky = f(t), y(0) = 0, y′(0) = 0.

The coefficients m, γ, and k describe the properties of some physical system, andf(t) is the input to the system. The solution y is the output at time t. Find thesystem transfer function.

Solution.By taking the Laplace transform and using the initial conditions we obtain

(ms2 + γs+ k)Y (s) = F (s).

Thus,

Φ(s) =Y (s)

F (s)=

1

ms2 + γs+ k(21.1)

Parameter IdentificationOne of the most useful applications of system transfer functions is for the parameteridentification of a system.

Example 21.5Consider a spring-mass system governed by

my′′ + γy′ + ky = f(t), y(0) = 0, y′(0) = 0. (21.2)

Suppose we apply a unit step force f(t) = h(t) to the mass, initially at equilibrium,and you observe the system respond as

y(t) = −1

2e−t cos t− 1

2e−t sin t+

1

2.

What are the physical parameters m, γ, and k?

163

Solution.Start with the model (21.2) with f(t) = h(t) and take the Laplace transformof both sides, then solve to find Y (s) = 1

s(ms2+γs+k). Since f(t) = h(t) we find

F (s) = 1s . Hence

Φ(s) =Y (s)

F (s)=

1

ms2 + γs+ k.

On the other hand, for the input f(t) = h(t) the corresponding observed output is

y(t) = −1

2e−t cos t− 1

2e−t sin t+

1

2.

Hence,

Y (s) =L[−1

2e−t cos t− 1

2e−t sin t+

1

2]

=− 1

2

s+ 1

(s+ 1)2 + 1− 1

2

1

(s+ 1)2 + 1+

1

2s

=1

s(s2 + 2s+ 2)

Thus,

Φ(s) =Y (s)

F (s)=

1

s2 + 2s+ 2.

By comparison, we conclude that m = 1, γ = 2, and k = 2


Practice Problems

Problem 21.1Find the Laplace transform of the periodic function whose graph is shown.




165

Problem 21.5State the period of the function f(t) and find its Laplace transform where

f(t) =

sin t, 0 ≤ t < π

f(t+ 2π) = f(t), t ≥ 0.0, π ≤ t < 2π.

Problem 21.6State the period of the function f(t) = 1 − e−t, 0 ≤ t < 2, f(t + 2) = f(t), andfind its Laplace transform.

Problem 21.7Using Example 21.3 find

L−1[s2 − ss3

+e−s

s(1− e−s)

].


ay′′ + by′ + cy = f(t), y(0) = y′(0) = 0, t > 0.

Suppose that the transfer function of this system is given by Φ(s) = 12s2+5s+2

.(a) What are the constants a, b, and c?(b) If f(t) = e−t, determine F (s), Y (s), and y(t).


ay′′ + by′ + cy = f(t), y(0) = y′(0) = 0, t > 0.

Suppose that an input f(t) = t, when applied to the above system produces theoutput y(t) = 2(e−t − 1) + t(e−t + 1), t ≥ 0.(a) What is the system transfer function?(b) What will be the output if the Heaviside unit step function f(t) = h(t) isapplied to the system?



y′′ + y′ + y = f(t), y(0) = y′(0) = 0,

where

f(t) =

1, 0 ≤ t ≤ 1

f(t+ 2) = f(t)−1, 1 < t < 2.

(a) Determine the system transfer function Φ(s).(b) Determine Y (s).


y′′′ − 4y = et + t, y(0) = y′(0) = y′′(0) = 0.

(a) Determine the system transfer function Φ(s).(b) Determine Y (s).


y′′ + by′ + cy = h(t), y(0) = y0, y′(0) = y′0, t > 0.

Suppose that L[y(t)] = Y (s) = s2+2s+1s3+3s2+2s

. Determine the constants b, c, y0, andy′0.

22 Convolution Integrals

We start this section with the following problem.

Example 22.1A spring-mass system with a forcing function f(t) is modeled by the followinginitial-value problem

mx′′ + kx = f(t), x(0) = x0, x′(0) = x′0.

Find solution to this initial value problem using the Laplace transform method.

Solution.Apply Laplace transform to both sides of the equation to obtain

ms2X(s)−msx0 −mx′0 + kX(s) = F (s).

Solving the above algebraic equation for X(s) we find

X(s) =F (s)

ms2 + k+

msx0ms2 + k

+mx′0

ms2 + k

(22.1)

=1

m

F (s)

s2 + km

+sx0

s2 + km

+x′0

s2 + km

Apply the inverse Laplace transform to obtain

x(t) =L−1[X(s)]

=1

mL−1

{F (s)

s2 + km

}+ x0L−1

{s

s2 + km

}+ x′0L−1

{1

s2 + km

}

=1

mL−1

{F (s) · 1

s2 + km

}+ x0 cos

(√k

m

)t+ x′0

√m

ksin

(√k

m

)t

167

168 22 CONVOLUTION INTEGRALS

Finding L−1{F (s) · 1

s2+ km

},i.e., the inverse Laplace transform of a product, re-

quires the use of the concept of convolution, a topic we discuss in this sectionConvolution integrals are useful when finding the inverse Laplace transform ofproducts H(s) = F (s)G(s). They are defined as follows: The convolution of twoscalar piecewise continuous functions f(t) and g(t) defined for t ≥ 0 is the integral

(f ∗ g)(t) =

∫ t

0f(t− s)g(s)ds.

Example 22.2Find f ∗ g where f(t) = e−t and g(t) = sin t.

Solution.Using integration by parts twice we arrive at

(f ∗ g)(t) =

∫ t

0e−(t−s) sin sds

=1

2

[e−(t−s)(sin s− cos s)

]t0

=e−t

2+

1

2(sin t− cos t)

Next, we state several properties of convolution product, which resemble those ofordinary product.

Theorem 22.1Let f(t), g(t), and k(t) be three piecewise continuous scalar functions defined fort ≥ 0 and c1 and c2 are arbitrary constants. Then(i) f ∗ g = g ∗ f (Commutative Law)(ii) (f ∗ g) ∗ k = f ∗ (g ∗ k) (Associative Law)(iii) f ∗ (c1g + c2k) = c1f ∗ g + c2f ∗ k (Distributive Law)

Example 22.3Express the solution to the initial value problem y′+αy = g(t), y(0) = y0 in termsof a convolution integral.

Solution.Solving this initial value problem by the method of integrating factor we find

y(t) = e−αty0 +

∫ t

0e−α(t−s)g(s)ds = e−αty0 + e−αt ∗ g(t)

The following theorem, known as the Convolution Theorem, provides a way forfinding the Laplace transform of a convolution integral and also finding the inverseLaplace transform of a product.

169

Theorem 22.2If f(t) and g(t) are piecewise continuous for t ≥ 0, and of exponential order a then

L[(f ∗ g)(t)] = L[f(t)]L[g(t)] = F (s)G(s).

Thus, (f ∗ g)(t) = L−1[F (s)G(s)].

Example 22.4Use the convolution theorem to find the inverse Laplace transform of

H(s) =1

(s2 + a2)2.

Solution.Note that

H(s) =

(1

s2 + a2

)(1

s2 + a2

).

So, in this case we have, F (s) = G(s) = 1s2+a2

so that f(t) = g(t) = 1a sin (at).

Thus,

(f ∗ g)(t) =1

a2

∫ t

0sin (at− as) sin (as)ds =

1

2a3(sin (at)− at cos (at))

Convolution integrals are useful in solving initial value problems with forcing func-tions.


4y′′ + y = g(t), y(0) = 3, y′(0) = −7

Solution.Take the Laplace transform of all the terms and plug in the initial conditions toobtain

4(s2Y (s)− 3s+ 7) + Y (s) = G(s)

or(4s2 + 1)Y (s)− 12s+ 28 = G(s).


Y (s) =12s− 28

4(s2 + 1

4

) +G(s)

4(s2 + 1

4

)=

3s

s2 +(12

)2 − 1412

s2 +(12

)2 +1

2G(s)

12

s2 +(12

)2


Hence,

y(t) = 3 cos

(t

2

)− 14 sin

(t

2

)+

1

2

∫ t

0sin(s

2

)g(t− s)ds.

So, once we decide on a g(t) all we need to do is to evaluate the integral and we’llhave the solution

Example 22.6Use Laplace transform to solve for y(t) :

y(t)−∫ t

0e(t−λ)y(λ)dλ = t.

Solution.Note that the given equation reduces to et ∗ y(t) = y(t)− t. Taking Laplace trans-

form of both sides we find Y (s)s−1 = Y (s)− 1

s2. Solving for Y (s) we find Y (s) = s−1

s2(s−2) .

Using partial fractions decomposition we can write

s− 1

s2(s− 2)=−1

4

s+

12

s2+

14

(s− 2).

Hence,

y(t) = −1

4+t

2+

1

4e2t, t ≥ 0

Example 22.7Use Laplace transform to solve for y(t) :

t ∗ y(t) = t2(1− e−t).

Solution.Taking Laplace transform of both sides we find Y (s)

s2= 2

s3− 2

(s+1)3. This implies

Y (s) = 2s −

2s2

(s+1)3. Using partial fractions decomposition we can write

s2

(s+ 1)3=

1

s+ 1− 2

(s+ 1)2+

1

(s+ 1)3.

Hence,

y(t) = 2− 2(e−t − 2te−t +t2

2e−t) = 2

(1− (1− 2t+

t2

2)e−t

), t ≥ 0

171

Example 22.8Solve the following initial value problem.

y′ − y =

∫ t

0(t− λ)eλdλ, y(0) = −1.

Solution.Note that y′ − y = t ∗ et. Taking Lalplace transform of both sides we find sY −(−1) − Y = 1

s2· 1s−1 . This implies that Y (s) = − 1

s−1 + 1s2(s−1)2 . Using partial

fractions decomposition we can write

1

s2(s− 1)2=

2

s+

1

s2− 2

s− 1+

1

(s− 1)2.

Thus,

Y (s) = − 1

s− 1+

2

s+

1

s2− 2

s− 1+

1

(s− 1)2=

2

s+

1

s2− 3

s− 1+

1

(s− 1)2.

Finally,y(t) = 2 + t− 3et + tet, t ≥ 0


Practice Problems

Problem 22.1Consider the functions f(t) = g(t) = h(t), t ≥ 0 where h(t) is the Heaviside unitstep function. Compute f ∗ g in two different ways.(a) By directly evaluating the integral.(b) By computing L−1[F (s)G(s)] where F (s) = L[f(t)] and G(s) = L[g(t)].

Problem 22.2Consider the functions f(t) = et and g(t) = e−2t, t ≥ 0. Compute f ∗ g in twodifferent ways.(a) By directly evaluating the integral.(b) By computing L−1[F (s)G(s)] where F (s) = L[f(t)] and G(s) = L[g(t)].

Problem 22.3Consider the functions f(t) = sin t and g(t) = cos t, t ≥ 0. Compute f ∗ g in twodifferent ways.(a) By directly evaluating the integral.(b) By computing L−1[F (s)G(s)] where F (s) = L[f(t)] and G(s) = L[g(t)].

Problem 22.4Compute and graph f ∗ g where f(t) = h(t) and g(t) = t[h(t)− h(t− 2)].

Problem 22.5Compute and graph f ∗g where f(t) = h(t)−h(t−1) and g(t) = h(t−1)−2h(t−2)].

Problem 22.6Compute t ∗ t ∗ t.

Problem 22.7Compute h(t) ∗ e−t ∗ e−2t.

Problem 22.8Compute t ∗ e−t ∗ et.

Problem 22.9

Suppose it is known that

n functions︷︸︸︷h(t) ∗ h(t) ∗ · · · ∗ h(t) = Ct8. Determine the constants C

and the positive integer n.

Problem 22.10Use Laplace transform to solve for y(t) :∫ t

0sin (t− λ)y(λ)dλ = t2.

173

174 CHEAT SHEETS

Cheat Sheets

Answer Key

Section 1

1.1 (a) 16 m/sec (b) 80 m.

1.2(a) 1.5 meters (b) 7 m/sec (c) −9.8 m/sec2.

1.3 The acceleration due to gravity is y′′(t) = −50 ft/sec2. The initial velocityis v(0) = y′(0) = 72 ft/sec. The initial height is y(0) = 40 ft.

1.4 The height of the object at any time t is given by y(t) = −16t2 + 400. Theobject hits the ground when y(t) = 0. Solving the equation −16t2 + 400 = 0 wefind t = 5 sec. The velocity of the object at the time of impact is v(5) = y′(5) =−32(5) = −160 ft/sec.

1.5 400 ft.

1.6 v(0) = 160 ft/sec and maximum height is 400 ft.

1.7 The impact time is t =√

2y0g . v(

√2y0g ) = −

√2gy0.

1.8 ε = π4 .

1.9 (a) Order 2 (b) Order 1 (c) Order 3.

1.10 (a) First order (b) , (c), and (d) second order.

1.11 u is the dependent variable whereas x and y are the independent variables.

1.12 (a) ODE (b) PDE (c) ODE.

181

182 ANSWER KEY

1.13 y(t) = t3

3 + C1t2 + C2t+ C3.

1.14 The solution is y(t) = 50e3t. The domain is the set of all real numbers,i.e., (−∞,∞).

1.15 λ = ±3.

1.16 m = −2 or m = −1.

1.17 Substituting y(t) = y′(t) = y′′(t) = et into the equation we find

y′′ −(

2 +2

t

)y′ +

(1 +

2

t

)y =et −

(2 +

2

t

)et +

(1 +

2

t

)et

=et − 2et − 2

tet + et +

2

tet = 0.

1.18 Finding the first and the second derivatives of y we obtain

y′(t) = −C1ω sinωt+ C2ω cosωt

andy′′(t) = −C1ω

2 cosωt− C2ω2 sinωt

Substituting this into the equation to obtain

d2y

dt2+ ω2y =− C1ω

2 cosωt− C2ω2 sinωt+ ω2(C1 cosωt+ C2 sinωt)

=0.

1.19 y(0) = 2 and k = 4.

1.20 n = 1 or n = 2.

1.21 (b) y(t) = e2t + e−2t (c) y(t) = 2e−2t.

1.22 y(0) = 1, m = −2, y(t) = −t+ 2.

1.23 t0 = 0, m = 1, y(t) = t2

2 − 1.

1.24 Finding the second derivative and substituting into the equation we find

y′′ + 4y = 4e2t + 4e2t = 8e2t 6= 0.

183

Thus, y(t) = e2t is not a solution to the given equation.

1.25 (a) Substituting y and y′ into the equation we find

−3Ce−3t + 2 + 3[Ce−3t + 2t+ 1] = −3Ce−3t + 3Ce−3t + 6t+ 5 = 6t+ 5.

(b) C = 2.

184 ANSWER KEY

Section 2

2.1 y0 = 3 and p(t) = −2t.

2.2 (a) (−∞,∞) (b) (−∞,∞).

2.3 (a) 3 < t <∞ (b) −2 < t < 2 (c) −∞ < t < −2.

2.4 (a) r = 3 and C = −1 (b) (−∞, 0) (c) (−∞,∞).

2.5 The interval of existence is 0 < t < ∞ if y(1) = 1 and −∞ < t < 0 ify(−1) = 1.

2.6 (a) −5 < t < 0 (b) 0 < t < 4 (c) −∞ < t < −5 (d) 4 < t <∞.

2.7 (a) 1 < t < 3 (b) No such solution since t > 0 (c) 3 < t <∞.

2.8 (a) If p(t) and g(t) are continuous functions in the open interval I = (a, b) andt0 a point inside I then the IVP

y′ + p(t)y = g(t), y(t0) = y0

has a unique solution y(t) defined on I.(b) Since p(t) = t2 and g(t) = et

3, the IVP has a unique solution.

2.9 (a) Non-linear (b) Linear and homogeneous (c) Linear and non-homogeneous (d)Linear and homogeneous.

2.10 0 < t < 5.

2.11 α = ln 2 and y0 = 8.

2.12(a) (iv) (b) (i) (c) (iii) (d) (ii).

2.13 A non-linear first order ordinary differential equation.

2.14 We have y′ = t+yt = 1 + y

t . Thus,

y′ − 1

ty = 1

This is a first-order linear non-homogeneous equation.

185

Section 3

3.1 y(t) = e1−t2

3.2 y(t) = 2e1−et.

3.3 (a)z′ + p(t)z = 0, z(t0) = y(t0)− α (b) y(t) = (y0 − α)e−

∫ tt0p(s)ds

+ α.

3.4 y(t) = e−t2

+ 3.

3.5 α = 2 and y0 = 0.25.

3.6 (a) n = 3 and y0 = 1 (b) y(−1) =4√e−1.

3.7 p(t) = 1t .

3.8 p(t) = −3t .

3.9 y(t) = t3.

3.10 y(t) = Cet2.

3.11 f(t) = 4e2t.

3.12 y(t) = Ce2t + C ′.

3.13 y = 12(1− e−t2).

3.14 y(t) = 3t−19 + e−2t + Ce−3t.

3.15 y(t) = 3 sin t+ 3 cos tt + C

t .

3.16 y(t) = 113(3 sin (3t) + 2 cos (3t)) + Ce−2t.

3.17 α = −2.

3.18 p(t) = 2 and g(t) = 2t+ 3.

3.19 y0 = −1 and g(t) = 2et + cos t+ sin t.

186 ANSWER KEY

3.20 limt→∞ y(t) = 1.

3.21

y(t) =

{2t+ 1 if 0 ≤ t ≤ 1t+ 2

t if 1 < t ≤ 2.

3.22

y(t) =

{1 + 2ecos t−1 if 0 ≤ t ≤ π

−1 + 2(1e + e

)ecos t if π < t ≤ 2π.

3.23

y(t) =

{1 + 2e−t if 0 ≤ t ≤ 1

(2 + e)e−t if t > 1.

3.24

y(t) =

3e−(t2−t) if 0 ≤ t ≤ 1

3 if 1 < t ≤ 3t if 3 < t ≤ 4.

The graph of y(t) is shown below

It follows that limt→∞ y(t) = ∞. The function y(t) is not differentiable at t = 1and t = 3 on the domain t > 0.

3.25 y(t) = (−tet + et)−1.

187

Section 4

4.1 (a) y′ = 13(1− 2t cos y) = f(t, y),

(b) D = {(t, y) : −∞ < t <∞, −∞ < y <∞}(c) R = {(t, y) : −∞ < t <∞, −∞ < y <∞}.

4.2 (a) y′ = 13t(1− 2 cos y) = f(t, y)

(b) D = {(t, y) : −∞ < t < 0, 0 < t <∞, −∞ < y <∞}(c) R = {(t, y) : 0 < t <∞, −∞ < y <∞}.

4.3 (a) y′ = − 2t1+y3

= f(t, y)

(b) D = {(t, y) : −∞ < t <∞, −∞ < y < −1, − 1 < y <∞}(c) R = {(t, y) : −∞ < t <∞, − 1 < y <∞}.

4.4 (a) y′ = t2−e−yy2−9 = f(t, y)

(b) D = {(t, y) : −∞ < t <∞, −∞ < y < −3, − 3 < y < 3, 3 < y <∞}(c) D = {(t, y) : −∞ < t <∞, − 3 < y < 3}.

4.5 (a) y′ = 2+tan tcos y = f(t, y)

(b) D = {(t, y) : t 6= (2n+ 1)π2 , y 6= (2m+ 1)π2 , where n and m are integers}(c) R = {(t, y) : −π

2 < t < π2 , −

π2 < y < π

2 }.

4.6 y′ = 1t(t−4)(y+1)(y−2) , y(2) = 0.

4.7 Since f(t, y) = y√t

and fy(t, y) = 1√t, both functions are continuous in the

regionD = {(t, y) : 0 < t <∞, −∞ < y <∞}

Since (0, 2) is not in D, Theorem 4.1 can not be applied in this case.

4.8 The function f(t, y) = t−yt+y is continuous everywhere except along the line

t + y = 0. (a) and (b): Since both (0, 0) and (1,−1) lie on this line, we cannotconclude existence from Theorem 4.1.(c) The point (−1,−1) is not on that line so we can find a small rectangle aroundthis point where Theorem 4.1 guarantees the existence of a solution. Furthermore,since fy(t, y) = − 2t

(t+y)2is continuous at (−1,−1), the solution is unique.

4.9 The equation is of the form y′ = f(t, y) = yt + 2. The function f is continuous

outside the line t = 0. The initial value point is (0,1), so there is no rectanglecontaining it in which f is continuous, and the conditions of Theorem 4.1 are notsatisfied.

188 ANSWER KEY

4.10 The equation is of the form y′ = f(t, y) = y sin y+ t. the function f is contin-uous in the whole plane, and so is its partial derivative fy(t, y) = sin y+ y cos y. Inparticular, any rectangle around the initial value point will satisfy the conditionsof Theorem 4.1.

189

Section 5

5.1 y3 = 32et2 + C.

5.2 y(t) = Cet2

2−2t.

5.3 y(t) = Ct2 + 4.

5.4 y(t) = 2Ce4t

1+Ce4tand y(t) ≡ 2.

5.5 y(t) =√

5− 4 cos (2t).

5.6 y(t) = −√

(−2 cos t+ 4).

5.7 y(t) =√−2t+ 3− 1.

5.8 y(t) = 2√−4t2+1

.

5.9 y(t) = tan(t− π

2

), − π

2 < t < π2 .

5.10 y(t) = 3−e−t2

3+e−t2.

5.11 y2 + cos y + cos t+ t2

2 = 2.

5.12 y0 = 12 , α = 1

2 , n = 3.

5.13 3y2y′ + (cos y)y′ + 2t = 0, y(2) = 0.

5.14 y0 = 18 .

5.15 y sin y − t3 + 1 = 0.

5.16 y(t) = Ce2t

(t+1)2.

5.17 tan y = t2 + 1

4 sin 2t+ 1.

5.18 y(t) = − ln(2− et), t < ln 2.

5.19 (a) e−y = − t3

3 −1t + C (b) e−y + t3

3 + 1t = e−1 + 4

3 .

190 ANSWER KEY

Section 6

6.1 ∂f∂t = ye−ty, ∂f

∂y = ln y + 1 + te−ty.

6.2 ∂f∂t = 1

t + 2ty−5 ,

∂f∂y = 1

y −t2+1

(y−5)2 .

6.3 dfdt = 24 sin2 t− 4 sin t− 12.

6.4 y(t) = −√

4t− t3 + 6.

6.5 y(t) = tan(t3 + t+ π

4

).

6.6 The given differential equation is not exact.

6.7 et+y + t3 + y2 = 1.

6.8 y sin (t+ y) + t2

2 + ty + y2

2 = 0.

6.9 m = 4, n = 3, α = 43 .

6.10 N(t, y) =∫

2y sin tdt = −2y cos t+ h(y).

6.11 y0 = ±2, M(t, y) = 3t2 + et, N(t, y) = t3 + 2y.

6.12 y(0) = −2, a = 1, b = 2.

6.13 The differential equation is exact.

6.14 t2 + 3t+ y2 − 2y = C.

6.15 B = 1, 12e

2ty + t2

2 = C.

6.16 ty2 − y2ey + 2yey − 2ey = C.

6.17 (a) ∂M∂y (t, y) = p(t) 6= 0 and ∂N

∂t (t, y) = 0 (b) y(t) = e−∫p(t)dt

∫e∫p(t)dtg(t)dt+

Ce−∫p(t)dt.

6.18 y(t) = t ln t+ 7t.

6.19 ty3 − t2y = C.

191

6.20 (a) 2t3y2 + 4ty − 8t+ 4y = 6 (b) 6t− 4ty + 8t2 + 5y2 + 2y = 30.

192 ANSWER KEY

Section 7

7.1 y2 = −t2 + Ct3.

7.2 y4 = 2e−t2

+ Ce−2t2.

7.3 y(t) = 1t(4.5−t) .

7.4 y(t) = 2tt2+1

.

7.5 y(t) = (1 + e−t)−1.

7.6 y(t) = (t+ 13 + Ce3t)−

13 and y(t) ≡ 0.

7.7 y(t) = (et2 ∫ t

0 e−s2ds+ Cet

2)−1 + t.

7.8 z′ − z = 1.

7.9 (a) z′ + 2εz = 2σ (b) y(t) = (σε + Ce−2εt)−12 and y(t) ≡ 0.

7.10 (a) z′ = f(z)−zt (b) 2 arctan

(yt

)= ln t2

(1 +

(yt

)2).

7.11 y(t) = (−3tet + Cet)−13 and y(t) ≡ 0.

7.12 y(t) = (2t+ Ct2)−12 and y(t) ≡ 0.

7.13 y(t) = (−t2 ln t+ t2 + Ct)−1 and y(t) ≡ 0.

7.14 y(t) = (−13 + Ce−3t)−1 + 2.

7.15 y(t) = (− t4 + Ct−3)−1 + 2

t .

193

Section 8

8.1 (a) We have

(b) We have

8.2 (A) y′ = sin t (B) y′ = y(2− y) (c) y′ = −t (D) y′ = 1− y

194 ANSWER KEY

8.3 (a) No (b) No (c) Yes (d) Yes

8.4 (a) y(t) ≡ 1, y(t) ≡ 2 (b) y(t) ≡ 1, y ≡ 2 (c) y ≡ 1, y ≡ 2, y ≡ 3

8.5 One answer is the differential equation: y′ = −(y − 1)2

8.6 y′ = C, where C > 0.

8.7 One answer is the DE y′ = sin (2πy)

8.8 An answer is y′ = y(2− y)

8.9 (a) y = 1 stable, y = −1 unstable (b) y = −1 is neither. This is a semi-stable equilibrium.

8 .10 The equilibrium solution at y = 1 is asymptotically stable where as theequilibrium solution at y = 0 is unstable.

8.11

8.12 The direction field is given below.

195

The equilibrium solution y = 1 is unstable; y = 0 is semi-stable; y = −1 is asymp-totically stable.

8.13 The direction field is given below.

The equilibrium solution y = 4 is unstable while y = 0 is asymptotically stable.If y(0) = 5 then limt→∞ y(t) = ∞. If y(0) = 4 then limt→∞ y(t) = 4. If y(0) = 3then limt→∞ y(t) = 0

8.14 y(t) = 0 and y(t) = π are two equilibrium solutions. According to thedirection field shown below we conclude that

limt→∞

y(t) = π

196 ANSWER KEY

197

Section 9

9.1 (a) Integrating y′ = 2t − 1 we find y(t) = t2 − t + C. Imposing the initialcondition y(1) = 0, we obtain y(t) = t2 − t.(b) yn+1 = yn + h

2 [(2tn − 1) + (2tn+1 − 1)]. (c) yn+1 = yn + h[2(tn + h2 )− 1]. (d)

n tn yn0 1.0000 01 1.1000 0.11002 1.2000 0.24003 1.3000 0.3900

(e)

n tn yn0 1.0000 01 1.1000 0.11002 1.2000 0.24003 1.3000 0.3900

(f)

n tn yn0 1.0000 01 1.1000 0.11002 1.2000 0.24003 1.3000 0.3900

9.2 (a) Integrating y′ = −y and imposing the initial condition we find y(t) = e−t.(b) Heun’s method takes the form yn+1 = (1− h+ 0.5h2)yn.(c) The modified Euler’s method takes the form yn+1 = (1− h+ 0.5h2)yn.(d)

n tn yn0 0.0000 1.00001 0.1000 0.90502 0.2000 0.81903 0.3000 0.7412

(e)

198 ANSWER KEY

n tn yn0 0.0000 1.00001 0.1000 0.90502 0.2000 0.81903 0.3000 0.7412

(f)

n tn yn0 0.0000 1.00001 0.1000 0.90482 0.2000 0.81873 0.3000 0.7408

9.3 (a) Solving the separable equation y2y′ + t = 0 we find y3 = −1.5t2 + C.

Imposing the initial condition y(0) = 1, we obtain y(t) = (1− 1.5t2)13 .

(b) yn+1 = yn + h2 [−tny−2n − tn+1(yn − htny−2n )−2].

(c) yn+1 = yn − h(tn + 0.5h)(yn − 0.5htny−2n )−2.

(d)

n tn yn0 0.0000 1.00001 0.1000 0.99502 0.2000 0.97963 0.3000 0.9529

(e)

n tn yn0 0.0000 1.00001 0.1000 0.99502 0.2000 0.97973 0.3000 0.9531

(f)

n tn yn0 0.0000 1.00001 0.1000 0.99502 0.2000 0.97963 0.3000 0.9528

9.4 (a) y(t) = tan(t− π

4

).

(b)

199

n tn Euler Heun Modified Euler0 0.0000 −1.0000 −1.0000 −1.00001 0.0500 −0.9000 −0.9047 −0.90492 0.1000 −0.8095 −0.8177 −0.81793 0.1500 −0.7267 −0.7375 −0.73784 0.2000 −0.6503 −0.6630 −0.6634

The errors at t = 1 are, respectively, | tan(1− π

4

)+0.6503| = 0.8683, | tan

(1− π

4

)+

0.6630| = 0.8809, and | tan(1− π

4

)+ 0.6634| = 0.8814.

9.5 (a) Solving the equation y′ + 2y = 4 by the method of integrating factor,we find y(t) = 2 +Ce−2t. Imposing the initial condition, we obtain C = 1. Hence,y(t) = 2 + e−2t.(b) Since f(t, y) = 4− 2y, we have:Euler’s method: yn+1 = yn + h(4− 2yn).Heuns’ Method: yn+1 = yn + 0.5h[8− 4yn − 2h(4− 2yn)].Modified Euler’s Method: yn+1 = yn + h[4− 2yn − h(4− 2yn)].

n tn Euler Heun Modified Euler0 0.0000 3.0000 3.0000 3.00001 0.0500 2.9000 2.9050 2.90502 0.1000 2.8100 2.8190 2.81903 0.1500 2.7290 2.7412 2.74124 0.2000 2.6561 2.6708 2.6708

The errors at t = 1 are, respectively, |3− 2.6561| = 0.3439, |3− 2.6708| = 0.3292,and |3− 2.6708| = 0.32929.6 Since tn = 2 + nh, h = 0.02, n = 0, 1, · · · , 49, it follows that t0 = 2 andN − 1 = 49. Thus, N = 50 and T = tN − t0 = 2 + Nh − 2 = 50(0.02) = 1. Fromthe form of the iteration, it must be Euler’s method. Therefore f(t, y) = y+ t2y3.

9.7 Since tn = nh, h = 0.01, n = 0, 1, · · · , 199, it follows that t0 = 0 andN − 1 = 199. Thus, N = 200 and T = tN − t0 = Nh = 200(0.01) = 2. Fromthe form of the iteration, it must be the modified Euler’s method. Therefore,f(t, y) = t sin2 y.

9.8 Since tn = 1 + nh, h = 0.05, n = 0, 1, · · · , 99, it follows that t0 = 1 andN − 1 = 99. Thus, N = 100 and T = tN − t0 = 1 +Nh− 1 = 100(0.05) = 5. Fromthe form of the iteration, it must be Heun’s method. Therefore f(t, y) = ty2 + 1.

200 ANSWER KEY

Section 10

10.1 π2 < t < 3π

2 .10.2 −∞ < t < −1.10.3 0 < t < 3.10.4 −∞ < t < 0.10.5 0 < t < π.10.6 e−

π2 < t < e

π2 .

10.7 y′′ + y′ + y = 0, y(0) = 0, y′(0) = 1.10.8 y′′ + 1

t−3y′ + y = 1, y(4) = 0, y′(4) = −1.

10.9 0 < t < 3.10.10 Let Y (t) = y(−t). Then Y ′′+ t2Y = y′′+ t2y = 0, Y (0) = Y ′(0) = 0 so thatY (t) is a solution to the given initial-value problem. By the existence and unique-ness theorem we must have Y (t) = y(t), i.e., y(−t) = y(t) for all real numbers t.

10.11 If p(t), q(t), and g(t) are continuous functions over an interval a < t < bcontaining t0 then the initial value problem

y′′ + p(t)y′ + q(t)y = g(t), y(t0) = y0, y′(t0) = y′0

has a unique solution in the interval a < t < b.

10.12 We have y′(t) = −2t sin (t2) and y′′(t) = −2 sin (t2)− 4t2 cos (t2). Hence,

ty′′ − y′ + 4t3y =t(−2 sin (t2)− 4t2 cos (t2))− (−2 sin (t2)) + 4t3 cos (t2)

=− 2t sin (t2)− 4t3 cos (t2) + 2 sin (t2) + 4t3 cos (t2) = 0.

10.13 −1 < t < 4.

10.14 0 < t < π.

10.15

L(αy1 + βy2) =(αy1 + βy2)′′ + p(t)(αy1 + βy2)

′ + q(t)(αy1 + βy2)

=αy′′1 + βy′′2 + p(t)αy′1 + p(t)βy′2 + q(t)αy1 + q(t)βy2

=α(y′′1 + p(t)y′1 + q(t)y2) + β(y′′2 + p(t)y′2 + q(t)y2)

=αL(y1) + βL(y2).

201

Section 11

11.1 (a) Yes (b) {y1, y2} is a fundamental set (c) y = e−2t.

11.2 (a) y1 is not a solution.

11.3 (a) Yes (b) {y1, y2} is a fundamental set (c) y(t) = 2e2t − 4te2t.

11.4 (a) Yes (b) {y1, y2} is a fundamental set (c) y(t) = 18 ln t− 9 ln (3t), t > 0.

11.5 (a) Yes (b) {y1, y2} is a fundamental set (c) y(t) = 12(t−1 − t3), t > 0.

11.6 (a) Yes (b) {y1, y2} is a fundamental set (c) y(t) = −t+ 5.

11.7 (a) y1 is not a solution.

11.8 (a) t2y′′− ty′+ y = t2t−1− t(3 + ln (3t)) + 2t+ t ln (3t) = 0 (b) c1 = 2 + ln 3and c2 = 1.

11.9 α = 0 and β = −9.

11.10 (a)p(t) = −tt−1 and q(t) = 1

t−1 (b) (−∞, 1) ∪ (1,∞) (c) W (y1(t), y2(t)) =et(t− 1) The Wronskian is nonzero for all t 6= 1 (d) Yes.

11.11 W (y1(2), y2(2)) = 4e−1.5.

11.12 α = 1 and β = −6.

11.13 p(t) = 2t.

11.14 Suppose for example that both functions have a same maximum at t0.Then y′1(t0) = y′2(t0) = 0. But

W (y1(t0), y2(t0)) = y1(t0)y′2(t0)− y′1(t0)y2(t0) = 0

Thus, {y1, y2} is not a fundamental set.

11.15 W (y1(t), y2(t)) = W (y1(t0), y2(t0))(1t −

1t0

).

11.16 y2(t) = tet.

202 ANSWER KEY

11.17 {y1, y2} is a fundamental set.

11.18 Suppose that t3 and t4 are both solutions. Since W (t) = t6 we findW (1) = 1 and so {y1, y2} is a fundamental set. By Abel’s Theorem, W (t) 6= 0 forall −∞ < t < ∞. But W (0) = 0, a contradiction. Hence, t3 and t4 can’t be bothsolutions for the differential equation for −∞ < t <∞.

11.19 p(t) = − 2tt2+1

.

203

Section 12

12.1 y(t) = et + 2e−2t, limt→−∞ y(t) =∞, limt→∞ y(t) =∞.

12.2 y(t) = −2et + e3t, limt→−∞ y(t) =∞, limt→∞ y(t) =∞.

12.3 y(t) = e−t, limt→−∞ y(t) =∞, limt→∞ y(t) = 0.

12.4 y(t) = 2e−2t − e−3t, limt→−∞ y(t) = −∞, limt→∞ y(t) = 0.

12.5 y(t) = 0, limt→−∞ y(t) = 0, limt→∞ y(t) = 0.

12.6 y(t) = 3, limt→−∞ y(t) = 3, limt→∞ y(t) = 3.

12.7 y(t) = −√

2e(−2−√2)t+

√2e(−2+

√2)t, limt→−∞ y(t) = −∞, limt→∞ y(t) = 0.

12.8 y(t) = −2e−√

22t, limt→−∞ y(t) = −∞, limt→∞ y(t) = 0.

12.9 y(t) =∫u(t)dt = C1e

2t + C2e3t + C3.

12.10 (a) x(t) = c1+c2e− kmt (b) x(t) =

(x0 + m

k v0)−m

k v0e− kmt (c) limt→∞ x(t) =

x0 + mk v0.

12.11 y′′ − y′ − 2y = 0.

12.12 y(t) = c1e−t + c2e

4t.

12.13 y(t) = c1et + c2e

−5t.

12.14 y(t) = c1et + c2e

−t3 .

12.15 y(t) = −15(3et + 2e−4t).

12.16 y(t) = 2et2 .

204 ANSWER KEY

Section 13

13.1 (a) y(t) = c1et3 +c2te

t3 (b) y(t) = e

t3−1(1−t) (c) limt→∞ y(t) = − limt→∞ e

t3−1(t−

1) = −∞.

13.2 (a) y(t) = c1e− 2t

5 + c2te− 2t

5 (b) y(t) = e−2t5 (t − 1) (c) limt→−∞ y(t) = −∞

and limt→∞ y(t) = 0.

13.3 (a)y(t) = c1e2t + c2te

2t (b) y(t) = 4e2t−2(2t − 3) (c) limt→−∞ y(t) = 0and limt→∞ y(t) =∞.

13.4 (a) y(t) = c1e−√2t+c2te

−√2t (b) y(t) = e−

√2t+√

2te−√2t = e−

√2t(1+

√2t) (c)

limt→−∞ y(t) = −∞ and limt→∞ y(t) = 0.

13.5 (a) y(t) = c1ert + c2te

rt (b) y(t) = e− t√

3 (5t + 2√

3) (c) limt→−∞ y(t) = −∞and limt→∞ y(t) = 0.

13.6 y(t) = c1e3t + c2te

3t.

13.7 y(t) = c1et2 + c2te

t2 .

13.8 y(t) = 2e−t2 + te−

t2 .

205

Section 14

14.1 Adding z and z we find 2α = z + z. Hence, α = 12(z + z). Next, subtracting

z from z we find 2iβ = z − z. Therefore, β = 12i(z − z).

14.2 1. 2eiπ3 = 2 cos (π3 ) + 2i sin (π3 ) = 1 + i

√3

2. (2− i)ei3π2 = −1− 2i

3. (√

2eiπ6 )4 = −2 + 2i

√3.

14.3 1. 2ei√2t = 2 cos

√2t+ 2i sin

√2t

2. −12e

2t+i(t+π) = 12e

2t cos t+ 12 ie

2t sin t

3. (√

3e(1+i)t)3 = 3√

3e3t cos (3t) + 3√

3ie3t sin (3t).

14.4 (a) r1 = −1− i and r2 = −1 + i(b) y(t) = e−t(c1 cos t+ c2 sin t)(c) y(t) = 3e−t cos t+ 2e−t sin t.

14.5 (a) r1 = 12(1− i) and r2 = 1

2(1 + i)

(b) y(t) = et2 (c1 cos t

2 + c2 sin t2)

(c) y(t) = −e12(t+π)(3 cos t

2 + sin t2).

14.6 (a) r1 = −2− i and r2 = −2 + i(b) y(t) = e−2t(c1 cos t+ c2 sin t)(c) y(t) = eπ−2t(cos t+ 1

2 sin t).

14.7 (a) r1 = −2πi and r2 = 2πi(b) y(t) = c1 cos 2πt+ c2 sin 2πt(c) y(t) = 2 cos 2πt+ (2π)−1 sin 2πt.

14.8 (a) r1 = −π3 i and r2 = π

3 i(b) y(t) = c1 cos π3 t+ c2 sin π

3 t(c) y(t) = −2 cos π3 t+ 3 sin π

3 t.

14.9 a = 0, b = 4, y0 = 2, y′0 = −2.

14.10 a = −2, b = 5, y0 = 12 −

√32 , y

′0 = −1

2 −3√3

2 .

14.11 y(t) =√

2 cos(t− π

4

).

The graph of y(t) is given below.

206 ANSWER KEY

14.12 y(t) = 2et cos(t− π

3

).


14.13 y(t) =√

2e−2t cos(2t− 7π

4

).


207

14.14 y(t) = c1 cos 3t+ c2 sin 3t.

14.15 y(t) = 5√

5 cos(7t+ arctan

(112

)).

14.16 y′′ − 4y′ + 13y = 0.

14.17 y′′ − 10y′ + 29y = 0.

14.18 y′′ + 4y = 0, y(0) = 3, y′(0) = −8.

14.19 Since x = ln t we find dxdt = 1

t . But dydt = dy

dxdxdt = 1

tdydx . Moreover, d2y

dt2=

− 1t2dydx + 1

t2d2ydx2

= 1t2

(d2ydx2− dy

dx

). Hence,

0 =t2y′′ + αty′ + βy

=t2(

1

t2

(d2y

dx2− dy

dx

))+ αt

(1

t

dy

dx

)+ βy

=d2y

dx2+ (α− 1)

dy

dx+ βy

14.20 y(t) = c1 cos (ln t) + c2 sin (ln t).

208 ANSWER KEY

Section 15

15.1 Since p(t) = t and q(t) = t2 + 2 are polynomials, they are analytic func-tions everywhere. Thus, every real number is an ordinary point.

15.2 We have p(t) = tt2−1 and q(t) = 1

t(t2−1) . Thus, the points 1, −1, and 0

are singular points of the equation, any other real number is an ordinary point ofthe equation.

15.3 We have p(t) = 3tt2−4 and q(t) = 1

t2−4 . Thus, the singular points are t = ±2and any other point is an ordinary point.

15.4 R = 2.

15.5 y(t) = a0(1− t2 + 1

4 t4 − 1

60 t6 + · · ·

)+ a1

(t− 1

2 t3 + 3

40 t5 − 1

1680 t7 + · · ·

).

15.6 y(t) = a0(1 + 1

12 t4 + 1

90 t6 + · · ·

)+ a1

(t+ 1

6 t3 + 3

40 t5 + 13

1008 t7 + · · ·

).

15.7 y(t) = a0(1− 1

2 t2 − 1

6 t3 + 1

12 t4 + 3

40 t5 + · · ·

)+a1

(t+ 1

2 t2 − 1

8 t4 − 1

40 t5 + · · ·

).

15.8∑

n=2[(n− 1)an−1 + n2an](t− 2)n.

15.9 y(t) = a0(1− 1

2 t2 + 1

8 t4 − 1

48 t6 + · · ·

)+ a1

(t− 1

3 t3 + 1

15 t5 − 1

105 t7 + · · ·

).

15.10 y(t) = a0(1 + t2 + 2

3 t3 + 2

3 t4 + 2

3 t5 + · · ·

)+ a1

(t+ 1

3 t3 + 1

3 t4 + 1

3 t5 + · · ·

).

209

Section 16

16.1 (a) y′p = −23e

2t, y′′p = −43e

2t, and y′′p − 2y′p − 3yp = −43e

2t + 43e

2t + e2t = e2t.(b) yh(t) = c1e

−t + c2e3t

(c) The general solution to the differential equation is y(t) = c1e−t + c2e

3t − 13e

2t

and the solution to the IVP is y(t) = 56e−t + 1

2e3t − 1

3e2t.

16.2 (a) y′p = y′′p = 0 and y′′p − y′p − 2yp = 0− 0− 2(−5) = 10.(b) yh(t) = c1e

−t + c2e2t.

(c) The general solution to the differential equation is y(t) = c1e−t + c2e

2t− 5 andthe solution to the IVP is y(t) = 3e−(t+1) + 2e2t+2 − 5.

16.3 (a) y′p = −2e−t + 2te−t, y′′p = 4e−t − 2te−t and y′′p + y′p = 4e−t − 2te−t −2e−t + 2te−t = 2e−t.(b) yh(t) = c1 + c2e

−t.(c) The general solution to the differential equation is y(t) = c1 + c2e

−t − 2te−t

and the solution to the IVP is y(t) = 6− 4e−t − 2te−t.

16.4 (a) y′p = y′′p = 2et−π and y′′p + 4y′p = 2et−π + 8et−π = 10et−π.(b) yh(t) = c1 cos 2t+ c2 sin 2t.(c) The general solution to the differential equation is y(t) = c1 cos 2t+ c2 sin 2t+2et−π and the solution to the IVP is y(t) = − sin 2t+ 2et−π.

16.5 (a) y′p = −2 sin t + cos t, y′′p = −2 cos t − sin t, and y′′p − 2y′p + 2yp =−2 cos t− sin t+ 4 sin t− 2 cos t+ 4 cos t+ 2 sin t = 5 sin t.(b) yh(t) = et(c1 cos t+ c2 sin t).(c) The general solution to the differential equation is y(t) = et(c1 cos t+c2 sin t)+2 cos t+ sin t and the solution to the IVP is y(t) = −2et−

π2 cos t+ 2 cos t+ sin t.

16.6 (a) y′p = 2t + 4 − sin t, y′′p = 2 − cos t, and y′′p − 2y′p + yp = 2 − cos t −4t− 8 + 2 sin t+ t2 + 4t+ 10 + cos t = t2 + 4 + 2 sin t.(b) yh(t) = c1e

t + c2tet.

(c) The general solution to the differential equation is y(t) = c1et + c2te

t + t2 +4t+10+cos t and the solution to the IVP y(t) = −10et+9tet+ t2 +4t+10+cos t.

16.7 u(t) = 12u1(t) + 2

3u3(t).

16.8 u(t) = 14u1(t) + 1

4u2(t) + 112u3(t).

16.9 g(t) = −14 t− 3

2 − 6− t−12 .

210 ANSWER KEY

16.10 g(t) = − 1(1+t)2

+ 11+t .

16.11 g(t) = t.

16.12 α = −2, β = 1, g(t) = 2et.

16.13 α = 0, β = 4, g(t) = 3 sin t− 4.

16.14 y(t) = c1 cos 2t+ c2 sin 2t+ et

5 .

16.15 y(t) = c1 cos 3t+ c2 sin 3t+ c3t+ c4 + t4.

211

Section 17

17.1 y(t) = c1 cos t+ c2 sin t+ ln | cos t| cos t+ t sin t.

17.2 y(t) = c1et + c2e

−t + 12 te

t.

17.3 y(t) = c1 cos 2t+ c2 sin 2t− 18 + 1

4 t2 + 2t sin 2t.

17.4 yp(t) = ( t2

2 ln t− 3t2

4 )e−t.

17.5 y = et

64(8t2 − 4t+ 15)− 14e−3t.

17.6 (a) Usually the first thing to do would be to check that y1(t) = e√t and

y2(t) = e−√t really are solutions to the equation. However, the question says that

this can be assumed and so we move on to the next step, which is to check thatthe Wronskian of the two solutions is non-zero on (0,∞). We have

y′1 = 12√te√t and y′2 = − 1

2√te−√t

and so

W (t) = y1y′2 − y′1y2 = − 1

2√t− 1

2√t

= − 1√t

This is indeed non-zero and so {e√t, e−

√t} is a fundamental set for the homoge-

neous equation.(b) y = (t−

√t)e√t.

17.7 y(t) = c1 cos t+ c2 sin t− 12 t cos t.

17.8 (a) r1 = 1 and r2 = −3.(b) y1(t) = t and y2(t) = t−3.(c) yp(t) = − 1

16 t−3 − 1

4 t−3 ln t.

17.9 y(t) = c1 cos t+ c2 sin t+ cos2 t− 13 cos4 t+ 1

3 sin4 t.

212 ANSWER KEY

Section 18

18.1∫∞0

11+t2

dt = π2 .

18.2∫∞0

t1+t2

dt =∞.

18.3∫∞0 e−t cos (e−t)dt = sin 1.

18.4 L[e3t] = 1s−3 , s > 3.

18.5 L[t− 5] = 1s2− 5

s , s > 0.

18.6 f(t) = e(t−1)2

does not have a Laplace transform.

18.7 F (s) = 4s + 2

s

(−2s + 1

s2

)= 4

s −4s2

+ 2s3, s > 0.

18.8 L[f(t)] = e−s

s2, s > 0.

18.9 L[f(t)] = − e−2s

s + 1s2

(e−s − e−2s), s 6= 0.

18.10 Let u′ = e−st and v = tn. Then u = − e−st

s and v′ = ntn−1. Hence,

∫tne−stdt = − t

ne−st

s+n

s

∫tn−1e−stdt, s > 0.

18.11 (a) limt→0 tne−st = 0 (b) limt→∞ t

ne−st = 0.

18.12 (a) Using the two previous problems we find

L[tn] = limT→∞

∫ T

0tne−stdt = lim

T→∞

{−[tne−st

s

]T0

+n

s

∫ T

0tn−1e−stdt

}

=n

slimT→∞

∫ T

0tn−1e−stdt =

n

sL[tn−1], s > 0

213

(b) We have

L[t] =1

s2

L[t2] =2

sL[t] =

2

s3

L[t3] =3

sL[t2] =

6

s4

L[t4] =4

sL[t3] =

24

s5

L[t5] =5

sL[t4] =

120

s5

(c) By induction, one can easily show that for n = 0, 1, 2, · · ·

L[tn] =n!

sn+1, s > 0.

18.13 L[cosωt] = ss2+ω2 , s > 0.

18.14 L[sinωt] = ωs2+ω2 , s > 0.

18.15 L[cosω(t− 2)] = s cos 2ω+ω sin 2ωs2+ω2 , s > 0.

18.16 L[e3t sin t] = 1(s−3)2+1

, s > 3.

18.17 (a) Neither (b) The graph of the function does not show that it can bebounded by exponential functions. Hence, no such numbers a and M.

18.18 (a) f(t) = et2

e2t+1is continuous on 0 ≤ t < ∞ (b) f(t) is not of expo-

nential order at infinity.

18.19 L−1(

3s−2

)= 3e2t, t ≥ 0.

18.20 L−1(− 2s2

+ 1s+1

)= −2t+ e−t, t ≥ 0.

214 ANSWER KEY

Section 19

19.1 L[2et + 5] = 2L[et] + 5L[1] = 2s−1 + 5

s , s > 1.

19.2 L[e3t−3h(t− 1)] = e−s

s−3 , s > 3.

19.3 L[sin2 ωt] = 12

(1s −

ss2+4ω2

), s > 0.

19.4 L[sin 3t cos 3t] = 3s2+36

, s > 0.

19.5 L[e2t cos 3t] = s−2(s−2)2+9

, s > 2.

19.6 L[e4t(t2 + 3t+ 5)] = 2(s−4)3 + 3

(s−4)2 + 5s−4 , s > 4.

19.7 L−1[ 10s2+25

+ 4s−3 ] = 2 sin 5t+ 4e3t, t ≥ 0.

19.8 L−1[ 5(s−3)4 ] = 5

6e3tt3, t ≥ 0.

19.9

L−1[ e−2s

s− 9] = e9(t−2)h(t− 2) =

{0, 0 ≤ t < 2

e9(t−2), t ≥ 2

19.10 L−1[ e−3s(2s+7)s2+16

] = 2 cos 4(t− 3)h(t− 3) + 74 sin 4(t− 3)h(t− 3), t ≥ 0.

19.11 Note that

f(t) =

0, 0 ≤ t < 11, 1 ≤ t < 32, t ≥ 3

The graph of f(t) is shown below. Using Table L we find

L[f(t)] = L[h(t− 1)] + L[h(t− 3)] =e−s

s+e−3s

s, s > 0.

215

19.12 Note that

f(t) =

0, 0 ≤ t < 1t, 1 ≤ t < 30, t ≥ 3


L[f(t)] =L[(t− 1)h(t− 1) + h(t− 1)− (t− 3)h(t− 3)− 3h(t− 3)]

=L[(t− 1)h(t− 1)] + L[h(t− 1)]− L[(t− 3)h(t− 3)]− 3L[h(t− 3)]

=e−s

s2+e−s

s− e−3s

s2− 3e−3s

s, s > 1.

19.13 Note that

f(t) =

0, 0 ≤ t < 13, 1 ≤ t < 40, t ≥ 4


L[f(t)] = 3L[h(t− 1)]− 3L[h(t− 4)] =3e−s

s− 3e−4s

s, s > 0.

19.14 Note that

f(t) =

0, 0 ≤ t < 1

|2− t|, 1 ≤ t < 30, t ≥ 3

216 ANSWER KEY


L[f(t)] =(2− t)h(t− 1) + 2(t− 2)h(t− 2)− (t− 2)h(t− 3)

=L[−(t− 1)h(t− 1) + h(t− 1) + 2(t− 2)h(t− 2)− (t− 3)h(t− 3)− h(t− 3)]

=− L[(t− 1)h(t− 1)] + L[h(t− 1)] + 2L[(t− 2)h(t− 2)]

−L[(t− 3)h(t− 3)]− L[h(t− 3)]

=− e−s

s2+e−s

s+

2e−2s

s2− e−3s

s2− e−3s

s, s > 0.

19.15 Note that

f(t) =

{1, 0 ≤ t ≤ 20, t > 2

The graph of f(t) is shown below. From this graph we see that f(t) = h(t)−h(t−2)h(t− 2). Using Table L we find

L[f(t)] = L[h(t)]− L[h(t− 1)h(t− 1)] =1− e−2s

s, s > 0.

19.16 Note that

f(t) =

1, 0 ≤ t < 12, 1 ≤ t ≤ 41, t > 4

217


L[f(t)] = L[h(t− 1)] + L[h(4− t)] =e−s

s+

∫ 4

0e−stdt =

1 + e−s − e−4s

s, s > 0.

19.17 From the graph we see that

f(t) = (t− 2)h(t− 2)− (t− 3)h(t− 3)− h(t− 4)

Thus,

L[f(t)] = L[(t−2)h(t−2)]−L[(t−3)h(t−3)]−L[h(t−4)] =e−2s − e−3s

s2−e−4s

s, s > 0.

19.18 From the graph we see that

f(t) = h(t− 1) + h(t− 2)− 2h(t− 3).

Thus,

L[f(t)] = L[h(t− 1)]− 2L[h(t− 3)] + L[h(t− 2)] =e−s − 2e−3s + e−2s

s, s > 0.

19.19 y(t) = 2e−4t+3[h(t−1)−h(t−3)]−3[e−4(t−1)h(t−1)−e−4(t−3)h(t−3)], t ≥ 0.

218 ANSWER KEY

Section 20

20.1 F (s) = A1(s−1)3 + A2

(s−1)2 + A3s−1 + B1

(s−2)2 + B2s−2 .

20.2 F (s) = A1s+A2s2+16

+ B1s−2 .

20.3 F (s) = A1s+A2(s2+1)2

+ A3s+A4s2+1

+ B1(s+4)2

+ B2s+4 .

20.4 F (s) = A1(s−2)2 + A2

s−2 + B1s+B2s2+8s+17

.

20.5 L−1[

1(s+1)3

]= 1

2e−tt2, t ≥ 0.

20.6 L−1[

2s−3(s−1)(s−2)

]= et + e2t, t ≥ 0.

20.7 L−1[4s2+s+1s(s2+1)

]= 1 + 3 cos t+ sin t, t ≥ 0.

20.8 L−1[s2+6s+8(s2+4)2

]= 3

2 t sin 2t+ 34 sin 2t− 1

2 t cos 2t, t ≥ 0.

20.9 y(t) = 9e−2t − 6 cos 3t+ 4 sin 3t, t ≥ 0.

20.10 y(t) = 4e−2t − 1 + 2t, t ≥ 0.

20.11 y(t) = 3e−2t − 2e−t + 6te−t, t ≥ 0.

20.12 y(t) = cos 2t+ 12 sin 2t+ t

4 sin 2t, t ≥ 0.

20.13 y(t) = −et − tet + e2t, t ≥ 0.

20.14 y(t) = cos 3t+ sin 3t+ 23(1− cos 3t)− 2

3(1 + cos 3t)h(t− π), t ≥ 0.

20.15 α = 1, β = 2, y0 = 2, and y′0 = −3.

219

Section 21

21.1 L[f(t)] = 3−2e−s−e−2s

s(1−e−2s).

21.2 L[f(t)] = 2−e−s−e−3s

s(1−e−4s).

21.3 L[f(t)] = e−s

s2(1−e−2s)[1− (s+ 1)e−s].

21.4 L[f(t)] = 1s2(1−e−2s)

[1− (2s+ 1)e−2s].

21.5 L[f(t)] = 1+e−πs

(1+s2)(1−e−2πs). T = 2π.

21.6 L[f(t)] = 1s −

1−e−2(s+1)

(s+1)(1−e−2s), T = 2.

21.7 Note first that

s2 − ss3

+e−s

s(1− e−s)=

1

s−(

1

s2− se−s

s2(1− e−s)

).

Using Example 21.3, we findg(t) = 1− f(t)

where f(t) is the sawtooth function shown below

21.8 (a) a = 2, b = 5, and c = 2. (b) F (s) = L[e−t] = 1s+1 , Y (s) = Φ(s)F (s) =

220 ANSWER KEY

1(s+1)(2s2+5s+2)

. y(t) = −e−t + 23e− t

2 + 13e−2t, t ≥ 0.

21.9 (a) Φ(s) = Y (s)F (s) = 1

(s+1)2(b) y(t) = L−1[Y (s)] = 1− e−t − te−t, t ≥ 0.

21.10 (a) Φ(s) = Y (s)F (s) = 1

s2+s+1(b) Y (s) = Φ(s)F (s) = (1−e−s)

s(1+e−s)(s2+s+1).

21.11 (a) Φ(s) = Y (s)F (s) = 1

s3−4 (b) Y (s) = s2+s−1s2(s−1)(s3−4) .

21.12 b = 3, c = 2, y0 = 1, y′0 = −1.

221

Section 22

22.1 (a) We have

(f ∗ g)(t) =

∫ t

0f(t− s)g(s)ds =

∫ t

0h(t− s)h(s)ds =

∫ t

0ds = t, t ≥ 0.

(b) Since F (s) = G(s) = L[h(t)] = 1s we have (f ∗ g)(t) = L−1[F (s)G(s)] =

L−1[ 1s2

] = t, t ≥ 0.

22.2 (a) We have

(f ∗ g)(t) =

∫ t

0f(t− s)g(s)ds =

∫ t

0e(t−s)e−2sds

=et∫ t

0e−3sds =

[e(t−3s)

−3

]t0

=et − e−2t

3, t ≥ 0.

(b) Since F (s) = L[et] = 1s−1 and G(s) = L[e−2t] = 1

s+2 we find (f ∗ g)(t) =

L−1[F (s)G(s)] = L−1[ 1(s−1)(s−2) ]. Using partial fractions decomposition we find

1

(s− 1)(s+ 2)=

1

3(

1

s− 1− 1

s+ 2).

Thus,

(f ∗ g)(t) = L−1[F (s)G(s)] =1

3

(L−1[ 1

s− 1]− L−1[ 1

s+ 2]

)=et − e−2t

3, t ≥ 0.

22.3 (a) Using the trigonometric identity 2 sin p cos q = sin (p+ q) + sin (p− q) wefind that 2 sin (t− s) cos s = sin t+ sin (t− 2s). Hence,

(f ∗ g)(t) =

∫ t

0f(t− s)g(s)ds =

∫ t

0sin (t− s) cos sds

=1

2[

∫ t

0sin tds+

∫ t

0sin (t− 2s)ds]

=t sin t

2+

1

4

∫ t

−tsinudu

=t sin t

2.t ≥ 0.

222 ANSWER KEY

(b) Since F (s) = L[sin t] = 1s2+1

and G(s) = L[cos t] = ss2+1

we find

(f ∗ g)(t) = L−1[F (s)G(s)] = L−1[ s

(s2 + 1)2] =

t

2sin t, t ≥ 0.

22.4 (f ∗g)(t) = t2

2 −(t−2)2

2 h(t−2)−2(t−2)h(t−2), t ≥ 0. The graph of (f ∗g)(t)is given below.

22.5 (f ∗ g)(t) = (t − 1)h(t − 1) − 3(t − 2)h(t − 2) + 2(t − 3)h(t − 3), t ≥ 0. Thegraph of (f ∗ g)(t) is given below.

22.6 t ∗ t ∗ t = t5

120 , t ≥ 0.

22.7 h(t) ∗ e−t ∗ e−2t = 12 − e

−t + 12e−2t, t ≥ 0.

22.8 t ∗ e−t ∗ et = −t+ et

2 −e−t

2 , t ≥ 0.

22.9 n = 9 and C = 18! .

22.10 y(t) = 2 + t2, t ≥ 0.

Index

Abel’s theorem, 84Amplitude, 106Analytic, 111attracting solution, 62Autonomous, 59

Bernoulli equation, 51

Characteristic equation, 91Characteristic polynomial, 91Complex exponential function, 103Convolution, 168Convolution Theorem, 168

Differential equations, 3Direction field, 59

Equilibrium solution, 61Euler equation, 110Euler’s function, 103Euler’s method, 69Exact DE, 45Exponential order a, 131Exponentially bounded, 131Extended chain rule, 43

First order linear differential equation,14

First Shifting Theorem, 143Fundamental set of solutions, 82

General solution, 6

Heaviside step function, 141

Heun’s Method, 70Homogeneous, 14

Initial conditions, 6Initial value problem, 6Integrating factor, 20Inverse Laplace transform, 135

Laplace transform, 129Linear combination, 81Linear time invariant, 161

Method of integrating factor, 19Method of Variation of Parameters, 123Modified Euler’s method, 72

Newton’s second law, 3Non-homogeneous, 14Non-linear, 14

Order, 4ordinary differential equation, 4ordinary point, 111

Parameters, 6Partial derivatives, 29Partial differential equation, 5Partial fractions, 151Particular solution, 6Periodic, 159Phase line, 63Phase shift, 106Piecewise continuous, 132Principle of superposition, 81

223

224 INDEX

Qualitative, 59

Recurrence formula, 114Repelling, 62Riccati Equation, 53

Second Shifting Theorem, 143Semi-stable, 62Separable, 35Singular point, 111Singular solution, 7, 39Slope field, 59Solution, 5Solution curve, 7Stable solution, 62Superposition principle, 14System transfer function, 162

Uniform motion, 3Unstable, 62

Wave equation, 5Wronskian, 82

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Introductory Notes in Ordinary Differential Equations for Physical