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Chapter 10

The Laplace Transform

A few worked examples should convince the reader that the Laplace transformfurnishes a useful technique for solving linear differential equations. (However, thestudy of differential equations is no more contained in a table of Laplace transformsthan the study of geometry in a set of trigonometric tables.)

T. W. Korner1

Sire, I had no need of that hypothesis.

Pierre-Simon de Laplace2

10.1 Introduction to the Laplace Transform

Many, if not most, engineers in academia and industry prefer solving linear differentialequations and systems with a method involving a certain type of improper integralknown as the Laplace transform. Although we will use this method for only first- andsecond-order linear equations, it can also be used to solve higher-order linear equations.The goal of this section is to introduce the notion of the Laplace transform and toformally define it.

We will introduce the Laplace transform as an alternative way of solving first-order linear equations. This will establish right away the effectiveness of the method.

1See [35, p. 378]. Thomas William Korner is a Cambridge University mathematician who has writtenresearch papers and books on Fourier analysis, including a popular textbook for engineers, physicists, andmathematicians entitled An Introduction to Fourier Analysis. This particular quote comes from another ofhis books, Fourier Analysis (Cambridge University Press), which consists of essays on the ideas and resultsof Fourier analysis and their applications to other disciplines.

2See [10, ch. 11]. Pierre-Simon de Laplace (1749–1827) was a French mathematician and astronomer.All of his mathematical research in celestial mechanics, such as applying the laws of Newton to model themotions of the planets, were bound together under the title of the Mecanique Celeste. He also held politicaloffices and was made a count by Napoleon Bonaparte. The above quote was in response to Napoleon’sremark: “You have written this huge book on the system of the world without once mentioning the author ofthe universe.” More on Laplace and other famous mathematicians can be found in the classic book Men ofMathematics by E. T. Bell [10], from which the above quote was taken.

313

314 CHAPTER 10. THE LAPLACE TRANSFORM

However, instead of introducing the Laplace transform in this way, we could begin withthe formal definition of the Laplace transform right away, after which we could stateand prove a number of its properties. Then we could determine the Laplace transformsof the most commonly occurring functions in engineering. But even after all of this,you probably would still fail to see any connection between the Laplace transformand differential equations. Eventually we would get around to showing how to usethe Laplace transform to solve differential equations. This is the approach of manytextbooks. However efficient this may be in getting students to solve equations, itmay still leave them wondering how anyone ever came up with the method in thefirst place. Actually the development and use of the Laplace transform was a lengthyprocess. It was not the work of any one person and did not occur over the courseof a few days, as a reader might construe from studying a textbook on the subject.The Laplace transform, an improper integral, was first introduced by Pierre-Simon deLaplace in Theorie Analytique des Probabilites, a treatise on probability published in1812. However, Laplace did not have the last word on the subject. Its application todifferential equations was still being developed as late as the early 1930s by G. Doetsch.

Our approach to the subject will be to ask whether there is another way to solvefirst-order linear differential equations besides the methods seen in earlier chapters.We begin our investigation with the initial value problem

dx

dtC 2x D 3et ; x.0/ D 1: (10.1.1)

Since the differential equation is already in standard form, its solutions are readilyfound by multiplying it by the integrating factor e 2t and then integrating the result:

e2t dx

dtC 2e2tx D 3e3t ) d

dt

e2tx

D 3e3t

)Z

d

dt

e2tx

dt D

Z3e3t dt C C

) e2t x D e3t C C:

(10.1.2)

Solving for x, we obtain the general solution of the differential equation:

x.t/ D et C Ce�2t

Setting t D 0 and using the initial condition x.0/ D 1 allows us to determine the valueof C :

x.0/ D 1 C C D 1 ) C D 0:

Therefore, the solution of the initial value problem is

x.t/ D et : (10.1.3)

However, as we have seen before, there is usually more than one way to attack a prob-lem. Suppose we take a different approach by ignoring that the left-hand side of thefirst equation in (10.1.2) is the derivative of a product. Instead let’s integrate each ofthe terms of (10.1.2) from t D 0 to t D T to see if somehowZ T

0

e2t x0.t/dt CZ T

0

2e2tx.t/dt DZ T

0

3e3t dt (10.1.4)

10.1. INTRODUCTION TO THE LAPLACE TRANSFORM 315

will yield the solution of (10.1.1). Let’s integrate the first integral on the left-hand sideby parts. First set u D e2t and dv D x0.t/dt ; then du D 2e2t dt and v D x.t/. Thus,Z

e2tx0.t/dt DZ

u dv D uv �Zv du D e2tx.t/ �

Z2e2tx.t/dt;

and so Z T

0

e2tx0.t/dt Dhe2tx.t/

iT

0�Z T

0

2e2tx.t/dt:

Thus, (10.1.4) becomes

e2T x.T /� x.0/� 2

Z T

0

e2tx.t/dt C 2

Z T

0

e2t x.t/dt D e3T � 1: (10.1.5)

Since the initial condition is x.0/ D 1, this simplifies to x.T / D e T . We obtain(10.1.3), the solution x.t/ D e t , by merely replacing T with t .

Now let’s look at a variation of the previous approach that is more general in thatwe multiply both sides of the differential equation by the exponential function e �st butdo not assign a value to s.3 Even though e�st is an integrating factor when s D �2,we ignore this fact to see if we can still obtain the solution of (10.1.1). As before, weintegrate each term of the resulting equation with respect to t from t D 0 to t D T :Z T

0

e�st x0.t/dt CZ T

0

2e�stx.t/dt DZ T

0

3e�stet dt: (10.1.6)

Note that this reduces to (10.1.4) when s D �2. Similar to our handling of (10.1.4), weintegrate the first integral on the left-hand side of (10.1.6) by parts by setting u D e �st

and dv D x0.t/dt obtaining

�e�stx.t/

�tDT

tD0�Z tDT

tD0

.�s/e�st x.t/dt CZ T

0

2e�stx.t/dt DZ T

0

3e�stet dt

which simplifies to

e�sT x.T / � x.0/C .s C 2/

Z T

0

e�stx.t/dt D � 3

s � 1

he�.s�1/T � 1

i: (10.1.7)

Note that this reduces to (10.1.5) when s D �2, but again we ignore the efficacy ofassigning s this value. Rather we take a different approach and that is to let T increasewithout bound; symbolically, to let T ! 1. If s < 1, the right-hand side of (10.1.7)diverges; but if s is restricted to values greater than 1, it converges to

3

s � 1;

3The reason for using e �st instead of est is dictated by tradition. We will be concerned shortly with theconvergence of improper integrals of the form

R10 e�st x.t/ dt . In this particular example, as it turns out,

these integrals converge for s > 1. If tradition had instead chosen e st , then that would have merely changedthe form of the improper integrals to

R10 est x.t/ dt ; and in this example, convergence would be for s < 1.


since e�.s�1/T ! 0 as T ! 1. In short, when we take the limit of both sides of(10.1.6) as T ! 1 restricting s to s > 1, the right-hand side of (10.1.6) becomes

limT !1

Z T

0

3e�stet dt D 3 limT !1

Z T

0

e�.s�1/t dt

D � 3

s � 1lim

T !1

he�.s�1/T � 1

iD 3

s � 1: (10.1.8)

This is what is meant by writingZ 1

0

3e�stet dt D 3

s � 1:

The integralR1

03e�stet dt is called an improper integral because 1 is not a real num-

ber in the usual sense; as a result, it must be treated with a little more care than theusual definite integrals both of whose limits of integration are real numbers. Recallfrom calculus that Z 1

a

f .t/dt

is really mathematical shorthand for

limT !1

Z T

a

f .t/dt:

Let’s also assume that e�sT x.T / ! 0 as T ! 1. This will be justified shortly. So,as T ! 1, (10.1.7) becomes

�x.0/C .s C 2/

Z 1

0

e�st x.t/dt D 3

s � 1: (10.1.9)

Setting x.0/ D 1 and solving for the improper integral yieldsZ 1

0

e�stx.t/dt D 3

.s � 1/.s C 2/C 1

s C 2;

which simplifies to Z 1

0

e�st x.t/dt D 1

s � 1: (10.1.10)

At first this result may seem interesting yet useless: we have not explicitly solvedfor x.t/; instead we have what is known as an integral transform of x.t/. But thenour hopes are rekindled when we recall that the quantity 1=.s � 1/ also appeared in aprevious integration, namely (10.1.8):Z 1

0

e�stet dt D 1

s � 1:

10.1. INTRODUCTION TO THE LAPLACE TRANSFORM 317

Even though we may not know how to find solutions of (10.1.10) directly, it does notreally matter, for we do know from (10.1.8) that the differentiable function x.t/ D e t

is a solution. Furthermore, the assumption that led to this result, namely that

e�sT x.T / ! 0 as T ! 1;

turns out to be valid since

e�sT x.T / D e�sT eT D e�.s�1/T

does approach 0 as T ! 1 for s > 1. And so we have managed to obtain the solutionof (10.1.1) for a third time!

Now the author must confess that this initial value problem was contrived so thatthe solution of (10.1.10) just happened to be around, namely, in (10.1.8). Normally wewon’t be so lucky. For example, if the initial condition is changed from x.0/ D 1 tox.0/ D 2, then (10.1.8) remains as it is; however, (10.1.10) becomesZ 1

0

e�stx.t/dt D 3

.s � 1/.s C 2/C 2

s C 2D 2s C 1

.s � 1/.s C 2/:

If this approach to solving linear differential equations is to be effective and an alter-native4 to the integrating factor method, we need some way of ascertaining functionsx.t/ that will satisfy Z 1

0

e�stx.t/dt D X.s/ (10.1.11)

when X.s/ is a known function of s. The function X.s/ is called an integral transformof x.t/. Actually, there are many well-known integral transforms5—this particular oneis called the Laplace transform. And right now we have our first example: the Laplacetransform of x.t/ D e t for t � 0 is

X.s/ D 1

s � 1

for s > 1.Before continuing, let us formally define what is meant by the Laplace transform.

The Laplace Transform

Definition 10.1. Let f .t/ be a function on Œ0;1/. The Laplace transform of fis the function F whose value at a real number s is defined by

F.s/ WDZ 1

0

e�stf .t/dt; (10.1.12)

provided the improper integral converges. The domain of F consists of all thevalues of s where the integral converges.

4The advantages over the integrating factor method will be discussed later after we have fully discussedthis new method of solving initial value problems.

5Another well-known integral transform that you may have already encountered is the Fourier transform.


A word about notation: note the use of the lowercase f and the uppercase F in(10.1.12). When Laplace transforms are involved, the convention is to use a lowercaseletter for the original function and the uppercase form of the same letter to denote itsLaplace transform.6 For instance, just as we used X to denote the Laplace transformof the function x.t/ D e t , we would use F and G to denote the Laplace transforms offunctions called f and g, respectively.

There is even another way of denoting the Laplace transform of a function f .t/and that is to write Lff .t/g. For example, either

Lfetg.s/ D 1

s � 1or L fx.t/g .s/ D 1

s � 1

expresses that the Laplace transform of x.t/ D et is X.s/ D 1=.s � 1/. It is evenacceptable to shorten this to

Lfetg D 1

s � 1:

This might be considered an abuse of notation: for just as we write

X.s/ D 1

s � 1and not X D 1

s � 1;

we should write

Lfetg.s/ D 1

s � 1instead of Lfe tg D 1

s � 1:

However, since writing the argument “s” in all of the steps of a problem is tedious andas it is already understood that Lff .t/g is a function of s, we usually omit it.

When taking the Laplace transform of a function, the original function f .t/ istransformed into a new function called Lff .t/g or F.s/. We portray this symbolicallyby writing

f .t/L��! F.s/:

Looking at it like this, we can say that the Laplace transform is a rule L that assigns tothe original function some other function—but only one other function. Recall that wehave seen rules like this before: they are called operators.

The operator L not only transforms a function f .t/ into another function F.s/, itis also accompanied by a change in domains. For example, we just determined that theLaplace transform of x.t/ D et for t � 0 is the function

Lfx.t/g D 1

s � 1for s > 1:

Associated with this transform is a change from the t-domain Œ0;1/ for f .t/ to thes-domain .1;1/ for F.s/. In applications involving time, where t denotes the time,the t-domain is called the time domain .

6A caveat: Not everyone follows this convention. In some books, it is reversed: the uppercase letterdenotes the original function and the corresponding lowercase letter denotes its transform.

10.2. FINDING LAPLACE TRANSFORMS WITH THE DEFINITION 319

The task of determining pairs of functions satisfying (10.1.12) that will be usefulto us is not as daunting as it might first seem. The reason for this is that the aim ofour investigation is not to find functions f .t/ satisfying (10.1.12) for every imaginablefunction F.s/: rather only for those F.s/ that correspond to functions f .t/ which areactual solutions of linear differential equations. From our prior experiences with linearequations, these are the exponential, sinusoidal, and algebraic functions, along withcertain combinations of these functions. So the game plan will be to substitute thesetypes of functions into (10.1.12) for f .t/ and then work out the resulting improperintegral to determine its Laplace transform F.s/. In this way, we can first create a listof useful Laplace transforms and then see where to go from there.

10.2 Finding Laplace Transforms with the Definition

In the first section, we found that Lfe tg D 1=.s � 1/, which is the succinct way ofsaying that the Laplace transform of the function f .t/ D e t is the function

F.s/ D 1

s � 1:

For our first example in using Definition 10.1, let’s generalize this result by finding theLaplace transform of the function f .t/ D ebt , where b represents any real number.

Example 10.1. Find Lfebtg, where b is a constant.

Solution. By (10.1.12), the definition of the Laplace transform,

Lfebtg DZ 1

0

e�stebt dt D limT !1

Z T

0

e.b�s/t dt

D limT !1

1

b � se.b�s/t

ˇˇtDT

tD0

D 1

b � slim

T !1

he.b�s/T � e0

i:

Implicit in carrying out the integration is that s ¤ b. As T ! 1,

e.b�s/T ! 0 if s > b

wherease.b�s/T ! 1 if s < b:

From this, we conclude that the integral converges to 1=.s � b/ if s > b and divergesif s < b. For the case s D b,

Lfebtg DZ 1

0

e�btebt dt D limT !1

Z T

0

1 dt D limT !1

T D 1:

Therefore,

Lfebtg D

8<:

1

s � b; for s > b

undefined; for s � b.

(10.2.1)


This is the same as saying that the Laplace transform of f .t/ D ebt is the functionF.s/ D 1=.s � b/ with domain .b;1/.

Example 10.2. Show that

Lf1g D 1

s

for s > 0.

Solution. The “1” really stands for a function whose value at every t � 0 is 1. In otherwords, this is a succinct way of saying that we are to find the Laplace transform of afunction f , where f .t/ D 1 for t � 0. By (10.1.12),

Lf1g DZ 1

0

e�st � 1 dt D limT !1

Z T

0

e�st dt

D limT !1

�1

se�st

ˇˇtDT

tD0

D �1

slim

T !1

he�sT � 1

i

D8<:

1

s; s > 0

undefined; s < 0:

If s D 0, the integral diverges. Therefore, for s > 0,

Lf1g D 1

s:

Note that Example 10.1 also implies this. Simply set b D 0 in (10.2.1):

Lf1g D 1

s � b

ˇˇbD0

D 1

s

for s > 0.

Example 10.3. Find the Laplace transform of the unit step function u.t � b/, whereb � 0 and

u.t � b/ D(

0; 0 � t < b

1; t > b:

Solution. By (10.1.12), Lfu.t � b/g D R10

e�stu.t � b/dt . Since the value of thefunction changes at t D b, we divide the improper integral into two integrals:

Lfu.t � b/g DZ b

0

e�st � 0 dt CZ 1

b

e�st � 1 dt D limT !1

Z T

b

e�st dt

D limT !1

�1

se�st

ˇˇtDT

tDb

D �1

slim

T !1

he�sT � e�sb

i:


We conclude

Lfu.t � b/g D e�bs

sfor s > 0:

Finally note that for b D 0, Lfu.t/g D 1=s, which agrees with Example 10.2.

The next two examples demonstrate the indispensability of the method of integra-tion by parts as a tool in determining Laplace transforms. The first one also demon-strates the importance of recognizing patterns.

Example 10.4. For n D 1; 2; 3; : : : , show that

Lftng D n!

snC1:

Solution. By Definition 10.1,

Lftng D limT !1

TZ0

e�st tn dt:

This is one of those classic integration-by-parts integrand. By the LIATE mnemonic,we set u D tn and dv D e�st dt . Then du D ntn�1 dt and v D � 1

se�st . Thus,

Z T

0

e�st tn dt D �1

se�st tn

ˇˇT0

�Z T

0

��1

se�st

�ntn�1 dt

D �1

se�sT T n C n

s

Z T

0

e�st tn�1 dt :

First let us determine the long-term behavior of the first term by using the Maclaurinseries expansion of ex:

1 C x C x2

2!C x3

3!C x3

4!C � � � C xn

n!C � � � :

Recall that this series converges to ex regardless of the value of x. Setting x D st wehave

T n

esTD T n

1 C sT C s2T 2

2!C s3T 3

3!C � � � C snT n

n!C snC1T nC1

.n C 1/!C snC2T nC2

.n C 2/!C � � �

D 1

1

T nC s

T n�1C s2

2!T n�2C � � � C sn

n!C snC1T

.n C 1/!C snC2T 2

.n C 2/!C � � �

:

Since snCkT k=.n C k/! ! 1 as T ! 1 for k D 1; 2; 3; : : : , we conclude that

T n

esT! 0 as T ! 1


for s > 0, regardless of the magnitude of n.7 So if the integral in the second term onthe right-hand side converges, its limit would be Lft n�1g, which in turn would implythat the integral on the left-hand side converges to what would be Lft ng. The end resultis the recursion formula

Lftng D n

sLftn�1g .s > 0/ (10.2.2)

provided Lft n�1g exists.Substituting n D 1 into (10.2.2) gives

Lft1g D 1

sLft0g D 1

sLf1g:

Now Lf1g does exist since we determined in Example 10.2 that Lf1g D 1=s. Itsexistence implies the validity of formula (10.2.2) for n D 1 and thereby the existenceof Lftg:

Lftg D 1

s2:

The existence of Lftg then validates (10.2.2) for n D 2; therefore,

Lft2g D 2

sLft1g D 2

s� 1

s2D 2

s3:

Continuing in this manner, we obtain

n D 3 W Lft3g D 3

sLft2g D 3

s� 2

s3D 3 � 2

s4

n D 4 W Lft4g D 4

sLft3g D 4

s� 3 � 2

s4D 4 � 3 � 2

s5

n D 5 W Lft5g D 5

sLft4g D 5

s� 4 � 3 � 2

s5D 5 � 4 � 3 � 2

s6

and so on. The pattern indicates that

Lftng D n!

snC1.s > 0/ (10.2.3)

for every positive integer n. Certainly, we have established that the formula is true forn D 1; 2; 3; 4; 5. Note that it is also true for n D 0 with the understanding that 0! D 1.What about for n > 5? Suppose (10.2.2) were true for an integer n D k; that is,

Lftkg D k!

skC1: (10.2.4)

The previous cases indicate that it then should be true for the next value of n, to wit,n D k C 1. Let’s check this. By recursion formula (10.2.2)

LftkC1g D k C 1

sLftkg:

7Another way to show this limit is with L’Hopital’s Rule.


This and (10.2.4) yields

LftkC1g D k C 1

s� k!

skC1D .k C 1/!

skC2:

So we have established that it is impossible for formula (10.2.3) to be true for someinteger but not for the next one. The validity of the formula for n D 1; 2; 3; 4; and 5

has already been established. Since it is valid for n D 5, it must be valid for n D 6.Its validity for n D 6 implies its validity for n D 7, which in turn implies its validityfor n D 8. This domino effect makes it valid for all positive integers. Some of youmay recognize that we have established the validity of formula (10.2.3) for all positiveintegers by using a method of proof known as mathematical induction.

Example 10.5. Find the Laplace transform of Lfsin btg, where b is a real constant.

Solution. By the definition of the Laplace Transform (10.1.12), for b ¤ 0,

Lfsin btg D limT !1

Z T

0

e�st sin bt dt:

Applying the integration-by-parts formula twice to the integral, we find that�1 C b2

s2

�Z T

0

e�st sin bt dt D �1

se�sT sin bT C b

s2e�sT cos bT C b

s2

for s ¤ 0. For s > 0, e�sT ! 0 as T ! 1. Sinceˇe�sT sin bT

ˇD e�sT jsin bT j � e�sT ;

we see that e�sT sin bT ! 0 as T ! 1. Likewise e�sT cos bT ! 0 as T ! 1. Weconclude that �

1 C b2

s2

�Lfsin btg D b

s2:

Solving for Lfsinbtg gives

Lfsin btg D b

s2 C b2

for s > 0. Note that this formula is valid for b D 0; so we can remove the stipulationthat b ¤ 0.

Example 10.6. Find the Laplace transform of the Dirac delta function at b for b � 0.

Solution. Once again by (10.1.12),

L fı.t � b/g DZ 1

0

e�stı.t � b/dt:

By the sifting property of the Dirac delta function,

L fı.t � b/g D limT !1

Z T

0

e�stı.t � b/dt D limT !1

e�sbu.T � b/:


Since u.T � b/ D 1 for T > b,

Lfı.t � b/g D e�bs:

In the next section, Table 10.1 lists all of the Laplace transforms that we workedout in this section. It also includes the transform of the cosine function:

Lfcos btg D s

s2 C b2

for s > 0. This can be verified using Definition 10.1.

10.3 Linearity of the Laplace Transform

Even though Laplace transforms can be computed with a CAS or looked up in a varietyof different handbooks or even in the appendices of many engineering textbooks, thetransforms of some of the basic functions should be memorized. Some of these arelisted in Table 10.1.

Table 10.1: Basic Laplace Transforms

f .t/ F.s/ D L ff .t/g

11

s; s > 0

tn; n D 0; 1; 2; : : :n!

snC1; s > 0

ebt1

s � b; s > b

sin btb

s2 C b2; s > 0

cos bts

s2 C b2; s > 0

u.t � b/; b � 0e�bs

s; s > 0

ı.t � b/; b � 0 e�bs

Observe that the Laplace transforms of sin bt and cos bt are the same except for thenumerators. The sentence

Some birds can swim.

10.3. LINEARITY OF THE LAPLACE TRANSFORM 325

can aid in remembering that the numerator of sin bt is b whereas the numerator ofcos bt is s.

Can we use this table to find the transform of a function that is a combinationof some of the functions listed in it? It turns out that the answer is yes for specialcombinations known as linear combinations. A linear combination of the functionsf1; f2; � � � ; fn is a function of the form

c1f1 C c2f2 C � � � C cnfn;

where c1; c2; � � � ; cn are constants. In integral calculus, we learned that the integrationof a linear combination of integrable functions can be carried out by first integratingeach of the functions and then by taking the same linear combination of the results;that is to say, in the case of n D 2:Z b

a

.c1f1.t/C c2f2.t// dt D c1

Z b

a

f1.t/dt C c2

Z b

a

f2.t/dt:

This property of integration is known as linearity: Since Laplace transforms are im-proper integrals, they inherit this linearity property, as we now state.

Linearity of the Laplace Transform

Theorem 10.1. Let f1 and f2 be functions whose Laplace transforms exist fors > b1 and s > b2, respectively. Let b be the larger of b1 and b2. Then for s > b,

Lfc1f1.t/C c2f2.t/g D c1Lff1.t/g C c2Lff2.t/g; (10.3.1)

where c1 and c2 are any constants.

Proof. To show this, we merely use the linearity property of integration: For all T � 0,Z T

0

e�st Œc1f1.t/ C c2f2.t/� dt D c1

Z T

0

e�stf1.t/dt C c2

Z T

0

e�stf2.t/dt:

Then letting T ! 1, we obtain (10.3.1).

Special versions of (10.3.1) are

(a) Lff1.t/C f2.t/g D Lff1.t/g C Lff2.t/g(b) Lfc1f1.t/g D c1Lff1.t/g

Obviously, (a) follows from setting both constants in (10.3.1) equal to 1, while (b)follows from setting c2 equal to zero. So we have just showed that (10.3.1) implies (a)and (b). You should convince yourself that the converse is also true: (a) and (b) takentogether imply (10.3.1). Therefore we can say that properties (a) and (b) are equivalent


to property (10.3.1). An operator O, such as L or d=dt or anything else for that matter,with the property

Ofc1f1.t/ C c2f2.t/g D c1Off1.t/g C c2Off2.t/gfor all functions f1 and f2 in its domain and all constants c1 and c2 is called a linearoperator. So stating that L is a linear operator is equivalent to stating that it satisfiesproperty (10.3.1). We leave it to the reader to show that (10.3.1) can be extended to alinear combination of more than two functions:

Lfc1f1.t/ C c2f2.t/ C � � � C cnfn.t/g Dc1Lf1.t/ C c2Lff2.t/g C � � � C cnLffn.t/g: (10.3.2)

Let’s demonstrate the use of the linearity property in determining Laplace trans-forms of some linear combinations of the functions listed in Table 10.1.

Example 10.7. Find Lft 3 C 2e5tg.

Solution. By (10.3.1), or (10.3.2) with n D 2,

Lft3 C 2e5tg D Lft3g C 2Lfe5tg:By Table 10.1, Lft3g D 3!

s4 for s > 0 and Lfe5tg D 1s�5

for s > 5. Therefore, fors > 5,

Lft3 C 2e5tg D 3!

s4C 2 � 1

s � 5D 6

s4C 2

s � 5:

Example 10.8. Find the Laplace transform of the function

f .t/ D 5e�3t � 1

2sin 6t C �:

Solution. It follows from (10.3.2) with n D 3 and Table 10.1 that

Lff .t/g D 5Lfe�3tg � 1

2Lfsin6tg C Lf� � 1g

D 5 � 1

s � .�3/� 1

2� 6

s2 C 62C �Lf1g D 5

s C 3� 3

s2 C 36C �

s

for s > 0.

10.4 First-Order Linear Equations

Recall that the primary objective of this chapter is to develop a method for solvinglinear differential equations using the Laplace transform, so we must bear this in mindas we search for more of its properties. Recall that we began this chapter with the initialvalue problem (10.1.1), namely,

dx

dtC 2x D 3et ; x.0/ D 1:

10.4. FIRST-ORDER LINEAR EQUATIONS 327

In our attempt to discover another way of solving (10.1.1), one that did not involveintegrating factors, we stumbled upon the idea of the Laplace transform. Now that wehave formally defined the Laplace transform, discovered that it is a linear operator, andfound the transform of some commonly occurring functions, let’s retrace the steps thatled to the solution of (10.1.1) and tidy them up so as to streamline the procedure ofusing Laplace transforms to find solutions. We began by tinkering with the differentialequation in (10.1.1). This eventually resulted in equation (10.1.6). Then we took thelimit of both of its sides as T ! 1. Admittedly all of this was done in a clumsy,roundabout way; however, it is really more straightforward than it seems. Observe thatthe process of obtaining (10.1.6) and then taking its limit amounts to nothing more thantaking the Laplace transform of both sides of the differential equation

Lfx0 C 2xg D Lf3etgand then using the linearity property

Lfx0g C 2Lfxg D 3Lfetg:From Table 10.1, Lfetg D 1=.s � 1/; thus,

Lfx0g C 2Lfxg D 3

s � 1: (10.4.1)

If Lfxg were the only unknown in the equation, then we could solve for it which wouldgive us the transform of the solution. However, the presence of Lfx 0g prevents this—unless there is a formula that relates Lfx 0g to Lfxg; if this were the case, then wecould use it to convert the previous equation to one involving only Lfxg. So the utilityof Laplace transforms for solving differential equations depends on the existence ofsuch a formula. Let’s look into this.

By Definition 10.1, the Laplace transform of the derivative of a function f is

Lff 0.t/g D limT !1

Z T

0

e�stf 0.t/dt:

Integrating by parts with u D e �st , dv D f 0.t/dt , du D �se�st dt , and v D f .t/,we obtain Z T

0

e�stf 0.t/dt D e�stf .t/ˇT0

C s

Z T

0

e�stf .t/dt (10.4.2)

D e�sT f .T / � f .0/C s

Z T

0

e�stf .t/dt:

For Lff 0g to exist, the right-hand side must converge as T ! 1. Now the lastterm converges to sLff g if Lff g exists. There is no problem with the second termsince it is a constant. But what about the first term? If you recall, we faced this sameproblem before in solving the initial value problem (10.1.1). There we assumed at theoutset that its solution x.t/ would behave in the following manner: e �st x.t/ ! 0 ast ! 1. After we found it, we verified that this was indeed the case for s > 1. This


behavior was also used in Example 10.4 to find the Laplace transform of f .t/ D t n;there we proved for any fixed s > 0 that e�stf .t/ D e�st tn ! 0 as t ! 1. Let’sdefine a class of functions whose long-term behavior for any fixed value s > willturn out to be

e�stf .t/ ! 0 as t ! 1; (10.4.3)

where the value of depends on the particular function f .

Exponential Order

Definition 10.2. A function f .t/ is said to be of exponential order as t ! 1if there are constants , M , and t1 such that

jf .t/j � Me� t for all t � t1: (10.4.4)

We can always choose the value of t1 to be nonnegative. At times, for the sake ofbrevity, we will say that f .t/ is of exponential order—omitting the constant “ ” andthe phrase “as t ! 1.” A concise way of writing this is

f .t/ D O�e� t�: (10.4.5)

We say that f .t/ is of the order of e� t . The use of “O” in this way is called big-ohnotation—so we can also say that “f .t/ is big oh of e � t .” In particular, when D 0,(10.4.4) becomes f .t/ D O

�e0t� D O .1/. Consequently,

f .t/ D O.1/

will mean that jf .t/j � M for all t � t1. This is a succinct way to say f .t/ is boundedon the interval Œt1;1/ for some t1 � 0.

Let’s look at some examples to become better acquainted with the notation andproficient at recognizing functions of exponential order. After that we can continuewith our search for a formula for the Laplace transform of the derivative.

Example 10.9. Let f .t/ D tn, for all t � 0, where n D 0; 1; 2; 3; : : : . From theMaclaurin series

et D 1 C t C t2

2!C t3

3!C � � � C tn

n!C � � � ;

it follows that e t > tn=n! and so

jf .t/j D tn � n! et for all t � 0:

In the notation of (10.4.4), M D n!, D 1, and t1 D 0. This shows that f .t/ D tn

is of exponential order 1. Equivalently, in terms of the big-oh notation, t n D O�et�.

Actually, tn is of exponential order for any > 0. To see this, let be any fixedpositive number, no matter how small or large. While working through Example 10.4,we proved that e�� t tn ! 0 as t ! 1. Since e�� t tn is a continuous function, this


implies that e�� t tn is bounded. In other words, there exists a constant K > 0 such thatˇe�� t tn

ˇ � K for all t � 0. Thus,

jtnj � Ke� t

for all t � 0, which is the same as saying that

tn D O�e� t�

for any positive number .

Example 10.10. The unit step function at b, u.t � b/, is of exponential order 0 as

ju.t � b/j � 1 D 1 � e0t :

Equivalently, we can writeu.t � b/ D O.1/

as t ! 1.

Example 10.11. Consider the function

g.t/ D �6te3t=2 sin t

for t � 0. As �1 � sin t � 1 for all t and t < e t for all t � 0,

jg.t/j � 6te3t=2 � 6ete3t=2 D 6e5t=2

for t � 0. Thus, g is of exponential order 2:5.

Example 10.12. The functionh.t/ D et2

is not of exponential order. For if it were, then constants M , , and t1 would exist suchthat jh.t/j D et2 � Me� t for all t � t1. Or, simply et2�� t � M , for all t � t1. Butthis contradicts that e t.t��/ ! 1 as t ! 1.

Example 10.13. As a final example, consider the function

p.t/ D 5pt:

Note that p.t/ ! 1 as t ! 0C. So the function is unbounded on the interval .0;1/.However, on Œ1;1/, 0 < p.t/ � 5 as p is a positive, decreasing function and p.1/ D 5.So we can certainly say that jp.t/j � 5, for all t � 1. Thus, by the remarks following(10.4.5), p.t/ D O.1/ as t ! 1.


Now that we know what the class of functions of exponential order are and lookedat some examples, let’s verify that the members of this class display the long-termbehavior that was claimed earlier in (10.4.3). Suppose f .t/ D O

�e� t�

as t ! 1.This means according to (10.4.4) that there are constants M > 0 and t1 � 0 such that

�Me� t � f .t/ � Me� t for all t � t1:

Multiplying both inequalities by e �st , we have

�Me�.s��/t � e�stf .t/ � Me�.s��/t :

If s > , ˙Me�.s��/t ! 0 as t ! 1. Since the middle quantity, e�stf .t/, is sand-wiched between two other quantities that approach 0 as t ! 1, it too must approach0. Thus, we have the result:

Long-Term Behavior of Functions of Exponential Order

Theorem 10.2. If f .t/ D O�e� t�

as t ! 1, then for s > ,

limt!1 e�stf .t/ D 0: (10.4.6)

At last we have all the necessary pieces in place to derive a formula relating theLaplace transform of the derivative of a function to the transform of the function itself.

Transform of the First Derivative

Theorem 10.3. Suppose a function f .t/ is differentiable on Œ0;1/ and of expo-nential order , its Laplace transform exists, and its derivative f 0.t/ is integrableon Œ0;T � for every T � 0. Then the Laplace transform of its derivative exists forall s > and is given by the formula

Lff 0.t/g D sLff .t/g � f .0/: (10.4.7)

Proof. For every T � 0, e�stf 0.t/ is integrable on Œ0;T � since both of its factors areintegrable on Œ0;T �.8 This, along with the differentiability of e �st and f .t/, allows usto integrate e�stf 0.t/ by parts,9 which was done earlier and resulted in (10.4.2):Z T

0

e�stf 0.t/dt D e�sT f .T / � f .0/C s

Z T

0

e�stf .t/dt:

It follows from (10.4.6) that for s > , e �sT f .T / ! 0 as T ! 1. By the hypotheses,the Laplace transform of f exists; hence,

R T

0e�stf .t/dt ! Lff .t/g as T ! 1.

8If the functions f and g are integrable on an interval Œa; b�, then their product fg is also integrable onŒa; b�. For a proof of this, see any advanced calculus or real analysis book, such as Ross [46, p. 197].

9See Ross [46, p. 200].


Example 10.14. Find the Laplace transform of cos bt , where b is a real constant, giventhat the Laplace transform of sin bt is

Lfsin btg D b

s2 C b2

Solution. Let f .t/ D sin bt . Then, f 0.t/ D b cos bt and f .0/ D 0. Then (10.4.7) andthe result of Example 10.5 yields

Lfb cos btg D sLfsin btg � 0 D sb

s2 C b2:

By linearity,

Lfcos btg D s

s2 C b2:

Example 10.15. Find Lfsin2 btg given that Lfsin btg D b=.s2 C b2/.

Solution. Let f .t/ D sin2 bt . Then f .0/ D 0. Substituting into (10.4.7), we obtain

L

�d

dtsin2 bt

D sLfsin2 btg � 0:

Carrying out the differentiation yields

Lnsin2 bt

oD 1

sLf2b.sin bt/.cos bt/g:

By the double-angle identity 2.sin bt/.cos bt/ D sin 2bt . Hence,

Lnsin2 bt

oD 1

sL fb sin 2btg D b

s� 2b

s2 C .2b/2D 2b2

s�s2 C 4b2

� :Now that we have a formula relating the transform of a function to the transform of

its derivative, namely (10.4.7), let’s go back to (10.4.1) to continue retracing the stepsthat led to the solution of the initial value problem (10.1.1). Recall it is our intention tostreamline the roundabout way in which we used Laplace transforms to solve (10.1.1).This will be accomplished through the L operator. As

Lfx0.t/g D sLfx.t/g � x.0/

andx.0/ D 1;

(10.4.1) becomes

sLfx.t/g � 1 C 2Lfx.t/g D 3

s � 1:


It is here that we should point out one of the advantages 10 in using the Laplace trans-form to solve initial value problems. Note that L has transformed (10.1.1), the differ-ential equation and the initial condition, into an algebraic equation:�

x0 C 2x D 3et

x.0/ D 1

L��! sLfx.t/g � 1 C 2Lfx.t/g D 3

s � 1:

Solving the algebraic equation for the unknown Lfx.t/g, we get

.s C 2/Lfx.t/g D 3

s � 1C 1 D s C 2

s � 1

and so

Lfx.t/g D 1

s � 1:

Recall this is (10.1.10) on page 316. Of course, this is the transform of the solutionof (10.1.1), not the solution. Let’s look for a continuous function with this transform.From Table 10.1, we see that

1

s � 1D Lfetg:

Thus,Lfx.t/g D Lfetg:

From this we surmise thatx.t/ D et :

Recall from earlier discussions that this is indeed the unique solution of the initial valueproblem (10.1.1).

When using transforms to solve an initial value problem, we will eventually ar-rive at the point where the transform of the solution x.t/ is expressed in terms of thetransform of some other function, say f .t/. That is,

Lfx.t/g D Lff .t/g:(For example, we found Lfx.t/g D Lfetg in the previous discussion.) So we ask ifx.t/ D f .t/? The answer to this is given by a special version of a result known asLerch’s theorem.11 For continuous functions, it says that different continuous func-tions have different laplace transforms. Therefore, there can only be one continuousfunction whose transform is equal to the transform of the solution.

Uniqueness of Inverse Laplace Transforms

Theorem 10.4 (Lerch’s theorem). Suppose f .t/ and g.t/ are continuous onŒ0;1/ and of exponential order. If Lff .t/g D Lfg.t/g for all s > for some � 0, then f .t/ D g.t/ for all t � 0.

10Other advantages will be pointed out later, such as the efficient manner in which the transform canhandle discontinuous functions.

11Mathias Lerch (1860–1922), born in Czechoslovakia, was a professor of mathematics at the Universityof Freiburg in Germany and at Masaryk University in Czechoslovakia. He published about 240 papers,mostly in analysis and number theory. Some of his results are important to the study of operators, such asthe Laplace transform operator.


This version of Lerch’s theorem is stated and proved in Marsden [46, p. 390; p. 398f.]. We do not prove it here because the proof depends on advanced results from realand complex analysis. A statement of Lerch’s theorem and where its proof can befound is given in Churchill’s Operational Mathematics [16, p. 14; p. 184].

Example 10.16. Let’s try out this method we are in the process of developing on

dx

dt� 2x D 4; x.0/ D 0: (10.4.8)

Solution. First take the Laplace transform of both sides of the above differential equa-tion:

L˚x0 � 2x

� D Lf4g:Next use the initial condition and properties of the transform to express the equationobtained in part (a) in terms of the transform of the solution. First linearity gives

Lfx0g � 2Lfxg D 4Lf1g:Then the transform of the constant function “1” and (10.4.7) yields

sLfxg � x.0/� 2Lfxg D 4 � 1

s:

Since the initial condition is x.0/ D 0, this simplifies to

sLfxg � 2Lfxg D 4

s:

Solving this equation for Lfx.t/g, we obtain the transform of the solution x.t/:

Lfx.t/g D 4

s.s � 2/: (10.4.9)

Now that we have Lfx.t/g, we can obtain the solution x.t/ itself by finding a con-tinuous function with this particular transform. But the right-hand side of (10.4.9) oranything resembling it does not appear in Table 10.1 on page 324. We could look for itin an engineering or math handbook that has a more extensive table of Laplace trans-forms. There we would find an entry for Œ.s � a/.s � b/��1. Even so, we can express(10.4.9) in terms of the functions listed in Table 10.1. We would run into this sameproblem if we had to integrate (10.4.9): in that case we would simply write it as asum of two rational functions before integrating—in other words, we would first applythe method of partial fractions. Let’s try that here. The form of the partial fractionexpansion of (10.4.9) is

4

s.s � 2/D A

s � 2C B

s:

To find A and B, we first multiply by s.s�2/ to clear the fractions of their denominatorsobtaining

As C B.s � 2/ D 4:


Then set s D 2 to find A:

s D 2 ) 2A D 4 ) A D 2:

And s D 0 to find B:

s D 0 ) �2B D 4 ) B D �2:

Therefore,4

s.s � 2/D 2

s � 2� 2

s

and so

Lfx.t/g D 2 � 1

s � 2� 2 � 1

s:

From Table 10.1 and the linearity of the transform, we see that

2 � 1

s � 2� 2 � 1

sD 2Lfe2tg � 2Lf1g D Lf2e2t � 2g:

Since Lfx.t/g D Lf2e2t � 2g, we conclude from Theorem 10.4 that the solution ofthe initial value problem (10.4.8) is

x.t/ D 2e2t � 2:

An examination of how we solved the initial value problems (10.1.1) and (10.4.8)provides the steps for solving any linear initial value problem with Laplace Transforms.

The Method of Laplace Transforms

To solve an initial value problem:

1. Take the Laplace transform of both sides of the differential equation.

2. Use the initial condition and properties of the transform to express the alge-braic equation obtained in Step 1 in terms of the transform of the solution.

3. Solve the equation obtained in Step 2 for the transform of the solution.

4. Determine the solution by finding a continuous function with a transformthe same as the one obtained in Step 3.

Before we work out any more examples, let’s point out the assumptions whichare unsaid but implicit when we employ the above method. In Step 1, we assume theexistence of a function x.t/ that is continuous on Œ0;1/ and satisfies the first-order dif-ferential equation and accompanying initial condition. This is a valid assumption sincefor any pair of continuous functions p and q on Œ0;1/ there is a unique continuoussolution x.t/ of

x0 C p.t/x D q.t/; x.0/ D x0


on Œ0;1/. (See the existence and uniqueness theorem in Chapter 5.) It follows fromthe continuity of the right-hand side of

x0.t/ D q.t/ � p.t/x.t/

that the derivative x 0.t/ is continuous on Œ0;1/. Consequently, x 0.t/ is integrable onŒ0;T � for every T � 0. So some of the hypotheses of Theorem 10.3 are automaticallyfulfilled. This still leaves open the question of the existence of the Laplace transformof x.t/ and whether it is of exponential order. What we do in practice is to go aheadwith the method of Laplace transforms assuming that all the hypotheses are satisfied.If Steps 3 and 4 can be carried out, then we can check if the result of Step 4 fulfills thehypotheses and is a solution.

In the next section we take a look at conditions that allow us to determine theexistence of the Laplace transform of a function and avoid the tedium that comes fromusing Definition 10.1. But first we illustrate the method of Laplace transforms bysolving a couple of initial value problems.

Example 10.17. Solve the initial value problem

dx

dt� x D �t; x.0/ D 1: (10.4.10)

Solution. First take the Laplace transform of both sides of the differential equation andapply the linearity property:

Lfx0g � Lfxg D �Lftg:Then use Theorem 10.3, the entry for tn with n D 1 in Table 10.1, and the alternativenotation X.s/ for Lfx.t/g to get

sX.s/ � x.0/� X.s/ D � 1

s2:

Next use the initial condition x.0/ D 1 and solve this equation for X.s/:

X.s/ D 1

s � 1� 1

.s � 1/s2:

Factor out 1=.s � 1/ and simplify:

X.s/ D 1

s � 1

�1 � 1

s2

�D 1

s � 1

�s2 � 1

s2

�D s C 1

s2D 1

sC 1

s2:

Thus, the transform of the solution x.t/ of the initial value problem (10.4.10) is

Lfx.t/g D 1

sC 1

s2:

To find the solution x.t/ itself, we see from Table 10.1 that

1

sC 1

s2D Lf1g C Lftg:


Consequently,Lfx.t/g D Lf1g C Lftg D Lf1 C tg:

Therefore, by Theorem 10.4,x.t/ D 1 C t:

Example 10.18. Solve

x0 C 4x D 2 � 25 cos 3t ; x.0/ D 2:

Solution. Taking the Laplace transform, we get

Lfx0g C 4Lfxg D 2 Lf1g � 25 Lfcos 3tg:As in the previous example, use Table 10.1 and the transform of the first derivativeproperty:

sX.s/ � x.0/C 4X.s/ D 2 � 1

s� 25 � s

s2 C 32;

where X.s/ D Lfx.t/g. Solving for X.s/, we get

X.s/ D 2

s C 4C 2

s.s C 4/� 25s

.s C 4/.s2 C 9/:

Now we need to determine the solution by finding a continuous function with this sametransform. Unfortunately, the second and third terms are not listed in Table 10.1. How-ever, we can still use the table by expressing these terms as sums of partial fractions.First let’s expand the second term by expressing it as the sum

2

s.s C 4/D A

sC B

s C 4:

To obtain the values of the coefficients, we multiply both of its sides by s.s C 4/:

A.s C 4/C Bs D 2:

Now we can readily obtain the values of A and B from this equation by setting s D 0

and s D �4:

s D 0 ) 4A D 2 ) A D 1

2

s D �4 ) �4B D 2 ) B D �1

2:

Thus,2

s.s C 4/D 1

2� 1

s� 1

2� 1

s C 4: (10.4.11)

The partial fraction expansion of the third term is

25s

.s C 4/.s2 C 9/D C

s C 4C Ds C E

s2 C 9:

10.5. EXISTENCE OF THE LAPLACE TRANSFORM 337

Multiplying by .s C 4/.s 2 C 9/ gives

C.s2 C 9/C .Ds C E/.s C 4/ D 25s:

Successively setting s equal to �4, 0, and 1 gives

s D �4 W 25C D �100 ) C D �4

s D 0 W 9C C 4E D 0 ) E D �9

4C D 9

s D 1 W 10C C 5.D C E/ D 25 ) D D 5 � 2C � E D 5 � 2.�4/ � 9 D 4:

Therefore,

25s

.s C 4/.s2 C 9/D �4 � 1

s C 4C 4s C 9

s2 C 9(10.4.12)

D �4 � 1

s C 4C 4 � s

s2 C 9C 3 � 3

s2 C 9:

Now let’s substitute the two partial fraction expansions, (10.4.11) and (10.4.12), for thesecond and third terms of X.s/ respectively and then simplify:

X.s/ D 1

2� 1

sC 11

2� 1

s � .�4/� 4 � s

s2 C 32� 3 � 3

s2 C 32

D 1

2Lf1g C 11

2Lfe�4tg � 4 Lfcos 3tg � 3 Lfsin 3tg:

This is the same as

Lfx.t/g D L

�1

2C 11

2e�4t � 4 cos 3t � 3 sin 3t

by the linearity property. Finally, from the uniqueness of transforms property, we con-clude that the solution is

x.t/ D 1

2C 11

2e�4t � 4 cos 3t � 3 sin 3t :

10.5 Existence of the Laplace Transform

In this section we state and prove some results guaranteeing the existence of Laplacetransforms for certain types of functions. The following theorem is one of those results.

Existence of the Laplace Transform I

Theorem 10.5. If a function f .t/ is bounded and integrable on every finite in-terval Œ0;T � and is of exponential order , then its Laplace transform exists fors > .


Proof. To prove existence, we have to show the improper integralR1

0e�stf .t/dt con-

verges for s > . That is, we must proveR T

0e�stf .t/dt exists for every T > 0 and

s > and that it approaches a finite limit as T ! 1. Clearly the integral exists onŒ0;T � since e�stf .t/ is bounded and integrable on Œ0;T �.12 And as f .t/ is of expo-nential order , there are constants M and t1 such that jf .t/j � Me� t for all t � t1.In fact, since f .t/ is bounded on Œ0;T � we can even say that

jf .t/j � Me� t

for all t � 0 by increasing the value of M if necessary. Now we can address thequestion of the convergence of

R10

e�stf .t/dt for s > . Observe that the integrandis dominated by the exponential function Me �.s��/t for t � 0 since

je�stf .t/j D e�st jf .t/j � e�stMe� t � Me�.s��/t : (10.5.1)

Then as s > ,

limT !1

Z T

0

Me�.s��/t dt D limT !1

�M

�.s � /e�.s��/t

�T

0

D M

s � : (10.5.2)

By a well-known comparison test for improper integrals, 13R1

0e�stf .t/dt converges.

That is, Lff .t/g exists for s > .

Example 10.19. In Example 10.11, we showed

g.t/ D �6te3t=2 sin t

is of exponential order 2:5. Since g is continuous everywhere, it is integrable on ev-ery finite interval Œ0;T �. Therefore, the theorem allows us to conclude that Lfg.t/gexists for s > 2:5 without having to go through the trouble of finding its transform.Sometimes this is all we need to know.

Example 10.20. The function tu.t � b/ for b > 0 is discontinuous at t D b; neverthe-less, it is integrable on every finite interval Œ0;T �. Since for t > 0,

jtu.t � b/j � t < et ;

the function is of exponential order 1. Therefore, its transform exists for s > 1.

The proof of Theorem 10.5 suggests:

12The product of two integrable functions is integrable. See Ross [46, p. 197] or Fulks [25, p. 120].13If g and h are integrable on Œt 1; T � for every T > t1 and if jg.t/j � h.t/ for all t � t1, then

R1

t1g.t/ dt

converges ifR1

t1h.t/dt converges. For more information, see a real analysis textbook, such as Bartle [4,

p. 262] or Fulks [25, p. 468, p. 472]. Weaker versions of the comparison test are stated in elementary calculustextbooks, such as Stewart [53, p. 572] or Thomas and Finney [55, p. 525].

10.5. EXISTENCE OF THE LAPLACE TRANSFORM 339

Long-Term Behavior of Laplace Transforms

Corollary 10.1. If a function f .t/ satisfies the hypotheses of Theorem 10.5, then

lims!1 Lff g.s/ D 0: (10.5.3)

Proof. We see from (10.5.1) and (10.5.2) that

jLff g.s/j DˇˇZ 1

0

e�stf .t/dt

ˇˇ �

Z 1

0

ˇe�stf .t/

ˇdt � M

s � for s > . Consequently, the transform is bounded:

0 � jLff g.s/j � M

s � :

Since the lower bound is 0 and the upper bound approaches 0 as t ! 1, (10.5.3)follows.

Take for example the function

F.s/ D s

s � 1:

This result implies that there is no bounded, integrable function of exponential orderthat has F.s/ as its Laplace transform. And that is because

lims!1 F.s/ D lim

s!1s

s � 1D 1:

Corollary 10.1 will be put to good use later on in Section 10.8.1 to help solvedifferential equations with variable coefficients.

10.5.1 Piecewise Continuous Functions

Continuous functions suffice for modeling many phenomena and situations. Neverthe-less, other models require a broader class of functions. One useful class is the set offunctions that are discontinuous at isolated points but whose Laplace transforms ex-ist. In order to fully describe these functions, we must first understand what is meantby a jump discontinuity. A function f .t/ has a jump discontinuity at t D � if it isdiscontinuous at � but the left-and right-hand limits

limt!��

f .t/ and limt!�C

f .t/

exist and are finite. If the point t D � is an endpoint of an interval, then we still sayf .t/ has a jump discontinuity at � if the appropriate one-sided limit exists and is finite.


Piecewise Continuous Functions

Definition 10.3.

A function f .t/ is piecewise continuous on a finite interval Œa; b� if it iscontinuous at each point in Œa; b� except for at most a finite number of jumpdiscontinuities..

A function f .t/ is piecewise continuous on the infinite interval Œ0;1/ ifit is piecewise continuous on every finite interval Œ0;T �.

Example 10.21. The unit step function at b > 0 is continuous on Œ0;1/ except at b.At this point, u.t �b/ has a jump discontinuity since the following one-sided limits arefinite:

limt!b�

u.t � b/ D 0 and limt!bC

u.t � b/ D 1:

Hence, u.t � b/ is piecewise continuous on Œ0;1/.

Existence of the Laplace Transform II

Corollary 10.2. If a function f .t/ is piecewise continuous on Œ0;1/ and of ex-ponential order , then its Laplace transform exists for s > .

Proof. Let Œ0;T � be any finite interval. It follows from the definition of piecewisecontinuity that f .t/ is bounded and has at most a finite number of jump discontinuitieson Œ0;T �. This implies f .t/ is integrable on Œ0;T �.14 By Theorem 10.5, Lff .t/g existsfor s > .

10.6 Properties of the Laplace Transform

We have already found two properties of the Laplace transform that are needed to solveinitial value problems:

the linearity property for transforms of linear combinations of functions:

Lfc1f1.t/ C c2f2.t/g D c1Lff1.t/g C c2Lff2.t/g

the property for taking the transform of the first derivative of a function:

Lff 0.t/g D sLff .t/g � f .0/:

Now we will derive some other very useful properties. The first two of these involveshifting (translating) along the s-axis and the t-axis.

14See Bartle and Sherbert [5, p. 252], Ross [46, p. 196], or Gaughan [26, p. 167].

10.6. PROPERTIES OF THE LAPLACE TRANSFORM 341

10.6.1 Shift in the s-Domain

Oftentimes we need the transform of a product of two functions, one of which is anexponential function while the other is a function whose Laplace transform we alreadyhave. In other words, the product has the form ebtf .t/, where b is a constant and Lff gis known or can easily be found. This particular product is important; one reason beingthat it may turn up as a solution of a linear equation. For example, you can check thate5t sin t is a solution of Py � 5y D e 5t cos t .

Now that we have established the importance of knowing the transform of e btf .t/,let’s get to the business of finding it. First, we assume that Lff .t/g exists and is equalto a function F.s/ for s > a. By (10.1.12),

Lfebtf .t/g DZ 1

0

e�st � ebt dt DZ 1

0

e�.s�b/tf .t/dt:

Since

F.s/ D Lff .t/g DZ 1

0

e�stf .t/dt;

it follows thatLfebtf .t/g D F.s � b/

for s � b > a. In other words, when f is multiplied by e bt , its Laplace transform canbe obtained by merely shifting the argument of the Laplace transform of f by b units.Since the shift takes place on the s-axis, we refer to this property as the shift in thes-domain. Shortly, we will take a look a shift property that takes place on the t-axis.

Shift in the s-Domain

Theorem 10.6. If the Laplace transform of f .t/ exists and is equal to the functionLff .t/g D F.s/ for s > a, then

Lfebtf .t/g D F.s � b/ (10.6.1)

for s > a C b.

Example 10.22. Find Lfe2t cos 3tg.

Solution. We begin with

Lfcos 3tg D s

s2 C 32D F.s/

for s > 0. In the notation of (10.6.1), a D 0 and b D 2. Thus,

Lfe2t cos 3tg D F.s � 2/ D s � 2

.s � 2/2 C 32

for s > a C b D 2.


Example 10.23. Find Lfe�t t5g.

Solution. Start with

Lft5g D 5!

s6D F.s/:

Now set b D �1 in (10.6.1):

Lfe�tt5g D F.s C 1/ D 5!

.s C 1/6

for s > �1.

Sometimes we use the shift property backwards as the next example illustrates.

Example 10.24. Find a function f with the Laplace transform

Lff .t/g D 1

s2 � s C 1:

Solution. Although 1=.s 2 � s C 1/ is not listed directly in Table 10.1, it is related tothe sin bt entry. To see this, complete the square of its denominator to write it in theform b=.s2 C b2/:

1

s2 � s C 1D 1

.s � 12/2 C 3

4

D 2p3

�p

32

.s � 12/2 C .

p3

2/2:

From Table 10.1, we seep

32

s2 C .p

32/2

D Lfsinp

32

tg:

By the shift property (10.6.1),

p3

2

.s � 12/2 C .

p3

2/2

D Lfe 12

t sin .p

32

t/g :

Thus,

f .t/ D 2p3

e12 t sin .

p3

2t/:

10.6.2 Shift in the t-Domain

Abrupt changes occur in a variety of physical situations. Examples that we consideredin Chapter 7 were:

culling a certain number of animals from a herd,


hitting a baseball with a bat, and

depositing money to an account at regular intervals.

There are other examples which you will most likely study in the future, such as turninga switch on and off in an electrical circuit.

Situations involving abrupt changes such as those cited above are often modeledwith linear nonhomogeneous differential equations where the nonhomogeneous termsinvolve functions with jump discontinuities. One such function is the unit step functionu.t � b/. Recall its Laplace transform is

Lfu.t � b/g D e�bs

s;

where b is a non-negative constant. With this transform we can solve linear equationswith nonhomogeneous terms that are constant multiples of u.t � b/. But what aboutequations with more complicated nonhomogeneous terms, such as:

f .t/u.t � b/ D(

0; if t < b

f .t/; if t > b?(10.6.2)

To answer this, let’s start off assuming the transform of f .t/ exists for s > a. That is,assume

Lff .t/g.s/ DZ 1

0

e�stf .t/dt

exists for s > a. Applying the definition of the Laplace transform to (10.6.2), we have

Lff .t/u.t � b/g DZ 1

0

e�stf .t/u.t � b/dt DZ 1

b

e�stf .t/dt;

which is the result of the integrand being equal to 0 when t < b and to e �stf .t/ whent > b. Now move the lower limit of integration back to 0 with a change of variablesin order to rewrite the integral as a Laplace transform. This is accomplished by letting� WD t � b so that � D 0 when t D b. Consequently,

Lff .t/u.t � b/g DZ 1

0

e�s.�Cb/f .� C b/d� D e�sb

Z 1

0

e�stf .t C b/dt:

Note that the integral on the right-hand side is the transform of the function f .t C b/.Thus, we have shown

Lff .t/u.t � b/g D e�bsLff .t C b/g:We can use this to find other Laplace transforms. Since it involves shifting the argumentof f , the domain of which is on the t-axis, we call it the shift in the t-domain property.Applying the foregoing steps to f .t � b/u.t � b/, we obtain the alternate form

Lff .t � b/u.t � b/g D e�bsLff .t/g:This form of the shift property is particularly useful when solving certain initial valueproblems.


Shift in the t-Domain

Theorem 10.7. Let b � 0. If the Laplace transform of f .t/ exists and is equalto the function Lff .t/g.s/ for s > a, then the Laplace transform of f .t/u.t � b/

exists for s > a and is given by the shift formula

Lff .t/u.t � b/g D e�bsLff .t C b/g: (10.6.3)

A variant of the shift formula suited for solving initial value problems is

Lff .t � b/u.t � b/g D e�bsLff .t/g: (10.6.4)

Example 10.25. Find the Laplace transform of e2t u.t � 1/.

Solution. Using the notation of (10.6.3), f .t/ D e 2t and b D 1. Thus,

f .t C b/ D f .t C 1/ D e2.tC1/ D e2 � e2t

and so

Lff .t C b/g D e2 � Lfe2tg D e2

s � 2:

Therefore, by (10.6.3)

Lfe2tu.t � 1/g D e�s � e2

s � 2D e2�s

s � 2:

Example 10.26. Find the Laplace transform of

g.t/ D8<:0; if t < �

sin.0:5t/; if t > � .

Solution. The function g has a jump discontinuity at t D � . In terms of the unit stepfunction at � , it is

g.t/ D sin.0:5t/ � u.t � �/:In terms of the notation of (10.6.3), f .t/ D sin.0:5t/ and b D � and so

f .t C �/ D sinŒ0:5.t C �/� D sin

�t

2C �

2

�D cos

t

2:

Hence,

Lfsin.0:5t/ � u.t � �/g D e��sL

�cos

t

2

D e��s � s

s2 C 14

:


Example 10.27. Find the solution of

Px C 9x D 2ı.t � 3/

satisfying the initial condition x.0/ D �5.

Solution. Taking the Laplace transform and using the linearity property as well as thetransform of the first derivative property, we obtain

sX.s/ � x.0/C 9X.s/ D 2e�3s;

where as usual we use the alternate notation X.s/ in lieu of Lfx.t/g. As x.0/ D �5,we have

.s C 9/X.s/ C 5 D 2e�3s

and so

X.s/ D � 5

s C 9C 2

e�3s

s C 9: (10.6.5)

The second term on the right-hand side of (10.6.5) involves e �bsLff .t/g where b D 3

and Lff .t/g D 1=.s C 9/. Hence, f .t/ D e�9t . According to shift property (10.6.4),

e�bsLff .t/g D Lff .t � b/u.t � b/g:Since f .t � 3/ D e�9.t�3/,

e�3s

s C 9D e�3sLfe�9tg D Lfe�9.t�3/u.t � 3/g:

It follows that (10.6.5) can be written as

Lfx.t/g D Lf�5e�9t C 2e�9.t�3/u.t � 3/g:Therefore, the solution of the initial value problem is

x.t/ D �5e�9t C 2e�9.t�3/u.t � 3/ D(

�5e�9t ; if t < 3

�5e�9t C 2e�9.t�3/; if t > 3:

10.6.3 Derivatives of Transforms

Is the derivative of the Laplace transform of a function of any use? Let’s look into this.Formally differentiating the Laplace transform F.s/ of a function f .t/ with respect tos, we obtain

F 0.s/ D d

dsF.s/ D d

dsLff .t/g.s/ D d

ds

Z 1

0

e�stf .t/dt (10.6.6)

DZ 1

0

@

@sŒe�stf .t/� dt D

Z 1

0

e�st .�t/f .t/dt D Lf�tf .t/g:


The step where we move d=ds after the integral sign and change it to @=@s reverses theorder of differentiation and integration. This seems reasonable but that in itself doesnot make it true. Although we are not in a position to prove it because of the advancedmathematics that would be needed, this step is valid for the integrand e �stf .t/ if thefunction f .t/ is piecewise continuous and of exponential order. 15 Consequently, if thatis the case, we conclude from (10.6.6) that

Lf�tf .t/g D F 0.s/: (10.6.7)

Are there similar formulas when higher powers of t are involved? To answer that,consider Lf.�t/2f .t/g. By (10.6.7),

Lf.�t/2f .t/g D Lf.�t/Œ�tf .t/�g D d

dsLf�tf .t/g D d

dsŒF 0.s/� D F 00.s/:

Similarly, it can be shown that Lf.�t/3f .t/g D F 000.s/. The pattern that is developingsuggests the formula

Lf.�t/nf .t/g D F .n/.s/ (10.6.8)

when n is a positive integer. We now prove this is the case by mathematical induction.Suppose (10.6.8) is true for an integer n D k (as it is for n D 1; 2; 3/. Then forn D k C 1, it follows from (10.6.7) and (10.6.8) that

Lf.�t/kC1f .t/g D Lf.�t/Œ.�t/kf .t/�gD d

dsLf.�t/kf .t/�g D d

dsF .k/.s/ D F .kC1/.s/:

So if (10.6.8) is true for n D k , logic demands that it also be true for n D k C 1.Therefore, since it has already been established that (10.6.8) is true for n D 1; 2; 3, itfollows that it also must be true for the rest of the positive integers.

In short, because of the validity of differentiating inside the integral sign and theinduction argument, we have the following property.

Derivatives of Transforms

Theorem 10.8. Suppose a function f .t/ is piecewise continuous on Œ0;1/ andof exponential order . Let F.s/ D Lff .t/g. Then

Lftnf .t/g D .�1/ndn

dsnF.s/ (10.6.9)

for s > and n D 1; 2; 3; : : : .

Example 10.28. Find Lft sin btg.

15Churchill [16, Sec. 12] states and proves such a result.


Solution. Clearly Theorem 10.8 applies here since sin bt is multiplied by t and weknow that Lfsin btg D b=.s2 C b2/. In the notation of the theorem, f .t/ D sin bt andF.s/ D b=.s2 C b2/. Setting n D 1 in (10.6.9), we have

Lft sin btg D � d

dsF.s/ D � �2bs

.s2 C b2/2:

Therefore,

Lft sin btg D 2bs

.s2 C b2/2:

Example 10.29. Find Lft 2e3tg.

Solution. Since Lfe3tg D 1=.s � 3/,

Lft2e3tg D Lf.�t/2e3tg D d2

ds2

�1

s � 3

�D d

ds

� �1

.s � 3/2

�D 2

.s � 3/3:

Another approach is to use the shift property in the s-domain; it is also less work. SinceLft2g D 2=s3, it immediately follows from shifting s that

Lft2e3tg D 2

.s � 3/3:

10.6.4 Integral of a Transform

We just saw that multiplying a function f by t corresponds to the negative of thederivative of the transform of f . Now we will see that dividing f by t corresponds toan integral of its transform–precisely what integral is the subject of the next theorem.

Integral of a Transform

Theorem 10.9. Suppose a function f .t/ is piecewise continuous on Œ0;1/ andof exponential order . Furthermore, suppose

limt!0C

f .t/

t(10.6.10)

exists and is finite. Then

L

�f .t/

t

D

1Zs

F.u/du; (10.6.11)

where F.s/ D Lff g.s/.


Proof. Integrating F.u/ D R10

e�utf .t/dt from u D s to u D T where T > s, weobtain Z T

s

F.u/du DZ T

s

�Z 1

0

e�utf .t/dt

�du:

The hypotheses of Theorem 10.9 allows the order of integration to be reversed:16

Z T

s

F.u/du DZ 1

0

f .t/

Z T

s

e�ut du

!dt: (10.6.12)

The piecewise continuity of f .t/ on Œ0;1/ and (10.6.10) implies that f .t/=t is piece-wise continuous on Œ0;1/. Since f .t/ is of exponential order, it is easy to show thatf .t/=t is of exponential order. Thus, the Laplace transform of f .t/=t exists. Becauseof this and as Z T

s

e�ut du D � 1

te�ut

ˇˇTs

D 1

te�st � 1

te�T t ;

we can write (10.6.12) asZ T

s

F.u/du DZ 1

0

f .t/

te�st dt �

Z 1

0

f .t/

te�T t dt

D L

�f .t/

t

.s/ � L

�f .t/

t

.T /:

This implies (10.6.11) since Lff .t/=tg.T / ! 0 as T ! 1 by (10.5.3).

Example 10.30. Find the Laplace transform ofet � 1

t.

Solution. In the notation of Theorem 10.9,

f .t/ D et � 1

and

F.s/ D Lff .t/g D 1

s � 1� 1

s:

By L’Hopital’s Rule,

limt!0C

et � 1

tD lim

t!0Cet D 1:

Consequently, the hypotheses of Theorem 10.9 are satisfied. Integrating F.s/, we ob-tain Z T

s

F.u/du DZ T

s

�1

u � 1� 1

u

�du D ln

�u � 1

u

�ˇˇTs

D ln

�T � 1

T

�� ln

�s � 1

s

�:

16The reason for this is found in Churchill [16, Sec. 18]. However, it may be difficult for a reader at thislevel to understand since uniform convergence is used, a concept that is introduced in advanced calculus orreal analysis textbooks, such as Bartle and Sherbert [5], Fulks [25], and Ross [46].


Thus, Z 1

s

F.u/du D limT !1

Z T

s

F.u/du D � ln

�s � 1

s

�D ln

s

s � 1

as

limT !1

ln

�T � 1

T

�D lim

T !1ln

�1 � 1

T

�D ln 1 D 0:

Therefore,

L

�et � 1

t

D ln

s

s � 1

:

Example 10.31. Find

L

�sin bt

t

;

where b is a constant.

Solution. The hypotheses of Theorem 10.9 are satisfied since sin bt is continuous,sin bt D O.1/, and (10.6.10) holds:

limt!0C

sin bt

tD lim

t!0C

�b � b3t2

3!C b5t4

5!� b7t6

7!C � � �

�D b:

By (10.6.11),

L

�sin bt

t

DZ 1

s

b

u2 C b2du D lim

T !1

�tan�1 T

b� tan�1 s

b

�D �

2� tan�1 s

b:

As tan�1 x C tan�1.x�1/ D �=2,

L

�sin bt

t

D tan�1 b

s: (10.6.13)

10.6.5 Transform of an Integral

Assuming a function f .t/ is piecewise continuous on Œ0;1/ and of exponential or-der , another useful property can be derived by applying the Transform of the FirstDerivative formula (10.4.7) to the integral function g.t/ D R t

0f .�/d� . Since f is

integrable on Œ0;1/, g is continuous on Œ0;1/; moreover, by one of the versions ofthe Fundamental Theorem of Calculus, g is differentiable at those places where f iscontinuous, in which case17

g0.t/ D d

dt

Z t

0

f .�/d� D f .t/:

17See Bartle and Sherbert [5, pp. 257-258], Ross [46, p. 200], or Gaughan [26, p. 160].


From this it follows that g 0 is piecewise continuous on Œ0;1/.Is g of exponential order? If it is, then this along with its piecewise smoothness

will guarantee the existence of its transform. Since f is of exponential order , thereare constants M1 > 0 and t1 � 0 such that jf .�/j � M1e�� for � � t1. If the constant is not already positive, we can certainly replace it with a positive number. Sincef is piecewise continuous, it is bounded on the finite interval Œ0; t 1�. In other words,there is a constant M2 such that jf .�/j � M2 for 0 � � � t1. Even more so then,jf .�/j � M2e�� on Œ0; t1�. Letting M WD maxfM1;M2g, we have jf .�/j � Me�� onŒ0;1/. Since �jf .�/j � f .�/ � jf .�/j,ˇ

ˇZ t

0

f .�/d�

ˇˇ �

Z t

0

jf .�/j d�:

Thus,

jg.t/j �Z t

0

jf .�/j d� � M

Z t

0

e�� d� � M

e� t

for all t � 0. Hence, g is of exponential order for s > .Generally there are points in Œ0;1 where g0 is not differentiable; that depends

on f and the location of its discontinuities. However, as we concluded above, g 0 ispiecewise continuous on Œ0;1/. This, that g.t/ D O.e� t /, and the continuity of g onŒ0;1/ guarantee that formula (10.4.7) still applies.18 Therefore,

Lfg0.t/g D sLfg.t/g � g.0/

for s > . Since g.0/ D 0,

Lff .t/g D sLfg.t/g D sL

�Z t

0

f .�/d�

:

As a result, we have the following property.

Transform of an Integral

Theorem 10.10. If f .t/ is piecewise continuous on Œ0;1/ and of exponentialorder , then

L

�Z t

0

f .�/d�

D 1

sLff .t/g (10.6.14)

for s > .

Example 10.32. For f .t/ in (10.6.14), let f .t/ � 1 on Œ0;1/. It clearly satisfies thehypotheses of Theorem 10.10. Hence,

L

�Z t

0

1 d�

D 1

sLf1g:

18See Churchill [16, p. 8].


SinceR t

0d� D t and Lf1g D 1=s, we conclude

Lftg D 1

s2for s > 0:

Even though we obtained this result many pages ago in Section 10.2, we chose it asour first example in order to become comfortable with Theorem 10.10 with a familiarresult. The next example is new and not as easy.

Example 10.33. Find the Laplace transform of the sine-integral function

Si .t/ DZ t

0

sin �

�d�:

Solution. The integrand t �1 sin t is piecewise continuous on Œ0;1/ as

limt!0C

sin t

tD 1:

And it is of exponential order D 0 becauseˇˇ sin t

t

ˇˇ � 1

t� 1 for t � 1:

Consequently, the conditions of Theorem 10.10 are satisfied, By (10.6.10),

L fSi.t/g D L

�Z t

0

sin �

�d�

D 1

sL

�sin t

t

D 1

stan�1 1

s

for s > 0, where we have used (10.6.13) in the last step.

10.6.6 Transform of a Periodic Function

In this section, we will learn how to find Laplace transforms of piecewise continuousfunctions that are also periodic. A function f is said to be periodic on Œ0;1/ if aconstant b exists such that

f .t C b/ D f .t/ (10.6.15)

at all values of t 2 Œ0;1/ for which f .t/ is defined. The smallest positive value of b

for which (10.6.15) holds is called the period of f . For example, the cosine and sinefunctions are periodic with period 2� .

For a given periodic function f with period b, let fE be the function that extendsthe rule for f on the interval Œ0; b/ to the entire interval Œ0;1/.

Example 10.34. Let f be the periodic function with period 1, where the rule for f onŒ0; 1/ is

f .t/ D t:

Then the function fE isfE.t/ D t

for 0 � t < 1. The graphs of f and fE are shown side by side in Figure 10.1.


Fig. 10.1: The graphs of f and fE

Transform of a Periodic Function

Theorem 10.11. If f .t/ is periodic with period b and piecewise continuous onŒ0; b/ , then

Lff .t/g D 1

1 � e�bs

Z b

0

e�stf .t/dt (10.6.16)

D 1

1 � e�bs

hLffE.t/g � e�bsLffE.t C b/g

i; (10.6.17)

where fE is the function that extends the rule for f on Œ0; b/ to Œ0;1/.

Proof. Piecewise continuity of f on Œ0; b/ and its periodicity of b imply f is boundedand piecewise continuous on all of Œ0;1/. Therefore, by Corollary 10.2, the transformof f exists, where

Lff .t/g DZ b

0

e�stf .t/dt CZ 1

b

e�stf .t/dt:

With the change of variable u D t � b, the improper integral becomesZ 1

b

e�stf .t/dt DZ 1

0

e�s.uCb/f .u C b/du D e�sb

Z b

0

e�suf .u/du

as the period of f is b. Therefore,

Lff .t/g DZ b

0

e�stf .t/dt C e�sbLff .t/g;


from which (10.6.16) follows. To obtain (10.6.17), first observe thatZ b

0

e�stf .t/dt DZ b

0

e�stfE.t/dt DZ 1

0

e�stfE.t/Œ1 � u.t � b/� dt

as fE.t/ � f .t/ on Œ0; b� and

fE.t/Œ1 � u.t � b/� D8<:fE.t/; if 0 � t < b

0; if t > b.

Thus,Z b

0

e�stf .t/dt DZ 1

0

e�stfE.t/dt �Z 1

0

e�stfE.t/u.t � b/dt

D LffE.t/g � LffE.t/u.t � b/g D LffE.t/g � e�bsLffE.t C b/gby the shift in the t-domain property.

Example 10.35. Find the Laplace transform of the square-wave function whose graphis shown in Figure 10.2.

Fig. 10.2: A periodic square wave

Solution. In order to apply Theorem 10.11, we must have a formula for the function.Let’s call it f . It is apparent from the graph that f is periodic with period 2 and

f .t/ D8<:1; 0 < t < 1

0; 1 < t < 2


for 0 < t < 2. By (10.6.16) with b D 2:

Lff .t/g D 1

1 � e�2s

Z 2

0

e�stf .t/dt

D 1

1 � e�2s

Z 1

0

e�st � 1 dt CZ 2

1

e�st � 0 dt

!

D 1

1 � e�2s

�1

s� 1

se�s

�D 1 � e�s

s.1 � e�2s/:

Example 10.36. Find the Laplace transform of the function f that is periodic withperiod 1 and with the values

f .t/ D t

for 0 � t < 1.

Solution 1. We have already seen the graph of f in Figure 10.1. By (10.6.16) withb D 1,

Lff .t/g D 1

1 � e�s

Z 1

0

e�st t dt:

Integration by parts yieldsZ 1

0

e�st t dt D 1

s2.1 � e�s/ � 1

se�s:

Thus,

Lff .t/g D 1

1 � e�s

�1 � e�s

s2� 1

se�s

�D 1

s2� 1

s.es � 1/:

Solution 2. Alternatively, we can use (10.6.17) and avoid integrating by parts. Sinceb D 1 and fE.t/ D t for 0 � t < 1,

fE.t C b/ D fE.t C 1/ D t C 1:

Thus,

Lff .t/g D 1

1 � e�sŒLftg � e�sLft C 1g� D 1

1 � e�s

�1

s2� e�s

�1

s2� 1

s

��

D 1

s2� 1

s.es � 1/:

Example 10.37. Find Lfg.t/g, where g.t/ is periodic with period � and

g.t/ D sin t

for 0 � t < � . Its graph is the full-rectified sine wave shown in Figure 10.3. 19


Fig. 10.3: A full-rectified sine wave

Solution. Applying (10.6.17) with b D � and g E.t/ D sin t , we have

Lfg.t/g D 1

1 � e��s

�LfgE.t/g � e��sLfgE.t C �/g�

D 1

1 � e��s

�Lfsin.t/g � e��sLfsin.t C �/g�:

As sin.t C �/ D � cos t ,

Lfg.t/g D 1

1 � e��s

�1

s2 C 1C e��s s

s2 C 1

�D 1 C e��s

.s2 C 1/.1 � e��s/:

10.6.7 Convolution

Consider the initial value problem

dx

dt� x D g.t/; x.0/ D 0; (10.6.18)

where g is piecewise continuous and of exponential order on Œ0;1/. Taking the trans-form of both sides of the equation in (10.6.18), using the initial condition, and solvingfor Lfx.t/g, we get

Lfx.t/g D 1

s � 1Lfg.t/g

orLfx.t/g D Lfetg Lfg.t/g (10.6.19)

19A full-rectified sine wave is obtained from a sine wave by inverting all of its negative (or positive)half-cycles.


as 1=.s � 1/ D Lfetg. This says the transform of the solution x.t/ of (10.6.18) is theproduct of the transform of e t and that of g.t/. At this point we do not yet know whatto do next to find x.t/ itself even if g.t/ were known. So let’s leave this for a momentand solve (10.6.18) by the integrating factor method or with the variation of parametersformula given in Chapter 5. Using either method, we find

x.t/ DZ t

0

et�ug.u/du: (10.6.20)

Comparing the Laplace transform of (10.6.19) to (10.6.20), we see

Lfetg Lfg.t/g D L

�Z t

0

et�ug.u/du

: (10.6.21)

The point to be made is that the solutions of some initial value problems, such as(10.6.18), can be expressed by integrals of the formZ t

0

f .t � u/g.u/du:

In this example, f .t/ D et . Integrals of this type are known as convolution integralsas we formally define next.

Convolution Integrals

Definition 10.4. Let f .t/ and g.t/ be piecewise continuous functions on Œ0;1/.The convolution of f and g is the integral function h defined by

h.t/ WDZ t

0

f .t � u/g.u/du: (10.6.22)

Example 10.38. Find the convolution of f .t/ D t and g.t/ D e t .

Solution. By (10.6.22) and integration by parts, we obtain

h.t/ DZ t

0

.t � u/eu du D �.t � u/eu C eu

�t0

D et � t � 1:

Instead of h or some other letter to denote the convolution integral, it is standardpractice to use the symbol f g:

.f g/.t/ WDZ t

0

f .t � u/g.u/du:

With this notation, the convolution of t and e t is written as follows:

t et DZ t

0

.t � u/eu du D et � t � 1:

Convolution has the properties of commutativity, associativity, and distributivity asset forth in the next theorem.


Properties of Convolution

Theorem 10.12. If f .t/, g.t/, and h.t/ are piecewise continuous functions onŒ0;1/, then the convolution operator has the following properties:

(i) Commutativity: f g D g f(ii) Associativity: f .g h/ D .f g/ h

(iii) Distributivity: f .g C h/ D f g C g h

(iv) 0 f D f 0 D 0

Proof. To prove (i), simply change variables: v D t � u. Then

.f g/.t/ DZ t

0

f .t � u/g.u/du D �Z t

0

f .v/g.t � v/dv

DZ t

0

g.t � v/f .v/dv D .g f /.t/:

The proof of (iv) is trivial:

.f 0/.t/ DZ t

0

f .t � u/ � 0 du DZ t

0

0 du D 0;

which together with (i) proves (iv). The proofs of (ii) and (iii) are left as exercises.

Even though it appears from Theorem 10.12 that the convolution operator has allof the properties of ordinary multiplication, it does not! For example, 1 �1 D 1 whereas1 1 ¤ 1 as

1 1 DZ t

0

1 � 1 du D t:

For the exponential function f .t/ D e t , we determined from (10.6.21) that

L f.f g/.t/g D Lff .t/g Lfg.t/g:In fact, as we prove next, this is true for any two piecewise continuous functions ofexponential order.

Convolution Theorem

Theorem 10.13. Let f .t/ and g.t/ be piecewise continuous functions on Œ0;1/

and of exponential order . Then

Lf.f g/.t/g D Lff .t/g Lfg.t/g (10.6.23)

for s > .


Proof. By Corollary 10.2, the Laplace transforms

Lff .t/g DZ 1

0

e�stf .t/dt DW F.s/

and

Lfg.t/g DZ 1

0

e�st g.t/dt DW G.s/

exist for s > . Starting with the left-hand side of (10.6.23), we have

Lff .t/g Lfg.t/g D F.s/

Z 1

0

e�stg.t/dt

DZ 1

0

e�stF.s/g.t/dt DZ 1

0

e�suF.s/g.u/du:

Note that we can move F.s/ inside the above integral because it does not depend onthe variable of integration. Consequently,

Lff .t/g Lfg.t/g DZ 1

0

e�su

�Z 1

0

e�stf .t/dt

�g.u/du

DZ 1

0

�Z 1

0

e�s.tCu/f .t/dt

�g.u/du:

With the change of variable v D t C u, this becomes


0

�Z 1

u

e�svf .v � u/dv

�g.u/du: (10.6.24)

The region of integration is shown in Figure 10.4. It is the shaded v-shaped region thatextends infinitely to the right.

Fig. 10.4: Region of integration for (10.6.24)

This is where the proof gets technical and goes beyond the mathematical level of thisbook. Suffice it to say that it can be argued that it is legitimate to change the order ofintegration in (10.6.24) (cf. [16, Sec. 13]). Thus,



0

e�sv

�Z v

0

f .v � u/g.u/du

�dv

DZ 1

0

e�sv.f g/.v/dv DZ 1

0

e�st .f g/.t/dt

D Lf.f g/.t/gfor s > .

One important application of the convolution theorem (10.6.23) is that it can beused to solve certain types of integral equations. An integral equation is an equa-tion with the unknown function appearing under an integral sign. One such integralequation has the form

x.t/ D f .t/ CZ t

0

k.t � u/x.u/du; (10.6.25)

where f and k are known functions. The function k is called the kernel. This particulartype of equation is referred to as a Volterra integral equation of convolution type. Thephrase “of convolution type” means that the kernel k depends only on the combinationt � u. In the following example, we illustrate how to solve such an equation using theconvolution theorem.

Example 10.39. Solve the Volterra integral equation

x.t/ D t CZ t

0

.t � u/x.u/du: (10.6.26)

Solution. First take the Laplace transform of (10.6.26) and apply (10.6.23):

Lfx.t/g D Lftg C L

�Z t

0

.t � u/x.u/du

D Lftg C Lftg Lfx.t/g D 1

s2C 1

s2Lfx.t/g:

Now solve for Lfx.t/g:

Lfx.t/g�

1 � 1

s2

�D 1

s2) Lfx.t/g D 1

s2 � 1:

From the partial fraction expansion

1

s2 � 1D 1

2

�1

s � 1

�� 1

2

�1

s C 1

�;

we see that1

s2 � 1D 1

2Lfetg � 1

2Lfe�tg:


Therefore,

Lfx.t/g D L

�1

2et � 1

2e�t

from which we conclude the solution is

x.t/ D et � e�t

2D sinh t:

There are also some types of integro-differential equations that can be solved withLaplace transforms. Integro-differential equations involves both derivatives and inte-grals of the unknown function. One type that we can solve with transforms is a Volterraintegro-differential equation of convolution type of the form

x0.t/ D f .t/ CZ t

0

k.t � u/x.u/du: (10.6.27)

Example 10.40. Find the solution of the Volterra integro-differential equation

x0.t/ D t �Z t

0

et�ux.u/du

satisfying the initial condition x.0/ D 0.

Solution. The Laplace transform of the integro-differential equation is

Lfx0.t/g D Lftg � Ln Z t

0

et�ux.u/duo

or

sLfx.t/g � x.0/ D 1

s2� Lfetg Lfx.t/g

by the convolution theorem. Using the initial condition x.0/ D 0 and writing this interms of X.s/ WD Lfx.t/g, we have

sX.s/ D 1

s2� 1

s � 1X.s/ :

Hence,

X.s/ D s � 1

s2.s2 � s C 1/:

A partial fraction expansion of X.s/ is

X.s/ D � 1

s2C 1

s2 � s C 1:

In Example 10.24, we determined that

1

s2 � s C 1D L

n 2p3

e12 t sin .

p3

2t/o:

Therefore, the solution is

x.t/ D �t C 2p3

e12

t sin .p

32

t/ :

10.7. INVERSE LAPLACE TRANSFORMS 361

10.7 Inverse Laplace Transforms

When solving a first-order initial value problem with Laplace transforms, we eventuallyarrive at the point where we have determined the Laplace transform of the solution. Forinstance, in Example 10.18, we considered the initial value problem

x0 C 4x D 2 � 25 cos 3t ; x.0/ D 2:

The transform of the solution, denoted by X.s/, turned out to be

X.s/ D 2

s C 4C 2

s.s C 4/� 25s

.s C 4/.s2 C 9/:

By Lerch’s theorem, there is only one continuous function x.t/ whose Laplace trans-form is equal to X.s/. This function x.t/ is known as the inverse Laplace transform ofthe function X.s/ and is the solution of the initial value problem. Even though we hadnot used this term before, the last step in solving a differential equation by the Methodof Laplace Transforms is to determine an inverse Laplace transform.

Inverse Laplace Transform

Definition 10.5. An inverse Laplace transform of a function X.s/ is a functionx.t/, if it exists, whose Laplace transform is X.s/. If Lfx.t/g D X.s/, then wewrite

L�1fX.s/g D x.t/:

Example 10.41. Find L�1

�s

s2 C 9

.

Solution. Since L fcos 3tg D s=.s2 C 32/,

L�1

�s

s2 C 9

D cos 3t:


�6

s4

.

Solution. Recall that Lft ng D n!=snC1. In particular, Lft3g D 3!=s4. Thus,

L�1

�6

s4

D L�1

�3!

s4

D t3:

Note that functions do not have unique inverse transforms. For example, the Laplacetransform of the continuous function t 3 and of the piecewise continuous function

h.t/ WD

8<:

t3; if t ¤ 2; 7

�1; if t D 2

10; if t D 7


is 3!=s4. So do we say

L�1

�3!

s4

D t3 or L�1

�3!

s4

D h.t/?

Moreover, there are infinitely many other possibilities too. Which of them is the inverseLaplace transform? The answer is found in our objective of solving linear differentialequations. When the function g.t/ is continuous, solutions of linear first-order initialvalue problems of the form

ax0.t/ C bx.t/ D g.t/; x.t0/ D x0

or of linear second-order initial value problems of the form

ax00.t/ C bx0.t/C cx.t/ D g.t/; x.0/ D x0; x0.0/ D x1

are continuous functions. Obtaining these solutions by the method of Laplace trans-forms necessitates choosing the inverse transform that is a continuous function. Lerch’stheorem implies that a function has a unique continuous inverse transform, if it has oneat all. For that reason, we stipulate the following convention when solving differentialequations with Laplace transforms.

Convention for Choosing Inverse Laplace Transforms

The inverse Laplace transform of a function X.s/ is the unique functionx.t/ that is continuous on the interval Œ0;1/ whose Laplace transform isX.s/; that is,

Lfx.t/g D X.s/; (10.7.1)

provided that such a continuous function x.t/ exists.

If every function satisfying (10.7.1) is discontinuous on Œ0;1/, the inverseLaplace transform of X.s/ is any piecewise continuous function x.t/ satis-fying (10.7.1).


�e�3s

s

.

Solution. Since Lfu.t � b/g D e�bs=s, it follows that

L�1

�e�3s

s

D u.t � 3/ D

(0; if t < 3

1; if t > 3:

The shift properties are indispensable, especially when it comes to finding inversetransforms, as illustrated by the following examples.


Example 10.44. Find L�1fX.s/g for X.s/ D s � 3

.s � 3/2 C 25.

Solution. Except for the shift of 3 units in s, X.s/ is s=.s 2 C 25/, which is the Laplacetransform of cos 5t :

s

s2 C 52D L fcos 5tg :

It follows from (10.6.1), the shift property in s, that

s � 3

.s � 3/2 C 52D Lfe3t cos 5tg

and soL�1fX.s/g D e3t cos 5t :

Example 10.45. Find L�1fY .s/g for Y .s/ D e�5ss

s2 C 4.

Solution. First observe that Y .s/ D e �5sLfcos 2tg. By (10.6.4), the alternate form ofthe shift property in the t-domain,

e�5sLfcos 2tg D Lfcos 2.t � 5/ � u.t � 5/g:Consequently,

L�1fY .s/g D cos.2t � 10/ � u.t � 5/:

Since the inverse Laplace transform inherits the linearity property from the Laplacetransform, it is also a linear operator as we state more formally in the next theorem.

Linearity of the Inverse Laplace Transform

Theorem 10.14. If F1.s/ and F2.s/ are functions whose inverse Laplace trans-forms exist for s > b, then for any constants c1 and c2, an inverse transform ofthe linear combination c1F1.s/C c2F2.s/ for s > b is given by

L�1fc1F1.s/C c2F2.s/g D c1L�1fF1.s/g C c2L

�1fF2.s/g:In other words, L�1 is a linear operator.


�5s

s2 C 9� 24

s4

.

Solution. By the above linearity property,

L�1

�5s

s2 C 9� 24

s4

D 5L�1

�s

s2 C 9

� L�1

�4 � 3!

s4

D 5L�1

�s

s2 C 32

� 4L�1

�3!

s4

D 5 cos 3t � 4t3:



�2s

s2 C 5

.

Solution. First note thats

s2 C 5D s

s2 C .p

5/2D Lfcos t

p5g. Thus, by the linearity

property

L�1

�2s

s2 C 5

D 2L�1

�s

s2 C .p

5/2

D 2 cos.t

p5/:

The previous examples illustrate that a function may have to be expressed in a dif-ferent form in order to determine its inverse Laplace transform. Some techniques thatare useful in this regard are the same ones that aid in carrying out certain integrations,such as

partial fraction expansion, and

completion of the square.

Example 10.48. Find L�1fF.s/g for

F.s/ D 26s � 13

2s2 C 7s � 15:

Solution. A good rule of thumb to follow is to first factor any reducible quadraticfunctions in the denominator. After this is done, we will know how to write the partialfraction expansion:

F.s/ D 26s � 13

.2s � 3/.s C 5/D A

2s � 3C B

s C 5:

Clearing the fractions of their denominators, we obtain

A.s C 5/C B.2s � 3/ D 26s � 13:

Set s D �5 to find B:

�13B D 26.�5/ � 13 ) B D �143

�13D 11:

Set s D 32

to find A:

13

2A D 26

�3

2

�� 13 ) A D 52

13D 4:

And so

L�1fF.s/g D L�1

�4

2s � 3

C L�1

�11

s C 5

D 4

2L�1

(1

s � 32

)C 11L�1

�1

s � .�5/

D 2e

3t2 C 11e�5t:


Example 10.49. Find the inverse Laplace transform of

G.s/ D 1

s2 C 4s C 8:

Solution. Unlike the previous example, partial fraction expansion will not work herebecause the quadratic function is irreducible. In a case as this, the rule of thumb is tocomplete the square:

G.s/ D 1

.s C 2/2 C 22:

Unfortunately, this does not quite have the form of any of the functions appearing inthe right-hand column of Table 10.1. On the other hand, it is clearly related to the formb=.s2 C b2/. All that is required is a little mathematical legerdemain:

1

.s C 2/2 C 22D 1

2� 2

.s C 2/2 C 22:

By linearity and the shift property in the s-domain, the inverse is

L�1fG.s/g D 1

2L�1

(2

.s C 2/2 C 22

)D 1

2e�2t sin 2t :

Example 10.50. Find the inverse transform of H.s/ D 3s

s2 C 4s C 13.

Solution. Again the quadratic function in the denominator is irreducible. Completingits square, we have

H.s/ D 3s

.s C 2/2 C 32:

H.s/ essentially has the form s=.s2 C b2/ except for the shift from s to s C 2 in thedenominator. This calls for another trick of the mathematical trade, namely, to shift thes in the numerator:

H.s/ D 3s

.s C 2/2 C 32D 3.s C 2/ � 3 � 2

.s C 2/2 C 32

D 3s C 2

.s C 2/2 C 32� 2

3

.s C 2/2 C 32:

Therefore, by linearity and the shift property in the s-domain,

L�1fH.s/g D 3 L�1

�s C 2

.s C 2/2 C 32

� 2 L�1

�3

.s C 2/2 C 32

D 3e�2t cos 3t � 2e�2t sin 3t :

The point of the next two examples is to demonstrate how the convolution theorem(Theorem 10.13) can be used to find inverse transforms of rational functions.


Example 10.51. Find the inverse transform of

P .s/ D 1

s2 .s2 C 1/:

Solution 1. By the convolution theorem,

P .s/ D 1

s2� 1

s2 C 1D Lftg Lfsin tg D Lft sin tg:

So the inverse transform of P .s/ is

L�1fP .s/g D t sin t DZ t

0

.t � u/ sin u du:

Integrating by parts, we get

L�1fP .s/g D �� .t � u/ cos u � sin u� t

0D t � sin t:

Solution 2. A partial fraction expansion for P .s/ is

P .s/ D A

sC B

s2C C

s2 C 1;

where A D 0, B D 1, and C D �1. Thus, So the inverse transform of P .s/ is

L�1fP .s/g D L�1

�1

s2

� L�1

�1

s2 C 1

D t � sin t:

Example 10.51 shows that the convolution theorem is an alternative to partial fractionexpansion for finding inverse transforms of rational functions.

Example 10.52. Find the inverse transform of

Q.s/ D s

.s2 C 1/2:

Solution. From

Q.s/ D 1

s2 C 1� s

s2 C 1D Lfsin tg Lfcos tg D Lfsin t cos tg;

we have

L�1fQ.s/g D sin t cos t DZ t

0

sin.t � u/ cos u du:

With the trigonometric identity

sin � cos� D 1

2Œsin.� � �/C sin.� C �/�;

10.8. SECOND-ORDER LINEAR EQUATIONS 367

we can rewrite the integrand as

sin .t � u/ cos u D 1

2Œ sin.t � 2u/C sin.t/ �;

enabling us to carry out the integration:Z t

0

sin.t � u/ cos u du D 1

2

Z t

0

Œ sin.t � 2u/C sin.t/ � du

D 1

4cos.t � 2u/

ˇt0

C 1

2t sin t D 1

2t sin t:

Therefore,

L�1fQ.s/g D 1

2t sin t:

10.8 Second-Order Linear Equations

The next theorem adds to the tools that enable us to solve second-order linear differen-tial equations with Laplace transforms.

Transform of the Second Derivative

Theorem 10.15. Suppose that the function f .t/ is twice differentiable on Œ0;1/

and both f .t/ and f 0.t/ are of exponential order . If the second derivative f 00.t/is integrable on Œ0;T � for every T � 0, then its Laplace transform exists for alls > and is given by the formula

Lff 00.t/g D s2Lff .t/g � sf .0/ � f 0.0/: (10.8.1)

Proof. Since f 00.t/ exists on Œ0;1/, both f .t/ and f 0.t/ are continuous on Œ0;1/.This, along with them being of exponential order , implies that their transforms existby Corollary 10.2. Thus, by Theorem 10.3,

Lff 0.t/g D sLff .t/g � f .0/: (10.8.2)

Again appealing to Theorem 10.3, we see that the Laplace transform of f 00.t/ exists bythe integrability of f 00.t/ on Œ0;T � for every T � 0 and by f 0.t/ being differentiableand of exponential order ; hence,

Lff 00.t/g D sLff 0.t/g � f 0.0/:

Using (10.8.2), we obtain

Lff 00.t/g D s�sLff .t/g � f .0/

� � f 0.0/;

which simplifies to (10.8.1).


The followingexamples illustrate Theorem 10.15 and the properties in Section 10.6.We list the primary properties or techniques in boldface that are used with each of theexamples.

Example 10.53. (linear factors; partial fraction expansion)

Solve the initial value problem

x00 C 6x0 C 5x D 36et; x.0/ D 0; x0.0/ D 10:

Solution. First take the Laplace transform of both sides of the differential equation anduse the linearity property:

Lfx00.t/g C 6Lfx0.t/g C 5Lfx.t/g D 36 Lfetg:Let X.s/ WD Lfx.t/g. Then use Table 10.1 and apply Theorem 10.3 and Theo-rem 10.15:

s2X.s/ � sx.0/� x0.0/C 6�sX.s/ � x.0/

�C 5X.s/ D 36 � 1

s � 1:

Now use the initial conditions and collect like terms:

.s2 C 6s C 5/X.s/ � s � 0 � 10 D 36 � 1

s � 1:

Then solve for X.s/:

X.s/ D36

s � 1C 10

.s C 1/.s C 5/D 36 C 10.s � 1/

.s � 1/.s C 1/.s C 5/D 10s C 26

.s � 1/.s C 1/.s C 5/:

Expand X.s/ in partial fractions:

10s C 26

.s � 1/.s C 1/.s C 5/D A

s � 1C B

s C 1C C

s C 5:

Clearing the fractions of their denominators, we have

A.s C 1/.s C 5/C B.s � 1/.s C 5/C C.s � 1/.s C 1/ D 10s C 26:

Now we can easily find A by letting s D 1:

.2/.6/A D 10.1/C 26 ) 12A D 36 ) A D 3:

To find B, let s D �1:

.�2/.4/B D 10.�1/C 26 ) B D �2:

Finally, let s D �5:

.�6/.�4/C D 10.�5/C 26 ) C D �1:


And so the Laplace transform is

X.s/ D 3

s � 1� 2

s C 1� 1

s C 5:

Taking the inverse Laplace transform of X.s/, we arrive at the solution

x.t/ D L�1fX.s/g D 3et � 2e�t � e�5t :

Example 10.54. (quadratic factors; partial fraction expansion; shift in the s-domain)


x00 C 2x0 C 10x D �25e3t ; x.0/ D 0; x0.0/ D 6:

Solution. Taking the Laplace transform of both sides of the differential equation, weget

Lfx00.t/g C 2Lfx0.t/g C 10Lfx.t/g D �25Lfe3tg:Letting X.s/ D Lfx.t/g, using Table 10.1, and applying the transforms of derivativesproperties, this becomes

s2X.s/ � sx.0/� x0.0/C 2�sX.s/ � x.0/

�C 10X.s/ D �25 � 1

s � 3:

Now solve for X.s/:

X.s/ D 6s � 43

.s � 3/�.s C 1/2 C 9

� :The partial fraction expansion of X.s/ is

6s � 43

.s � 3/�.s C 1/2 C 9

� D A

s � 3C B.s C 1/C 3C

.s C 1/2 C 32:

Clearing the fractions of their denominators, we have

A�.s C 1/2 C 9

�C �B.s C 1/C 3C

�.s � 3/ D 6s � 43:

Setting s equal to 3, �1, and 0, we find A D �1, B D 1, and C D 10=3. Thus,

X.s/ D � 1

s � 3C s C 1

.s C 1/2 C 32C 10

3��

3

.s C 1/2 C 32

�:

Therefore, the solution is given by the inverse Laplace transform of X.s/:

x.t/ D �e3t C e�t cos 3t C 10

3e�t sin 3t :


Example 10.55. (both shift properties)


x00 C 6x0 C 13x D 5ı.t � 2/; x.0/ D 0; x0.0/ D 2:

Solution. As is always the case, the first step is to take the Laplace transform of theequation:

Lfx00.t/g C 6Lfx0.t/g C 13Lfx.t/g D 5Lfı.t � 2/g:Applying the transforms of derivatives properties and using Table 10.1, we find that

s2X.s/ � sx.0/� x0.0/C 6 ŒsX.s/ � x.0/�C 13X.s/ D 5e�2s

where X.s/ D Lfx.t/g. Thus,

.s2 C 6s C 13/X.s/ D 2 C 5e�2s:

Solving for X.s/, we have

X.s/ D 2

s2 C 6s C 13C 5e�2s � 1

s2 C 6s C 13:

Observing that s2 C 6s C 13 is irreducible over the reals, we complete the squareobtaining

X.s/ D 2

.s C 3/2 C 22C 5

2e�2s � 2

.s C 3/2 C 22:

To find the solution, we must take the inverse Laplace transform of X.s/. It is clearthat

2

s2 C 22D Lfsin2tg

is involved. By the shift property in the s-domain,

Lfe�3t sin 2tg D F.s � .�3// D F.s C 3/ D 2

.s C 3/2 C 22:

To find the inverse transform of the second term, we apply the shift property in thet-domain:

e�bsLff .t/g D Lff .t � 2/u.t � b/g:With b D 2, we have

e�2s � 2

.s C 3/2 C 22D e�2sL

˚e�3t sin 2t

�D L ff .t � 2/u.t � 2/g D L

˚e�3.t�2/ sin Œ2.t � 2/� u.t � 2/

�;

where f .t/ D e�3t sin 2t . And so

L

�e�2s � 2

.s C 3/2 C 22

D e6�3t sin.2t � 4/u.t � 2/:


Therefore, the solution is

x.t/ D e�3t sin 2t C 5

2e6�3t sin.2t � 4/u.t � 2/

D(

e�3t sin 2t ; if t < 2

e�3t sin 2t C 52e6�3t sin .2t � 4/; if t > 2:

Example 10.56. (convolution theorem)

Find a formula for the solution of the initial value problem

x00 C 4x D g.t/; x.0/ D 0; x0.0/ D 2:

Assume g.t/ is piecewise continuous on Œ0;1/ and of exponential order.

Solution. Taking the Laplace transform of the differential equation, we get

s2X.s/ � sx.0/ � x0.0/C 4X.s/ D G.s/;

where X.s/ D Lfx.t/g and G.s/ D Lfg.t/g. Because of the initial conditions, thisbecomes

.s2 C 4/X.s/ � s � 0 � 2 D G.s/

and so

X.s/ D 2

s2 C 4C 1

s2 C 4� G.s/:

That is,

Lfx.t/g D Lfsin 2tg C 1

2Lfsin 2tg Lfg.t/g:

By the convolution theorem,

Lfx.t/g D Lfsin 2tg C 1

2Lfsin .2t/ g.t/g:

Consequently, the solution is

x.t/ D sin 2t C 1

2

Z t

0

sin Œ2.t � u/�g.u/du:

10.8.1 Equations with Variable Coefficients

Differential equations with variable coefficients are usually insolvable with Laplacetransforms. Nevertheless there are exceptions as the following examples will reveal.The Laplace transform of a second-order linear equation with coefficients that are con-stants or polynomials of the first degree (but no higher) is a first-order linear equation.The reason is due to the formulas below, which we derive using the derivative of a


transform property (Theorem 10.8) and the transform of derivatives properties (Theo-rems 10.3 and 10.15). For X.s/ D Lfx.t/g,

Lftx.t/g D � d

dsLfx.t/g D �X 0.s/; (10.8.3)

Lftx0.t/g D � d

dsLfx0.t/g D � d

ds

�sX.s/ � x.0/

� D �sX 0.s/ � X.s/; (10.8.4)

and

Lftx00.t/g D � d

dsLfx00.t/g D � d

ds

�s2X.s/ � sx.0/� x0.0/

�(10.8.5)

D �s2X 0.s/ � 2sX.s/C x.0/:

Example 10.57. Show that the transform of the second-order linear equation

tx00 C 2x0 C tx D f .t/

is a first-order linear equation. Assume the transform of f .t/ exists.

Solution. By (10.8.3) and (10.8.5), the Laplace transform of the differential equation is

�s2X 0.s/ � 2sX.s/ C x.0/C 2ŒsX.s/ � x.0/�� X 0.s/ D F.s/;

where F.s/ WD Lff .t/g. Simplifying we obtain the first-order equation

X 0.s/ D � F.s/

s2 C 1� x.0/

s2 C 1: (10.8.6)


tx00 C 2x0 C tx D 0; x.0/ D 1; x0.0/ D 0: (10.8.7)

Solution 1. For x.0/ D 1 and Lf0g D 0, (10.8.6) is

X 0.s/ D � 1

s2 C 1: (10.8.8)

Thus,

� d

dsX.s/ D 1

s2 C 1D Lfsin tg:

By the derivative of a transform property,

� d

dsX.s/ D � d

dsLfx.t/g D Lftx.t/g:


Consequently,Lftx.t/g D Lfsin tg:

Therefore,tx.t/ D sin t:

At first glance, it seems that we are done and

x.t/ D sin t

t

is the solution. However, there is the initial condition x.0/ D 0 but sin t=t is notdefined at t D 0. But it does have a removable discontinuity at t D 0 as

limt!0C

sin t

tD 1:

Therefore,

x.t/ D

8<:

sin t

t; t > 0

1; t D 0

(10.8.9)

is a continuous solution of (10.8.7) for t � 0.

Solution 2. Another approach is to first solve the transformed equation (10.8.8):

X.s/ D �Z

ds

s2 C 1D � tan�1 s C C:

Taking the limit of X.s/ as s ! 1, we have

lims!1 X.s/ D ��

2C C:

On the other hand, assuming X.s/ is the transform of a function that is piecewisecontinuous on Œ0;1/ and of exponential order, its long-term behavior is

lims!1 X.s/ D 0

by Corollary 10.1. Thus,

C D �

2:

As a result,

X.s/ D �

2� tan�1 s D tan�1 1

s:

By (10.6.13),

L�1n

tan�1 1

s

oD sin t

t:

Therefore, as argued earlier, (10.8.9) is the continuous solution for t � 0.



x00 C tx0 � 2x D 1; x.0/ D x0.0/ D 0: (10.8.10)

Solution. The transformed equation is the linear first-order equation

s2X.s/ � sx.0/� x0.0/ � sX 0.s/ � X.s/ � 2X.s/ D 1

s:

See (10.8.4). Using the initial conditions and rearranging terms to write this in standardform, we obtain

X 0.s/ ��

3

s� s

�X.s/ D � 1

s2: (10.8.11)

An integrating factor for (10.8.11) is

�.s/ D eR

. 3s

�s/ ds D s3e�s2=2:

Multiplying (10.8.11) by �.s/ and integrating, we obtain

d

ds

hs3e�s2=2X.s/

iD �se�s2=2

s3e�s2=2X.s/ D �Z

se�s2=2 ds C C D e�s2=2 C C:

Thus,

X.s/ D 1

s3C C

es2=2

s3:

By Corollary 10.1,lim

s!1 X.s/ D 0;

assuming X.s/ is the transform of a piecewise continuous function of exponential or-der. But this can only be the case if C D 0 as

lims!1

es2=2

s3D 1:

Therefore,

X.s/ D 1

s3:

Consequently, the solution is

x.t/ D L�1fX.s/g D 1

2L�1

n 2!

s3

oD 1

2t2:

A quick check verifies that this does solve the differential equation and satisfy the initialconditions of the initial value problem (10.8.10)

PROBLEMS 375

Problems

You see, in every job that must be done, there is an element of fun. You find the fun andsnap! The job’s a game.

from Walt Disney’s 1964 film Mary Poppins(Mary Poppins’ advice to Jane and Michael)

Laplace Transforms from the Definition

Use Definition 10.1 on page 317 to find the Laplacetransforms of the functions in Problems 1 through6.

1. cos bt , where b is any real constant

2. f .t/ D(

e2t ; if t < 5

0; if t > 5

3. g.t/ D(

0; if t < 5

e2t ; if t > 5

4. sinh t WD .et � e�t /=2 (the hyperbolicsine function)

5. h.t/ D(

1 � t; if t < 7

10; if t > 7

6. tu.t � 2/

Linearity of the Laplace Transform

Use Table 10.1 and Theorem10.1 to find the Laplacetransforms of the functions in Problems 7 through12.

7. sin 2t C e6t

8. 3e�2t C 2t7 � 12

9. 5u.t � 2/

10. ı.t � 2/

11. cosh .t=2/, where cosh t WD .e t C e�t /=2

12. 34ı.t � 5/� 2 cos 4

7t

Equivalent Properties

13. Show that the linearity property (10.3.1) isequivalent to the following two properties:

(a) Lff .t/C g.t/g D Lff .t/g C Lfg.t/g(b) Lfcf .t/g D cLfg.t/g,

where c is a constant. In other words, show that(10.3.1) implies (a) and (b) and that conversely (a)and (b) imply (10.3.1).

Properties of the Laplace Transform

In Problems 14 through 30, determine the Laplacetransform of each of the following functions usingTable 10.1 and the properties of the transform.

14. e5t t3

15. e�2t sin t3

16. 10t4 � sin 2t

17. t cos 2t

18. 3t4 � sin 2t

19. t3e�t

20. 1 C 32

e�5t � ı.t � 3/

21. te2t sin 3t

22. t2 sin t

23. 2 cos t2

C(

0; if t < 5

10; if t > 5

24. t sin 5t C 4.t � 1/2

25. t10 C 3ı.t � 2/

26. 5 C 8t sin 23

t � cosh5t

27. h.t/ D(

3 C e5t ; if t < 1

e5t ; if t > 1

28. 413

t2 sin 7t � 3t4e�2t C tu.t � 4/

29. cos .3t/u.t � �=6/

30. sin .3t C 2/

Potpourri

31. Explain why cos bt is of exponential order 0.

32. Explain why sin bt D O.1/ as t ! 1.


33. Explain why et is of exponential order 1.

34. Show that f .t/ D O.e� t / as t ! 1 isequivalent to the statement: e�� t jf .t/j isbounded for t � t1 for some t1 � 0.

35. Use the Laplace Transform of the DerivativeProperty to find Lfcos btg from

Lfsin btg D b

s2 C b2:

36. Find Lfsin2 btg by using an appropriate half-angle identity. (If you do not know this iden-tity, look it up in any trigonometry book,find out how it is derived, and memorizeit.) Compare your answer with the result ob-tained in one of the examples.

37. Let f .t/ be continuous on Œ0;1/ and of ex-ponential order � 0. Generalize the trans-form of an integral property by determiningthe transform of

R1a f .�/d� , where a � 0.

38. Suppose f .t/ is continuous on Œ0;1/ andof exponential order . If Lff .t/g D 0

for s > , show that Lff .t/P .t/g D 0 fors > for all polynomials P .t/.

First-Order Linear Equations

In Problems 39 through 47, solve the initial valueproblems using the Method of Laplace Transforms.Then check your answers by solving with integrat-ing factors.

39. x0 � 5x D e2t ; x.0/ D 1

40. y0 C 2y D 4e�3t ; y.0/ D �2

41. y0 C 12

y D 0; y.0/ D 4

42. x0 C 3x D cos 2t ; x.0/ D 1

43. x0 � 2x D 10 cos t; x.0/ D 3

44. x0 � 10x D 29 cos 5t; x.0/ D 1

45. x0 C x D 8u.t � 10/; x.0/ D �5

46. x0 � 3x D 6ı.t � 5/; x.0/ D 2

47. x0 � x D 4ı.t � 5/; x.0/ D �8

Inverse Laplace Transforms

Find the inverse Laplace transform of each of thefollowing functions.

48.3

s2 C 2

49.s C 2

s2 C 4s � 5

50.2s � 4

s2 � 4s C 18

51.s

s2 C 4s C 20

52.5s2 C 3s C 39

.s � 2/.s2 C 9/

53.8

.s � 2/5

54.12s

4s2 C 8s C 5

55.7

s � 5� 2s

s2 C 16

56.6

s C 4� 5s

s2 C 7

57.2s2 � s C 5

s3 C s2 � 6s

Second-Order Linear Equations

Use the method of Laplace Transforms to solve thefollowing linear second-order initial value prob-lems.

58.d2x

dt2C 4

dx

dtD 8;

x.0/ D 0; x0.0/ D 0

59. Rx � 4 Px C 5x D 6e3t ;

x.0/ D 2; Px.0/ D �5

60. x00 C 9x D 2et ;

x.0/ D 1; x0.0/ D 0

61.d2x

dt2C 5

dx

dt� 14x D 0;

x.0/ D 5; x0.0/ D 1

62. Rx C 2 Px C 10x D 12

t2 � 5e3t ;

x.0/ D 0; Px.0/ D 6

63. x00 C 4x0 C 5x D 6ı.t � 2/;

x.0/ D 0; x0.0/ D 1

PROBLEMS 377

64. Rx C 4x D 1 � 2u.t � 5/;

x.0/ D �1; Px.0/ D 0

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