the laplace transform - uhpwalker/chap4f09.pdfthe laplace transform provides an eﬀective method of...

CHAPTER 4

The Laplace Transform

4.1 Introduction

The Laplace transform provides an effective method of solving initial-value problems forlinear differential equations with constant coefficients. However, the usefulness of Laplacetransforms is by no means restricted to this class of problems. Some understanding of thebasic theory is an essential part of the mathematical background of engineers, scientists andmathematicians.

The Laplace transform is defined in terms of an integral over the interval [0,∞). In-tegrals over an infinite interval are called improper integrals, a topic studied in CalculusII.

DEFINITION Let f be a continuous function on [0,∞). The Laplace transform of f ,denoted by L[(f(x)], or by F (s), is the function given by

L[f(x)] = F (s) =∫ ∞

0e−sxf(x) dx. (1)

The domain of F is the set of all real numbers s for which the improper integral converges.

In more advanced treatments of the Laplace transform the parameter s assumes com-plex values, but the restriction to real values is sufficient for our purposes here. Note thatL transforms a function f = f(x) into a function F = F (s) of the parameter s. Thecontinuity assumption on f will hold throughout the first three sections. It is made forconvenience in presenting the basic properties of L and for applying the Laplace transformmethod to solving initial-value problems. In the last two sections of this chapter we extendthe definition of L to a larger class of functions, the piecewise continuous functions on[0,∞). There we will apply L to the problem of solving nonhomogeneous equations inwhich the nonhomogeneous term is piecewise continuous. This will involve some extensionof our concepts of differential equation and solution.

As indicated above, the primary application of Laplace transforms of interest to us issolving linear differential equations with constant coefficients. Referring to our work inChapter 3, the functions which arise naturally in the treatment of these equations are:

p(x)erx, p(x) cos βx, p(x) sin βx, p(x)erx cos βx, p(x)erx sin βx

where p is a polynomial.

We begin by calculating the Laplace transforms of some simple cases of these functions.

115

Example 1. Let f(x) = 1 · e0x ≡ 1 on [0,∞). By the Definition,

L[1] =∫ ∞

0

e−sx · 1 dx = limb→∞

∫ b

0

e−sx dx

= limb→∞

[e−sx

−s

∣∣∣∣b

0

]= lim

b→∞

[e−sb

−s

]+

1s

= limb→∞

[−1sesb

]+

1s.

Now, limb→∞

−1/sesb exists if and only if s > 0, and in this case

limb→∞

−1sesb

= 0.

Thus,

L[1] =1s, s > 0. �

Example 2. Let f(x) = erx on [0,∞). Then,

L[erx] =∫ ∞

0e−sx · erx dx = lim

b→∞

∫ b

0e−(s−r)x dx

= limb→∞

e−(s−r)x

−(s − r)

∣∣∣∣∣

b

0

= lim

b→∞

[e−(s−r)b

−(s − r)

]+

1s − r

.

The limit exists (and has the value 0) if and only if s − r > 0. Therefore

L[erx] =1

s − r, s > r.

Note that if r = 0, then we have the result in Example 1. �

Example 3. Let f(x) = cos βx on [0,∞). Then,

L[cos βx] =∫ ∞

0e−sx · cos βx dx = lim

b→∞

∫ b

0e−sx cos βx dx

= limb→∞

e−sx[−s cos βx − β sin βx]s2 + β2

∣∣∣∣b

0

.

(Note the integral was calculated using integration by parts; also, it is a standard entry ina table of integrals.)

Now,

L[cos βx] = −[

limb→∞

1e−sb

· s cos βb + β sin βb

s2 + β2+

s

s2 + β2

].

Since [s cos βb+β sin βb]/(s2 +β2) is bounded, the limit exists (and has the value 0)if and only if s > 0. Therefore,

L[cos βx] =s

s2 + β2, s > 0. �

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The following table gives a basic list of the Laplace transforms of functions that we willencounter in this chapter. While the entries in the table can be verified using the Definition,some of the integrations involved are complicated. The properties of the Laplace transformpresented in the next section provide a more efficient way to obtain many of the entries in thetable. Handbooks of mathematical functions, for example the CRC Standard MathematicalTables, give extensive tables of Laplace transforms.

Table of Laplace Transforms

f(x) F (s) = L[f(x)]

11s, s > 0

eαx 1s − α

, s > α

cos βxs

s2 + β2, s > 0

sin βxβ

s2 + β2, s > 0

eαx cos βxs − α

(s − α)2 + β2, s > α

eαx sin βxβ

(s − α)2 + β2, s > α

xn, n = 1, 2, . . .n!

sn+1, s > 0

xn erx, n = 1, 2, . . .n!

(s − r)n+1, s > r

x cos βxs2 − β2

(s2 + β2)2, s > 0

x sin βx2βs

(s2 + β2)2, s > 0

Exercises 4.1

Use the definition of the Laplace transform to find the Laplace transform of the givenfunction.

1. f(x) = x.

2. f(x) = x2.

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3. f(x) = sin x.

4. f(x) = xerx.

5. f(x) = sinh x.

6. f(x) = cosh x.

7. Use the fact that∫

eax cos bx dx =eax

a2 + b2[a cos bx+b sin bx] to find L[erx cos βx]

by the definition.

8. Use the fact that∫

eax sin bx dx =eax

a2 + b2[a sin bx− b cos bx] to find L[erx sin βx]

by the definition.

9. Show that L[sin x] =

∫ 2π

0e−sx sin x dx

1− e−2πs.

10. Let f be a continuous function on [0,∞) and suppose that f is periodic withperiod p. That is f is continuous and f(x + p) = f(x), p > 0, for all x. Showthat

L[f(x)] =

∫ p

0e−sxf(x) dx

1 − e−ps.

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4.2 Basic Properties of the Laplace Transform

In the preceding section we defined the Laplace transform and calculated the Laplace trans-forms of some of the functions that occur in solving linear differential equations with con-stant coefficients. In this section we consider the basic question of the existence of theLaplace transform of a function f , and we develop the properties of the Laplace transformthat will be used in solving initial value problems.

To motivate the material in this section, consider the differential equation

y′′ + ay′ + by = f(x) (2)

where a and b are constants and f is a continuous function on [0,∞). If we assumethat y = y(x) is a solution of (1) and formally apply L, we obtain

L[y′′(x) + ay′(x) + by(x)

]= L[f(x)]. (3)

The right-hand side of this equation suggests the basic question of the existence of L[f(x)].That is, for what functions f does L[f ] exist?

DEFINITION A function f , continuous on [0,∞), is said to be of exponential orderλ, λ a real number, if there exists a positive number M and a nonnegative number A

such that| f(x)| ≤ Meλx

on [A,∞).

Example 1. (a) If f is a bounded function on [0,∞) [for example, f(x) = cos βx

or f(x) = sin βx], then f is of exponential order 0.

f bounded implies that there exists a positive number M such that | f(x)| ≤ M

for all x ∈ [0,∞). Therefore,

| f(x)| ≤ M = Me0x on [0,∞).

[Note: if f(x) = cos βx or f(x) = sin βx, then we could take M = 1.]

(b) Let f(x) = x on [0,∞). For any positive number λ,

limx→∞

x

eλx= 0

by L’Hopital’s rule. Therefore, there exists a nonnegative number A such thatx

eλx≤ 1 on [A,∞).

This implies thatx ≤ eλx = 1 · eλx on [A,∞)

and f(x) = x is of exponential order λ. The same argument can be used to showthat f(x) = xα, α any real number, is of exponential order λ for any positivenumber λ. In general, if p = p(x) is a polynomial, then p is of exponential orderλ for any positive number λ.

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(c) If f(x) = erx, then f is of exponential order λ for any λ ≥ r.

(d) Consider the function f(x) = ex2. If f is of exponential order λ for some λ, then

there exists a positive number M and a nonnegative number A such that

ex2 ≤ Meλx on [A,∞) which implies e−λxex2 ≤ M on [A,∞).

But,lim

x→∞e−λxex2

= limx→∞

ex(x−λ) = ∞,

a contradiction. Thus f(x) = ex2is not of exponential order λ for any positive

number λ. �

Our first property of L is a sufficient condition for L[f(x)] to exist. We shall omitthe proof.

THEOREM 1. Let f be a continuous function on [0,∞). If f is of exponential orderλ, then the Laplace transform L[f(x)] = F (s) exists for s > λ.

We now turn to the left-hand side of equation (2) where we have L applied to thelinear combination y′′(x) + ay′(x) + by(x).

THEOREM 2. The operator L is a linear operator. That is, if g and h are continuousfunctions on [0,∞), and if each of L[g(x)] and L[h(x)] exists for s > λ, thenL[g(x) + h(x)] and L[c g(x)], c constant, each exist for s > λ, and

L[g(x) + h(x)] = L[g(x)] + L[h(x)]

L[c g(x)] = cL[g(x)].

Proof:

L[g(x) + h(x)] =∫ ∞

0e−sx[g(x) + h(x)] dx = lim

b→∞

∫ b

0e−sx[g(x) + h(x)] dx

= limb→∞

[∫ b

0e−sxg(x) dx +

∫ b

0e−sxh(x) dx

]

= limb→∞

∫ b

0e−sxg(x) dx + lim

b→∞

∫ b

0e−sxh(x) dx

=∫ ∞

0

e−sxg(x) dx +∫ ∞

0

e−sxh(x) dx

= L[g(x)] + L[h(x)]

The proof thatL[c g(x)] = cL[g(x)] for any constant c

is left as an exercise. �

120

COROLLARY Let g1(x), g2(x), . . . , gn(x) be continuous functions on [0,∞). IfL[g1(x)], L[g2(x)], . . . ,L[gn(x)] all exist for s > λ, and if c1, c2, . . . , cn are realnumbers, then

L [c1g1(x) + c2g2(x) + · · ·+ cngn(x)]

exists for s > λ and

L [c1g1(x) + c2g2(x) + · · ·+ cngn(x)] = c1L[g1(x)] + c2L[g2(x)] + · · ·+ cnL[gn(x)].

According to the Corollary, if L[y′′], L[y′], and L[y] all exist for s > λ for some λ,then

L[y′′ + ay′ + by] = L[y′′] + aL[y′] + bL[y].

The next property gives a relationship between the Laplace transform of the derivativeof a function and the Laplace transform of the function itself.

THEOREM 3. Let g be a continuously differentiable function on [0,∞). If g is ofexponential order λ, then L[g′(x)] exists for s > λ and

L[g′(x)] = sL[g(x)]− g(0).

Proof: By the definition of L, we have

L[g′(x)] =∫ ∞

0e−sxg′(x) dx = lim

b→∞

∫ b

0e−sxg′(x) dx.

Now, using integration by parts,∫ b

0e−sxg′(x) dx = e−sxg(x)

∣∣b0+ s

∫ b

0e−sxg(x) dx

= e−sbg(b)− g(0) + s

∫ b

0e−sxg(x) dx

Therefore,

L[g′(x)] = limb→∞

[e−sbg(b)− g(0) + s

∫ b

0

e−sxg(x) dx

]

= limb→∞

e−sbg(b)− g(0) + s limb→∞

∫ b

0e−sxg(x) dx

= limb→∞

e−sbg(b)− g(0) + sL[g(x)]

provided limb→∞

e−sbg(b) exists.

Since g is of exponential order λ, there exist constants M and A such that

| g(x)| ≤ Meλx

121

on [A,∞). Therefore,∣∣∣ e−sbg(b)

∣∣∣ ≤∣∣∣ e−sbMeλb

∣∣∣ = Me−(s−λ)b

for all b > A, and it follows that

limb→∞

e−sbg(b) = 0 for s > λ.

We can now conclude that

L[g′(x)] = sL[g(x)]− g(0). �

COROLLARY Let g be a function which is n-times continuously differentiable on[0,∞). If each of the functions g, g′, . . . , g(n−1) is of exponential order λ, then L[g(n)]exists for s > λ and

L[g(n)(x)] = snL[g(x)]− sn−1g(0)− sn−2g′(0)− · · · − g(n−1)(0).

In particular, if g′′ is continuous on [0,∞), and if g and g′ are of exponential order λ,then L[g′′(x)] exists for s > λ and

L[g′′(x)] = s2L[g(x)]− sg(0)− g′(0).

The proof of the Corollary follows by induction.

Application to Initial-Value Problems

Consider the first order initial-value problem

y′ + ay = f(x); y(0) = α

where α is a real number and the function f = f(x) has Laplace transform F = F (s).

Let y = y(x) be the solution of the problem. Applying the properties of L establishedabove, we have

L[y′(x) + ay(x)

]= L[f(x)]

L[y′(x)] + aL[y(x)] = F (s) (by linearity)

sL[y(x)] − y(0) + aL[y(x)] = F (s) (by Theorem 3)

(s + a)L[y(x)]− y(0) = F (s)

Applying the initial condition y(0) = α, and solving for L[y(x)] = Y (s), we get

Y (s) =F (s) + α

s + a.

122

Next consider the second order initial-value problem

y′′ + ay′ + by = f(x); y(0) = α, y′(0) = β.

where α and β are real numbers, and f is a function with Laplace transform L[f(x)] =F (s).

Let y = y(x) be the solution of this problem. Then

L[y′′(x) + ay′(x) + by(x)

]= L[f(x)] (equation (2))

L[y′′(x)] + aL[y′(x)] + bL[y(x)] = F (s) (by linearity)

s2L[y(x)]− sy(0) − y′(0) + a {sL[y(x)]− y(0)}+ bL[y(x)] = F (s) (by the Corollary to Theorem 3)

(s2 + as + b)L[y(x)]− sy(0)− y′(0)− ay(0) = F (s)

Applying the initial conditions y(0) = α, y′(0) = β and solving for L[y(x)] = Y (s), weget

Y (s) =F (s) + αs + β + aα

s2 + as + b.

Implicit in these derivations is the assumption that the Laplace transform of y andits derivatives exist. Assuming that this is the case, the importance of these results is thatit gives us the Laplace transform of the solution of an initial-value problem directly. Thequestion now is: Knowing Y (s), what is y(x)? This is the topic of the next section,“inverting” the Laplace transform.

Example 2. Find the Laplace transform L[y(x)] = Y (s) of the solution of the initial-valueproblem

y′ − 2y = 2e−3x; y(0) = −2.

SOLUTION If y = y(x) is the solution, then

L[y′(x)− 2y(x)] = L[2e−3x] = 2L[e−3x] =2

s + 3

L[y′(x)]− 2L[y(x)] =2

s + 3

sL[y(x)] − y(0)− 2L[y(x)] =2

s + 3

(s − 2)L[y(x)] + 2 =2

s + 3

(s − 2)L[y(x)] =2

s + 3− 2

Therefore,

L[y(x)] = Y (s) =2

(s − 2)(s + 3)− 2

s − 2. �

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Example 3. Find the Laplace transform L[y(x)] = Y (s) of the solution of the initial-valueproblem

y′′ + 4y = 3e2x; y(0) = 5, y′(0) = −2.


L[y′′(x) + 4y(x)] = L[3e2x] = 3L[e2x] =3

s − 2

L[y′′(x)] + 4L[y(x)] =3

s − 2

s2L[y(x)]− sy(0) − y′(0) + 4L[y(x)] =3

s − 2

(s2 + 4)L[y(x)]− 5s − (−2) =3

s − 2

(s2 + 4)L[y(x)] =3

s − 2+ 5s − 2

Therefore,

L[y(x)] = Y (s) =3

(s − 2)(s2 + 4)+

5s − 2s2 + 4

. �

In the next section we will see how to go from Y (s) to y(x).

Additional Applications and Properties

Although the main use of Theorem 3 and its Corollary is in solving initial-value problems,the results can also be used to determine entries in a table of Laplace transforms. Forexample, if f is a continuously differentiable function and L[f(x)] is known, then L[f ′(x)]can be determined: L[f ′(x)] = sL[f(x)]− f(0).

Example 4. In Example 3, Section 4.1, we showed that

L[cos βx] =s

s2 + β2.

We could use essentially the same calculations to obtain L[sin βx], but recall that theintegrations involved are a little “messy.”

Here is a simpler way to obtain L[sin βx]. Since [cos βx]′ = −β sin βx, we can write

L[−β sin βx] = L[(cos βx)′] = sL[cos βx] − cos(0) =s2

s2 + β2− 1

−βL[sin βx] =−β2

s2 + β2.

Therefore,

L[sin βx] =β

s2 + β2. �

124

Here are two more properties of the Laplace transform that are useful in determiningentries in a table of transforms.

THEOREM 4. Let f be a continuous function on [0,∞), and be of exponential orderλ. Then F (s) = L[f(x)] has derivatives of all orders, and for s > λ,

dnF (s)dsn

= (−1)nL [xnf(x)] , n = 1, 2, . . . .

Justification By the definition of L[f(x)] = F (s), we have

F (s) =∫ ∞

0e−sxf(x) dx.

Differentiation of this equation with respect to s can be justified and yields

dF

ds=

∫ ∞

0

d

ds[e−sxf(x)] dx =

∫ ∞

0e−sx[−xf(x)] dx

= L[−xf(x)] = −L[xf(x)]

Therefore, L[xf(x)] = −dF

ds.

Differentiating a second time, we have

d2F

ds2=

d

ds

[dF

ds

]=

∫ ∞

0

d

ds

(e−sx[−xf(x)]

)dx

=∫ ∞

0

e−sx[x2f(x)] dx = L[x2f(x)]

The result stated in the theorem follows by mathematical induction. �

Example 5. From the table at the end of Section 4.1, L[erx] =1

s − r, s > r. Therefore,

L[xerx] = −F ′(s) = − d

ds

(1

s − r

)=

1(s − r)2

, s > r,

L[x2erx] = F ′′(s) =d2

ds2

(1

s − r

)=

2(s − r)3

, s > r,

L[x3erx] = −F ′′′(s) = − d3

ds3

(1

s − r

)=

6(s − r)4

, s > r,

and so on. �

The final property we’ll consider is called the translation property of L.

THEOREM 5. If f is a continuous function on [0,∞), and if L[f(x)] = F (s) existsfor s > λ, then for any real number r,

L[erxf(x)] = F (s − r) for s > λ + r.

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Proof: From the definition of the Laplace transform,

F (s − r) =∫ ∞

0

e−(s−r)xf(x) dx =∫ ∞

0

e−sxerxf(x) dx

= L[erxf(x)]. �

Example 6.

(a) Since L[x] = 1/s2, s > 0, we have

L[xerx] = F (s − r) =1

(s − r)2, s > r (c.f. Example 5)

(b) Since L[cos βx] =s

s2 + β2, s > 0, we have

L[erx cos βx] = F (s−r) =s − r

(s − r)2 + β2, s > r (as indicated in the Table, Section 4.1). �

Summary of the Properties of L

1. If f is continuous on [0,∞), and of exponential order λ, then L[f(x)] = F (s)exists for s > λ. (Theorem 1)

2. L is a linear operator (Theorem 2): L[f + g] = L[f ] + L[g]; L[cf ] = cL[f ].

3. If f is continuously differentiable on [0,∞), and of exponential order λ, thenL[f ′(x)] exists for s > λ and

L[f ′(x)] = sL[f(x)]− f(0).

If f is twice continuously differentiable on [0,∞), and of exponential order λ, thenL[f ′′(x)] exists for s > λ and

L[f ′′(x)] = s2L[f(x)]− sf(0) − f ′(0).

And so on. (Theorem 3)

4. If f is continuous on [0,∞), and of exponential order λ, then F (s) = L[f(x)]has derivatives of all orders, and for s > λ,

dnF (s)dsn

= (−1)nL [xnf(x)] , n = 1, 2, . . . . (Theorem 4)

5. If f is continuous on [0,∞), and if L[f(x)] = F (s) exists for s > λ, then for anyreal number r,

L[erxf(x)] = F (s − r) for s > λ + r. (Theorem 5)

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Exercises 4.2

Use the properties of the Laplace transform and the Table to find L[f ].

1. f(x) = 3 − 2x + x2.

2. f(x) = 2e−x − 3 sin 4x.

3. f(x) = 3 + 4e3x − 2 cos 2x.

4. f(x) = 2xe−2x + 5e2x cos 3x.

5. f(x) = 5x2 − 2e−3x sin 2x.

6. f(x) = 2 − 3x + 4x2e2x.

7. f(x) = x sin x + 2x cos 2x.

8. Show that L[cosh βx] =s

s2 − β2.

9. Show that L[sinh βx] =β

s2 − β2.

10. Find L[3 cosh 2x − 5 sinh 3x].

11. Find L[e2x sinh x + e−x cosh 3x].

12. Find L[x cosh 2x].

13. Show that L[xerx] =1

(s − r)2by:

(a) Using Property 4.

(b) Using Property 5.

14. Use Property 4 to show that L[x sin βx] =2βs

(s2 + β2)2.

Find the Laplace transform Y (s) of the solution of the given initial-value problem.

15. y′ − 2y = 0; y(0 = 1.

16. y′ − 2y = x; y(0) = 1.

17. y′ + 4y = 2e2x − 3 sin 3x; y(0) = −3.

18. y′′ + 2y′ − 8y = 0; y(0) = 4, y′(0) = −2.

19. y′′ + 6y′ + 9y = 0; y(0) = 0, y′(0) = 2.

20. y′′ − 2y′ + 5y = 2x + e−x; y(0) = −2, y′(0) = 0.

127

21. y′′ − 2y′ − 15y = 3 + 4e−3x; y(0) = 1, y′(0) = −3.

22. y′′ − 4y′ + 4y = 5e2x; y(0) = −3, y′(0) = 2.

Here is another property of the Laplace transform.

23. Let f be a continuous function on [0,∞) and assume that both f and∫ x

0f(t) dt

are of exponential order λ on [0,∞). Show that if F (s) = L[f(x)], then

L[∫ x

0f(t) dt

]=

1sF (s).

24. Find L[∫ x

0sin 2t dt

]by using:

(a) The property given in Exercise 23.

(b) By first calculating the integral and then taking the Laplace transform of theresult.

128

4.3 Inverse Laplace Transforms and Initial-Value Problems

In Section 4.2 we saw that the Laplace transform of the solution y = y(x) of the initial-valueproblem

y′′ + ay′ + by = f(x); y(0) = α, y′(0) = β

is given by

L[y(x)] = Y (s) =F (s) + αs + β + aα

s2 + as + b

where F (s) = L[f(x)] is the Laplace transform of f .

Now that we know L[y(x)], the obvious question is: What is y(x)? The general problemof finding a function with a given Laplace transform is called the inversion problem. Theinversion problem and its application to solving initial-value problems is the topic of thissection.

If f is continuous on [0,∞), and if the Laplace transform, L[f(x)] = F (s) existsfor s > λ, then the function F is uniquely determined by f ; that is, the operator Lis itself a function. Our first result states that L is a one-to-one function. A proof of thisresult is beyond the scope of this introductory treatment.

THEOREM 1. If f and g are continuous functions on [0,∞), and if L[f(x)] = L[g(x)],then f ≡ g; that is f(x) = g(x) for all x ∈ [0,∞).

The following definition gives the terminology and notation used in treating the inversionproblem.

DEFINITION If F (s) is a given transform and if the function f , continuous on [0,∞),has the property that L[f(x)] = F (s), then f is called the inverse Laplace transform ofF (s), and is denoted by

f(x) = L−1[F (s)].

The operator L−1 is called the inverse operator of L.

There is a general formula for the inverse operator L−1 corresponding to (1), Section4.1, but use of the formula requires a knowledge of complex-valued functions, a topic whichis treated in more advanced courses.

The relationship between L and L−1 is given by the following equations:

L−1 {L[f(x)]} = f(x)

L{L−1[F (s)]

}= F (s)

for all functions f , continuous on [0,∞), such that L[f(x)] = F (s).

For convenience here, we reproduce the table of Laplace transforms given at the end ofSection 4.1.

129

Table of Laplace Transforms

f(x) F (s) = L[f(x)]

11s, s > 0

eαx 1s − α

, s > α

cos βxs

s2 + β2, s > 0

sin βxβ

s2 + β2, s > 0

eαx cos βxs − α

(s − α)2 + β2, s > α

eαx sin βxβ

(s − α)2 + β2, s > α

xn, n = 1, 2, . . .n!

sn+1, s > 0

xn erx, n = 1, 2, . . .n!

(s − r)n+1, s > r

x cos βxs2 − β2

(s2 + β2)2, s > 0

x sin βx2βs

(s2 + β2)2, s > 0

A simple way to interpret Theorem 1 is that the table can be read either from left toright or from right to left. That is, the table is simultaneously a table of Laplace transformsand of inverse Laplace transforms.

Example 1. (a) If L[f(x)] = F (s) =1

s − 4, then f(x) = e4x.

(b) If L[f(x)] = F (s) =s

s2 + 9, then f(x) = cos 3x.

(c) If L[f(x)] = F (s) =s + 2

s2 + 4s + 13=

s + 2(s + 2)2 + 9

=s − (−2)

[s − (−2)]2 + 9, then f(x) =

e−2x cos 3x.

The properties of the Laplace transform operator L can be used to derive correspondingproperties of its inverse operator L−1. For our purposes, the most important property isthat of linearity.

THEOREM 2. The operator L−1 is linear; that is

L−1[F (s) + G(s)] = L−1[F (s)] + L−1[G(s)], and

130

L−1[cF (s)] = cL−1[F (s)], c any constant.

The proof is left as an exercise.

Example 2. Find L−1[F (s)] if

F (s) =7

s + 2− 6

s2 + 4.

SOLUTION Since L−1 is a linear operator,

L−1[F (s)] = L−1

[7

s + 3− 6

s2 + 4

]

= 7L−1

[1

s + 3

]− 3L−1

[2

s2 + 4

]

Now, reading the table from right to left, we see that

L−1

[1

s + 3

]= e−3x and L−1

[2

s2 + 4

]= sin 2x.

Therefore,

L−1

[7

s + 3− 6

s2 + 4

]= 7e−3x − 3 sin 2x. �

The translation property of L (Theorem 5, Section 4.2) is also useful in finding inversetransforms. The analog of Theorem 5 for inverse transforms is:

THEOREM 3. If f is continuous on [0,∞), and if L[f(x)] = F (s) exists for s > λ,then, for any real number r,

L−1[F (s − r)] = erxf(x).

The following examples illustrate the kinds of manipulations that typically occur incalculating inverse Laplace transforms. The basic strategy is to try to re-write a givenexpression F (s) as sum of terms which appear in the table.


F (s) =4

(s − 3)2+

1s2 − 2s + 10

.

SOLUTION

L−1[F (s)] = 4L−1

[1

(s − 3)2

]+ L−1

[1

s2 − 2s + 10

]

From the table,

L−1

[1

(s − 3)2

]= xe3x.

131

To put1

s2 − 2s + 10in a form in the table, we complete the square in the denominator

and “adjust” the numerator:

1s2 − 2s + 10

=1

s2 − 2s + 1 + 9=

1(s − 1)2 + 9

=13

3(s − 1)2 + 9

.

From the table,

L−1

[1

s2 − 2s + 10

]=

13L−1

[3

(s − 1)2 + 9

]= 1

3 ex sin 3x.

Putting the two results together, we have

L−1

[4

(s − 3)2+

1s2 − 2s + 10

]= 4xe3x + 1

3 ex sin 3x. �


F (s) =2s + 1

s2 − 2s − 8.

SOLUTION By factoring the denominator, we can write

F (s) =2s + 1

s2 − 2s − 8=

2s + 1(s + 2)(s − 4)

.

Now, by partial fraction decomposition,

2s + 1(s + 2)(s− 4)

=32

s − 4+

12

s + 2.

Therefore

L−1

[2s + 1

s2 − 2s − 8

]=

32L−1

[1

s − 4

]+

12L−1

[1

s + 2

]= 3

2 e4x + 12 e−2x. �


F (s) =2s + 4

(s − 2)(s2 − 4s + 8).

SOLUTION The quadratic factor in the denominator cannot be factored into linear factors.

By partial fraction decomposition

F (s) =2s + 4

(s − 2)(s2 − 4s + 8)=

2s − 2

+−2s + 6

s2 − 4s + 8.

Next, we complete the square in the denominator of the second term:

2s − 2

+−2s + 6

s2 − 4s + 8=

2s − 2

+−2s + 6

s2 − 4s + 4 + 4=

2s − 2

+−2s + 6

(s − 2)2 + 4.

132

Finally, we “adjust” the numerator of the second term so that we can use the Table:

2s − 2

+−2s + 6

(s − 2)2 + 4=

2s − 2

+−2(s − 2) + 2(s − 2)2 + 4

=2

s − 2+

−2(s − 2)(s − 2)2 + 4

+2

(s − 2)2 + 4.

Now

L−1

[2s + 4

(s − 2)(s2 − 4s + 8)

]= L−1

[2

1s − 2

− 2s − 2

(s − 2)2 + 4+

2(s − 2)2 + 4

]

= 2L−1

[1

s − 2

]− 2L−1

[s − 2

(s − 2)2 + 4

]+ L−1

[2

(s − 2)2 + 4

]

= 2 e2x − 2 e2x cos 2x + e2x sin 2x. �

Solution of Initial-Value Problems

Here we complete the application of Laplace transforms to the solution of initial-valueproblems. For our first example, we finish Example 2, Section 4.2.

Example 6. Find the solution of the initial-value problem

y′ − 2y = 2e−3x; y(0) = −2.

SOLUTION From the Example, if y = y(x) is the solution, then

L[y′(x)− 2y(x)] = L[2e−3x] = 2L[e−3x] =2

s + 3

L[y′(x)]− 2L[y(x)] =2

s + 3

sL[y(x)] − y(0)− 2L[y(x)] =2

s + 3

(s − 2)L[y(x)] + 2 =2

s + 3

(s − 2)L[y(x)] =2

s + 3− 2

Therefore,

L[y(x)] = Y (s) =2

(s − 2)(s + 3)− 2

s − 2.

Now, by partial fraction decomposition

2(s − 2)(s + 3)

=2/5s − 2

− 2/5s + 3

.

Therefore,

Y (s) =2/5

s − 2− 2/5

s + 3− 2

s − 2=

−8/5s − 2

− 2/5s + 3

and

y(x) = L−1

[−8/5s − 2

− 2/5s + 3

]= −8

5 e2x − 25 e−3x. �

133

Next, we finish Example 3 of Section 4.2.


y′′ + 4y = 3e2x; y(0) = 5, y′(0) = −2.

SOLUTION From the Example, if y = y(x) is the solution, then

L[y′′(x) + 4y(x)] = L[3e2x] = 3L[e2x] =3

s − 2

L[y′′(x)] + 4L[y(x)] =3

s − 2

s2L[y(x)]− sy(0) − y′(0) + 4L[y(x)] =3

s − 2

(s2 + 4)L[y(x)]− 5s − (−2) =3

s − 2

(s2 + 4)L[y(x)] =3

s − 2+ 5s − 2

Therefore,

L[y(x)] = Y (s) =3

(s − 2)(s2 + 4)+

5s − 2s2 + 4

.


3(s − 2)(s2 + 4)

=38

s − 2−

38 s + 3

4

s2 + 4.

Therefore,

Y (s) =38

s − 2−

38 s + 3

4

s2 + 4+

5s − 2s2 + 4

=38

1s − 2

+378

s

s2 + 4− 11

82

s2 + 4

and

y(x) = L−1[Y (s)] =38L−1

[1

s − 2

]+

378

L−1

[s

s2 + 4

]− 11

8L−1

[2

s2 + 4

]

= 38 e2x + 37

8 cos 2x − 118 sin 2x. �


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

134


L[y′′(x) − 5y′(x) + 6y(x)] = L[x − 1] = L[x] − L[1] =1s2

− 1s

L[y′′(x)]− 5L[y′(x)] + 6L[y(x)] =1 − s

s2

s2L[y(x)]− sy(0) − y′(0)− 5 {sL[y(x)]− y(0)}+ 6L[y(x)] =1 − s

s2

(s2 − 5s + 6)L[y(x)]− 1 =1 − s

s2

(s2 − 5s + 6)L[y(x)] =1 − s

s2+ 1.

Therefore,

L[y(x)] = Y (s) =1 − s

s2(s2 − 5s + 6)+

1s2 − 5s + 6

=1 − s

s2(s − 2)(s − 3)+

1(s − 2)(s− 3)

.


1 − s

s2(s − 2)(s− 3)= − 1

36

(1s

)+

16

(1s2

)+

14

(1

s − 2

)− 2

9

(1

s − 3

)

and1

(s − 2)(s− 3)=

−1s − 2

+1

s − 3.

Therefore,

Y (s) = − 136

(1s

)+

16

(1s2

)− 3

4

(1

s − 2

)+

79

(1

s − 3

)

andy(x) = − 1

36 + 16 x − 3

4 e2x + 79 e3x. �

In many applications of differential equations it is not required to determine the solutionsexplicitly. Instead what is needed is information about the solutions. Often such informationcan be obtained by analyzing their Laplace transforms. The next example illustrates thistype of application.

Example 9. Consider the differential equation

y′′ − y′ − 6y = 2e−x on [0,∞)

together with the condition y(0) = −1. From Chapter 3, we know that the general solutionof the differential equation has the form

y(x) = C1e−2x + C2e

3x + Ae−x (∗)

where C1, C2 are arbitrary constants and A is a constant which can be determined

135

The question we want to examine is: Can we choose a value for α = y′(0) so thatsolution of the resulting initial value-problem

y′′ − y′ − 6y = 2e−x; y(0) = −1, y′(0) = α

has limit 0 as x → ∞? Since e−2x → 0 and e−x → 0 as x → ∞, we want to chooseα so that the coefficient of the e3x term is 0.

If y = y(x) is the solution of the initial-value problem, then

L[y′′(x) − y′(x)− 6y(x)] = L[2e−x] =2

s + 1

L[y′′(x)]− L[y′(x)]− 6L[y(x)] =2

s + 1

s2L[y(x)]− sy(0)− y′(0)− {sL[y(x)]− y(0)} − 6L[y(x)] =2

s + 1

(s2 − s − 6)L[y(x)] + s − α − 1 =2

s + 1

(s2 − s − 6)L[y(x)] =2

s + 1+ 1 + α − s.

Therefore,

L[y(x)] = Y (s) =2

(s + 1)(s2 − s − 6)+

1 + α − s

s2 − s − 6

=2

(s + 1)(s + 2)(s − 3)+

1 + α − s

(s + 2)(s − 3).


2(s + 1)(s + 2)(s− 3)

=A

s + 1+

B

s + 2+

C

s − 3=

−12

s + 1+

25

s + 2+

110

s − 3

and1 + α − s

(s + 2)(s − 3)=

D

s − 2+

E

s − 3=

−α+35

s + 2+

α−25

s − 3.

Combining these results, we have

Y (s) = −α + 15

(1

s + 2

)+

2α − 310

(1

s − 3

)− 1

2

(1

s + 1

).

Clearly, if 2α − 3 = 0, that is, if α = 3/2, then the e3x term in (∗) is eliminated. Theresulting solution is:

y(x) = −12 e−2x − 1

2e−x.

This solution has initial values: y(0) = −1, y′(0) = 32 , and lim

x→∞y(x) = 0. �

136

Exercises 4.3

Find L−1[F (s)]

1. F (s) =6

s + 7.

2. F (s) =1

2s + 2.

3. F (s) =1

s2 + 25.

4. F (s) =4s− 3

s − 4.

5. F (s) =s + 4

s2 + 8s + 17.

6. F (s) =4

s2 − 6s + 13.

7. F (s) =s + 4

s2 + 4s + 8.

8. F (s) =2s− 5

s2+

1s3

.

9. F (s) =2

(s + 2)2− s

s2 − 2s + 2.

10. F (s) =s + 3

s2 − 2s + 9.

Use partial fraction decomposition to find the inverse Laplace transform.

11. F (s) =1

(s + 1)(s2 + 1).

12. F (s) =s + 3

s2 − s − 2.

13. F (s) =1

s(s2 + 4).

14. F (s) =1

s2 − 1.

15. F (s) =s2 − 3s − 1s3 + s2 − 2s

.

16. F (s) =s

s4 − 1.

17. F (s) =4

s2(s − 1)(s− 2).

Find the solution of the initial-value problem.

137

18. y′ − 4y = 0; y(0) = −2.

19. y′ + 2y = ex; y(0) = 1.

20. y′ − 3y = e2x; y(0) = 2.

21. y′ + y = sin x; y(0) = 1.

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

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

24. y′′ − y = sin x; y(0) = 1, y′(0) = 1.

25. y′′ − y = ex; y(0) = 1, y′(0) = 0.

26. y′′ − y′ − 2y = sin 2x; y(0) = 1, y′(0) = 1.

27. y′′ + 2y′ + y = x + ex; y(0) = −74 , y′(0) = 9

4 .

28. y′′ + 2y′ + y = 4e−x; y(0) = 2, y′(0) = −1.

29. y′′ + 3y′ + 2y = 6ex; y(0) = 2, y′(0) = −1.

30. y′′ − 2y′ + 5y = 3e−2x; y(0) = 1, y′(0) = 1.

31. Given the initial-value problem

y′′ − y′ − 6y = 2e−x; y(0) = α, y′(0) = −1. (See Example 9)

What value should be assigned to α so that the resulting solution will have limit 0as x → ∞?

32. What initial conditions should be assigned with the differential equation

y′′ + y = e−x

so that limx→∞

y(x) = 0 where y = y(x) is the solution.

33. Consider the differential equation: y′′ + y′ − 2y = 3 sin 2x together with the initialvalue y(0) = 2. For what value(s) of β = y′(0) will the resulting solution(s) bebounded?

34. Consider the differential equation: y′′ + 3y′ + 2y = x together with the initial valuey′(0) = −2. For what value(s) of α = y(0) will the resulting solution(s) be bounded?

The Laplace transform method applies to initial-value problems in which the initialvalues are specified at x = 0. Actually, the method can be applied when the initial

138

conditions are specified at some point a 6= 0. All that is required is a simple changeof independent variable; a translation. For example, the initial-value problem

d2y

dx2− 3

dy

dx+ 2y = x; y(1) = 0, y′(1) = 1

is transformed into

d2y

dt2− 3

dy

dt+ 2y = t + 1, y(0) = 0, y′(0) = 1

by setting t = x−1. With this transformation, we have: t = 0 when x = 1; x = t+1;and

dy

dx=

dy

dt

dt

dx=

dy

dt· 1 =

dy

dt

d2y

dx2=

d

dx

[dy

dx

]=

d

dt

[dy

dx

]dt

dx=

d

dt

[dy

dt

]· 1 =

d2y

dt2.

35. Find the solution of the initial-value problem: y′′− 3y′ + 2y = x; y(1) = 0, y′(1) = 1by first solving the transformed problem.

36. Use the Laplace transform method to solve the initial-value problem

y′ − 2y = 1 + x; y(2) = −1.

139

4.4 Applications to Discontinuous Functions

In the preceding sections we assumed that the functions under consideration were continuouson [0,∞). In this section we consider Laplace transforms of certain types of discontinuousfunctions that occur in various applications.

A particular example of the type of discontinuous function that we will be consideringin this section is the unit step function, or Heaviside function u. This function is definedon (−∞,∞) by

u(x) =

{0 x < 01 x ≥ 0.

(1)

The graph of u is

-10 -5 5 10x

-1

1

2

3

y

It is clear that u is continuous on (−∞,∞) except at x = 0. At x = 0, limx→0

u(x)does not exist. However, note that the left-hand and right-hand limits of u at x = 0 doexist:

limx→0−

u(x) = 0 and limx→0+

u(x) = 1.

Recall from calculus that a function f is continuous at a point c if and only iflimx→c

f(x) = f(c), and limx→c

f(x) exists if and only if the left-hand and right-hand limits off at c both exist and are equal.

DEFINITION 1. (Jump Discontinuity) Let the function f = f(x) be defined onan interval I and continuous except at a point c ∈ I , c not an endpoint of I . If theleft-hand and right-hand limits of f at c both exist but are not equal, then f is said tohave a jump (or finite ) discontinuity at c.

Example 1. (a) The unit step function u is continuous on (−∞,∞) except at x = 0where it has a jump discontinuity.

(b) Let f be defined on [−5, 5] by

f(x) =

x + 5 −5 ≤ x < −2

− 1x

−2 ≤ x < 0

2x 0 ≤ x ≤ 2−(x − 5) 2 < x ≤ 5

140

The graph of f is

-5 -2 2 5x

-1

1

3

4

y

Clearly f is discontinuous at x = −2, x = 0, and x = 2. Since

limx→−2−

f(x) = 3 and limx→−2+

f(x) = 12 ,

andlim

x→2−f(x) = 4 and lim

x→2+f(x) = 3,

f has jump discontinuities at x = −2 and x = 2.

The discontinuity at x = 0 is not a jump discontinuity because limx→0−

f(x) does not

exist (f(x) → ∞ as x → 0−).

(c) Let g be defined on [0,∞) by

g(x) =

x 0 ≤ x < 13 x = 12 1 ≤ x < 5e−(x−5) x ≥ 5

The graph of g is

1 5x

-1

1

2

3

y

This function has jump discontinuities at x = 1 and x = 5. Note that the value of g

at x = 1 is independent of the left-hand and right-hand limits of g at x = 1. �

DEFINITION 2. (Piecewise Continuous) A function f defined on an interval I ispiecewise continuous on I if it is continuous on I except for at most a finite number ofpoints c1, c2, . . . , cn of I at which it has a jump discontinuity.

141

Remark A continuous function is also piecewise continuous. �

Referring to Example 1, the unit step function u is piecewise continuous on (−∞,∞);the function g in (c) is piecewise continuous on [0,∞). the function f in (b) isnot piecewise continuous on [−5, 5] because the discontinuity at x = 0 is not a jumpdiscontinuity.

In this section we want to apply Laplace transform methods to piecewise continuousfunctions. Before we can calculate the Laplace transform of a piecewise function we haveto know how to integrate such a function.

Suppose that f is piecewise continuous on [a, b] with jump discontinuities at c1 <

c2 < · · · < cn. It follows from the definition of the definite integral that∫ b

af(x) dx =

∫ c1

af(x) dx +

∫ c2

c1

f(x) dx + · · ·+∫ b

cn

f(x) dx.

Example 2. Let f be defined on [0, 3] by

f(x) =

{x 0 ≤ x ≤ 14 − (x − 1)2 1 < x ≤ 3

The graph of f is

1 2 3x

-1

1

4

y

We have∫ 3

0f(x) dx =

∫ 1

0x dx +

∫ 3

1(4− (x − 1)2) dx

=[x2

2

]1

0

+[4x − (x − 1)3

3

]3

1

=356

. �

The integration of piecewise continuous functions can be extended to include (improper)integrals on an infinite interval.

142

Example 3. Consider the function g of Example 1:∫ ∞

0g(x) dx =

∫ 1

0x dx +

∫ 5

12 dx +

∫ ∞

5e−(x−5) dx

=[x2

2

]1

0

+ [2x]51 + limb→∞

∫ b

5e−(x−5) dx

=12

+ 8 + limb→∞

[−e−(x−5)

]b

5

=172

+ [0 + 1] =192

. �

The following theorem is an extension of Theorem 1, Section 4.2.

THEOREM 1. If the function f is piecewise continuous on [0,∞), and of exponentialorder λ, then the Laplace transform L[f(x)] exists for s > λ.

Let c be a real number. The translation of the unit step function u by c is thefunction uc = u(x − c) defined on (−∞,∞) by

u(x− c) =

{0 x < c

1 x ≥ c.(2)

The graph of uc for c > 0 is

cx

-1

1

2

y

Example 4. The Laplace transform of uc, c > 0 is

L[u(x− c)] =e−cs

s, s > 0. (3)

Proof: By definition

L[u(x − c)] =∫ ∞

0

e−sxu(x − c) dx

=∫ c

0e−sx · 0 dx +

∫ ∞

ce−sx · 1 dx

= limb→∞

∫ b

ce−sx dx = lim

b→∞

[e−sx

−s

]b

c

= limb→∞

e−sb

−s+

e−sc

s=

e−cs

s, s > 0.

143

Note that if c = 0, then u0 = u(x − 0) = u(x) ≡ 1 on [0,∞), and L[u(x)] = L[1] =1/s, s > 0 as we saw in Section 4.1. �

Example 5. (a) Let f be the function defined on [0,∞) by

f(x) =

{1 0 ≤ x < 52 5 ≤ x < ∞.

1 3 5 7x

-1

1

2

3

y

Then

L[f(x)] =∫ ∞

0e−sxf(x) dx =

∫ 5

0e−sx · 1 dx +

∫ ∞

5e−sx · 2 dx

=[−e−sx

s

]5

0

+ limb→∞

∫ b

02e−sx dx

=−e−5s

s+

1s

+ limb→∞

[2e−sx

−s

]b

5

=−e−5s

s+

1s

+ limb→∞

2e−sb

−s+

2e−5s

s

=−e−5s

s+

1s

+2e−5s

s=

1s

+e−5s

s, s > 0.

(b) We calculate L[g(x)] where g is the function given in Example 1(c).

L[g(x)] =∫ ∞

0e−sxg(x) dx =

∫ 1

0e−sx · x dx +

∫ 5

1e−sx · 2 dx +

∫ ∞

5e−sxe−(x−5) dx

=[−xe−sx

s− e−sx

s2

]1

0

+[−2e−sx

s

]5

1

+ limb→∞

∫ b

0e−[x(s+1)−5] dx

=−e−s

s+

−e−s

s2+

1s2

+−2e−5s

s+

2e−s

s+ lim

b→∞

[−e−[x(s+1)−5]

s + 1

]∞

5

= −e−s

s− e−s

s2+

1s2

− 2e−5s

s+

2e−s

s+

e−5s

s + 1

=1s2

+e−s

s− e−s

s2− 2e−5s

s+

e−5s

s + 1. �

144

The unit step function and its translations can be used to obtain translations of arbitraryfunctions. For example, if f is defined on [0,∞) and c > 0, then the function

fc(x) = f(x − c)u(x − c) =

{0 x < c

f(x − c)u(x− c) x ≥ c

is the function f shifted c units to the right as illustrated in the figure below.

cx

y

cx

y

THEOREM 2. Let f be defined on [0,∞) and suppose L[f(x)] = F (s) exists fors > λ. If c > 0, then L[fc(x)] exists for s > λ and is given by

L[fc(x)] = L[f(x − c)u(x− c)] = e−csF (s).

Proof: By the definition,

L[fc(x)] = L[f(x − c)u(x− c)] =∫ ∞

0e−sxf(x− c)u(x− c) dx

= limb→∞

∫ b

ce−sxf(x − c) dx.

Now let t = x − c. Then

x = t + c, dx = dt, and t = 0 when x = c.

With this change of variable,

limb→∞

∫ b

c

e−sxf(x − c) dx = limb→∞

∫ b−c

0

e−s(t+c)f(t) dt

= e−cs limb→∞

(∫ b−c

0e−stf(t) dt

)

= e−cs

∫ ∞

0e−stf(t) dt = e−csF (s)

since b− c → ∞ as b → ∞. �

145

This theorem can be expressed in an equivalent manner using the inverse Laplace trans-form.

COROLLARY If L−1[F (s)] = f(x) and c > 0, then

L−1[e−csF (s)] = f(x − c)u(x− c) = fc(x).

We now illustrate how to use translations of the unit step function and Theorem 2 tocalculate Laplace transforms of piecewise continuous functions.

Example 6. We calculate L[f(x)] where f is the function given in Example 4:

f(x) =

{1 0 ≤ x < 52 5 ≤ x < ∞.

Since f has a jump discontinuity at x = 5, we’ll write f in terms of u(x−5). Definef1(x) by

f1(x) =

{1 0 ≤ x < 50 5 ≤ x < ∞

= 1 − u(x− 5)

and f2(x) by

f2(x) =

{0 0 ≤ x < 52 5 ≤ x < ∞

= 2u(x− 5).

Thenf(x) = f1(x) + f2(x) = 1 − u(x − 5) + 2u(x − 5) = 1 + u(x − 5)

and

L[f(x)] = L[1 + u(x− 5)] = L[1] + L[u(x − 5)] =1s

+e−5s

s, s > 0. �

Example 7. Let h(x) =

{x2 0 ≤ x < 23 2 ≤ x < ∞.

. Calculate L[h(x)].

SOLUTION The graph of h is

1 2 3x

-1

1

2

3

4

y

Let h1(x) = x2 − x2 u(x− 2) and h2(x) = 3u(x− 2). Then

h(x) = h1(x) + h2(x) = x2 − x2 u(x − 2) + 3u(x − 2)

146

and

L[h(x)] = L[x2] − L[x2 u(x − 2)] + 3L[u(x − 2)] =2s3

− L[x2 u(x− 2)] + 3e−2s

s.

Before we can apply Theorem 2 to calculate L[x2 u(x − 2)] we must have the coefficientx2 of u(x − 2) expressed as a function of (x − 2). Since

x2 = [(x − 2) + 2]2 = (x − 2)2 + 4(x − 2) + 4

we have

L[x2 u(x − 2)] = L[(x− 2)2u(x − 2) + 4(x − 2)u(x− 2) + 4u(x− 2)

]

= L[(x − 2)2u(x − 2)] + 4L[(x− 2)u(x− 2)] + 4L[u(x − 2)]

= e−2s 2s3

+ 4e−2s 1s2

+ 4e−2s 1s.

It now follows that

L[h(x)] =2s3

−[e−2s 2

s3+ 4e−2s 1

s2+ 4e−2s 1

s

]+ 3

e−2s

s=

2s3

− 2e−2s

s3− 4e−2s

s2− e−2s

s. �

Example 8. Let

g(x) =

2 0 ≤ x < 2x 2 ≤ x < 44 x ≥ 4

.

The graph of g is

2 4x

1

2

3

4

y

Set

g1(x) = 2 − 2 u(x− 2)

g2(x) = x u(x− 2)− x u(x− 4) = [(x − 2) + 2)]u(x− 2) − [(x − 4) + 4]u(x− 4)

= (x − 2)u(x− 2) + 2u(x− 2)− (x− 4)u(x− 4) − 4u(x − 4)

g3(x) = 4 u(x− 4).

Theng(x) = g1(x) + g2(x) + g3(x) = 2 + (x − 2)u(x− 2) − (x − 4)u(x − 4).

147

Therefore,

L[g(x)] =2s

+ e−2s 1s2

− e−4s 1s2

�

The Corollary to Theorem 2 is used to calculate inverse Laplace transforms.

Example 9. Suppose

F (s) =1s2

+e−4s

s − 2.

Then

L−1

[1s2

+e−4s

s − 2

]= L−1

[1s2

]+ L−1

[e−4s

s − 2

]

From the Table of Laplace transforms L−1[1/s2

]= x. To calculate L−1

[e−4s/(s − 2)

],

let F (s) = 1/(s − 2). Then L−1[F (s)] = e2x. Therefore, by the Corollary to Theorem 2,

L−1

[e−4s

s − 2

]= e2(x−4)u(x− 4).

It now follows that

L−1[F (s)] = x + e2(x−4)u(x − 4) =

{x 0 ≤ x < 4

x + e2(x−4) 4 ≤ x < ∞. �

Example 10. Suppose

F (s) =e−s

(s + 1)2+

e−πs

s2 + 4.

Then

L−1[F (s)] = L−1

[e−s

(s + 1)2

]+ L−1

[e−πs

s2 + 4

].

To calculate L−1[e−s/(s + 1)2], let F1(s) = 1/(s + 1)2. Then, from the Table of Laplacetransforms, L−1[F1(s)] = xe−x. Therefore,

L−1

[e−s

(s + 1)2

]= (x − 1)e−(x−1) u(x− 1).

To calculate L−1[e−πs/(s2 + 4)], let

F2(s) =1

s2 + 4=

12

2s2 + 4

.

From the Table of Laplace transforms, L−1[F2(s)] = 12 sin 2x. Therefore,

L−1

[e−πs

s2 + 4

]= 1

2 sin 2(x− π) u(x − π) = 12 sin 2x u(x − π). (sin 2(x− π) = sin 2x)

Finally,L−1[F (s)] = (x− 1)e−(x−1) u(x − 1) + 1

2 sin 2x u(x − π)

=

0 0 ≤ x < 1(x − 1)e−(x−1) 1 ≤ x < π

(x − 1)e−(x−1) + 12 sin 2x x ≥ π

148

Exercises 4.4

Use the definition of the Laplace transform to find L[f ].

1. f(x) =

x 0 ≤ x < 1

x − 1 1 ≤ x < 2

0 x ≥ 2

.

2. f(x) =

{−1 0 ≤ x < 1

x − 1 x ≥ 1.

3. f(x) =

{0 0 ≤ x < 5

2 x ≥ 5.

4. f(x) =

{2x 0 ≤ x < 3

1 x ≥ 3.

5. f(x) =

1 0 ≤ x < 2

x − 2 2 ≤ x < 4

e−(x−4) x ≥ 4

.

6. f(x) =

{1 0 ≤ x < 1

−1 1 ≤ x < 2

and extended periodically on [2,∞). See Exercises 4.1, Problem 10.

7. f(x) =

{2x 0 ≤ x < 2

2 2 ≤ x < 4

and extended periodically on [4,∞). (Sketch the graph of f .)

8. f(x) = x − n on [n, n + 1), n = 0, 1, 2, . . .. (Sketch the graph of f .)

Express the function f in terms of the unit step function and its translations. Then findL[f ].

9. f(x) =

{x2 0 ≤ x < 3

3x x ≥ 3.

10. f(x) =

sin x 0 ≤ x < π

sin 2x π ≤ x < 2π

sin 3x x ≥ 2π

. Sketch the graph of f .

149

11. f(x) =

0 0 ≤ x < π

1 + cos x π ≤ x < 2π

2 cos x x ≥ 2π

.

12. f(x) =

x 0 ≤ x < 2

0 2 ≤ x < 4

(x − 4)2 x ≥ 4

.

13. f(x) =

x 0 ≤ x < 1

2 − x 1 ≤ x < 3

x − 4 3 ≤ x < 4

0 x ≥ 4

. Sketch the graph of f .

Calculate the inverse Laplace transform of F (s).

14. F (s) =e−s

s+

2e−3s

s− 6e−4s

s.

15. F (s) =1 + e−πs

s2 + 1.

16. F (s) =1

s + 1− e−2s

s + 1.

17. F (s) =s + (s − 1)e−πs

s2 + 1.

18. F (s) =2s2

+e−2s

s2+

2e−2s

s3− 4e−3s

s.

19. F (s) =e−2s

s(s + 1).

20. F (s) =1 − e−2s

s2 + π2.

150

4.5 Initial-Value Problems with Piecewise Continuous Nonhomogeneous

Terms

In this section we illustrate the application of Laplace transforms to the solution of nonho-mogeneous initial-value problems in which the forcing function f is piecewise continuous.


y′ + 2y = f(x); y(0) = 4

where f(x) =

{x 0 ≤ x < 35 x ≥ 3.

SOLUTION The function f has a jump discontinuity at x = 3. Therefore, the first stepis to express f in terms of u3(x) = u(x− 3). You should verify that

f(x) = x−x u(x−3)+5u(x−3) = x−[(x−3)+3]u(x−3)+5u(x−3) = x−(x−3)u(x−3)+2u(x−3).

Now, taking the Laplace transform of the equation, we get

L[y′ + 2y] = sL[y] − y(0) + 2L[y] =1s2

− e−3s

s2+ 2

e−3s

s

Applying the initial condition and solving for L[y], we find that

L[y] = Y (s) =1

s2(s + 2)+

e−3s

s2(s + 2)+ 2

e−3s

s(s + 2)+

4s + 2

=4s2 + 1

s2(s + 2)+

(2s + 1)e−3s

s2(s + 2).

By partial fraction decomposition

4s2 + 1s2(s + 2)

= −14

1s

+12

1s2

+174

1s + 2

(2s + 1)e−3s

s2(s + 2)=

34

e−3s

s+

12

e−3s

s2− 3

4e−3s

s + 2

Therefore,

Y (s) = −14

1s

+12

1s2

+174

1s + 2

+34

e−3s

s+

12

e−3s

s2− 3

4e−3s

s + 2

and

y(x) = L−1[Y (s)] = −14 + 1

2 x + 174 e−2x + 3

4 u(x− 3) + 12 (x − 3) u(x− 3) − 3

4 e−2(x−3) u(x − 3)

=

{−1

4 + 12x + 17

4 e−2x 0 ≤ x < 3

x − 1 + 174 e−2x − 3

4e−2(x−3) x ≥ 3

151

The graph of y is

3x

y


y′′ + 4y = f(x); y(0) = 1, y′(0) = 0

where f(x) =

{1 0 ≤ x < π/2−2 x ≥ π/2.

SOLUTION The function f has a jump discontinuity at x = π/2. Therefore, we expressf in terms of u(x − π/2):

f(x) = 1 − 3u(x − π/2).

Applying the Laplace transform to the equation and using the initial conditions, we get

L[y′′ + 4y] = s2L[y] − sy(0) − y′(0) + 4L[y] =1s− 3e−πs/2

s

(s2 + 4)L[y]− s =1s− 3e−πs/2

s.

Solving for L[y] = Y (s), we have

Y (s) =1

s(s2 + 4)− 3e−πs/2 1

s(s2 + 4)+

s

s2 + 4.

By partial fraction decomposition,

1s(s2 + 4)

=14

1s− 1

4s

s2 + 4.

Therefore,

Y (s) =14

1s− 1

4s

s2 + 4− 3

4e−πs/2

s+

34

e−πs/2 s

s2 + 4+

s

s2 + 4

and

y(x) = L−1[Y (s)] = 14 − 1

4 cos 2x − 34 u(x− π/2) + 3

4 cos 2(x− π/2)u(x− π/2) + cos 2x

=

14 + 3

4 cos 2x 0 ≤ x < π/2

−12 x ≥ π/2.

152

The graph of y is

Π2

x

-1

-0.5

1

y

Exercises 4.5

Solve the initial-value problem.

1. y′ − 2y = f(x); y(0) = 2 where

f(x) =

{1 0 ≤ x < 1

2 x ≥ 1.

2. y′ + 3y = f(x); y(0) = 1 where

f(x) =

{sin x 0 ≤ x < π

0 x ≥ π.

3. y′′ + y = f(x); y(0) = 0, y′(0) = 1 where

f(x) =

{1 0 ≤ x < 1

0 x ≥ 1.

4. y′′ + 4y = f(x); y(0) = 1, y′(0) = 0 where

f(x) =

{0 0 ≤ x < 2

1 x ≥ 2.

5. y′′ + 2y′ + y = f(x); y(0) = 0, y′(0) = 0 where

f(x) =

{1 0 ≤ x < 2

x − 1 x ≥ 2.

6. y′′ + 4y = f(x); y(0) = 0, y′(0) = 0 where

f(x) =

{sin x 0 ≤ x < 2π

0 x ≥ 2π.

153

7. y′′ − 4y′ + 3y = f(x); y(0) = 0, y′(0) = 0 where

f(x) =

{−1 0 ≤ x < 1

1 x ≥ 1.

8. y′′ − 3y′ + 2y = f(x); y(0) = 0, y′(0) = 0 where

f(x) =

{0 0 ≤ x < 2

2x − 4 x ≥ 2.

9. y′′ + 2y′ + y = f(x); y(0) = 3, y′(0) = −1 where

f(x) =

{ex 0 ≤ x < 1

ex − 1 x ≥ 1.

10. y′′ − 4y′ + 4y = f(x); y(0) = 1, y′(0) = −1 where

f(x) =

{e2x 0 ≤ x < 2

−e2x x ≥ 2.

154

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