Laplace Transformation PDF.

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Laplace Transforms: Theory, Problems, andSolutions

Marcel B. FinanArkansas Tech University

c All Rights Reserved

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Contents

41 The Laplace Transform: Basic Denitions and Results 3

42 Further Studies of Laplace Transform 15

43 The Laplace Transform and the Method of Partial Fractions 29

44 Laplace Transforms of Periodic Functions 36

46 Convolution Integrals 46

47 The Dirac Delta Function and Impulse Response 55

48 Solutions to Problems 64

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41 The Laplace Transform: Basic Denitions

and ResultsLaplace transform is yet another operational tool for solving constant coeffi-cients linear differential equations. The process of solution consists of threemain steps:

The given hard problem is transformed into a simple equation. This simple equation is solved by purely algebraic manipulations. The solution of the simple equation is transformed back to obtain the so-lution of the given problem.In this way the Laplace transformation reduces the problem of solving a dif-ferential equation to an algebraic problem. The third step is made easier bytables, whose role is similar to that of integral tables in integration.The above procedure can be summarized by Figure 41.1

Figure 41.1In this section we introduce the concept of Laplace transform and discusssome of its properties.

The Laplace transform is dened in the following way. Let f (t) be denedfor t 0. Then the Laplace transform of f, which is denoted by L[f (t)]or by F (s), is dened by the following equation

L[f (t)] = F (s) = limT T 0 f (t)e st dt =

0f (t)e st dt

The integral which dened a Laplace transform is an improper integral. Animproper integral may converge or diverge , depending on the integrand.When the improper integral in convergent then we say that the function f (t)possesses a Laplace transform. So what types of functions possess Laplacetransforms, that is, what type of functions guarantees a convergent improper

integral.Example 41.1Find the Laplace transform, if it exists, of each of the following functions

(a) f (t) = eat (b) f (t) = 1 ( c) f (t) = t (d) f (t) = et2

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Solution.

(a) Using the denition of Laplace transform we see that

L[eat ] =

0e (s a)t dt = lim

T T 0 e (s a)t dt.But

T 0 e (s a)t dt = T if s = a1 e ( s a ) T s a if s = a.For the improper integral to converge we need s > a. In this case,

L[eat ] = F (s) =1

s a, s > a.

(b) In a similar way to what was done in part (a), we nd

L[1] =

0e st dt = lim

T T 0 e st dt = 1s , s > 0.(c) We have

L[t ] =

0te st dt =

te st

s e st

s2

0=

1s2

, s > 0.

(d) Again using the denition of Laplace transform we nd

L[et 2

] =

0 et2 st

dt.

If s 0 then t 2 st 0 so that et2 st 1 and this implies that

0 et 2 st dt

0 . Since the integral on the right is divergent, by the comparison theoremof improper integrals (see Theorem 41.1 below) the integral on the left is alsodivergent. Now, if s > 0 then

0 et (t s)dt

s dt. By the same reasoningthe integral on the left is divergent. This shows that the function f (t ) = et2

does not possess a Laplace transform

The above example raises the question of what class or classes of functionspossess a Laplace transform. Looking closely at Example 41.1(a), we noticethat for s > a the integral

0 e (s a)t dt is convergent and a critical compo-

nent for this convergence is the type of the function f (t). To be more specic,if f (t) is a continuous function such that

|f (t)| Me at , t C (1)4


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where M 0 and a and C are constants, then this condition yields

0f (t)e st dt

C

0f (t)e st dt + M

C e (s a)t dt.

Since f (t) is continuous in 0 t C, by letting A = max {|f (t)| : 0 t C }we have

C 0 f (t)e st dt A C 0 e st dt = A 1s e sC

s< .

On the other hand, Now, by Example 41.1(a), the integral

C e (s a)t dt is

convergent for s > a. By the comparison theorem of improper integrals (seeTheorem 41.1 below) the integral on the left is also convergent. That is, f (t)possesses a Laplace transform.We call a function that satises condition (1) a function with an exponentialorder at innity. Graphically, this means that the graph of f (t) is containedin the region bounded by the graphs of y = Me at and y = Me at for t C.Note also that this type of functions controls the negative exponential in thetransform integral so that to keep the integral from blowing up. If C = 0then we say that the function is exponentially bounded .

Example 41.2

Show that any bounded function f (t) for t 0 is exponentially bounded.Solution.Since f (t) is bounded for t 0, there is a positive constant M such that|f (t )| M for all t 0. But this is the same as (1) with a = 0 and C = 0 .Thus, f (t ) has is exponentially boundedAnother question that comes to mind is whether it is possible to relax thecondition of continuity on the function f (t ). Lets look at the following situ-ation.

Example 41.3Show that the square wave function whose graph is given in Figure 41.2possesses a Laplace transform.

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(a) f (t) = tn (b) f (t) = tn sin at

Solution.(a) Since et = n =0

t n

n ! tn

n ! we have tn n !et . Hence, tn is piecewise con-tinuous and exponentially bounded.

(b) Since |tn sin at | n !et , tn sin at is piecewise continuous and exponentiallyboundedNext, we would like to establish the existence of the Laplace transform forall functions that are piecewise continuous and have exponential order atinnity. For that purpose we need the following comparison theorem fromcalculus.

Theorem 41.1Suppose that f (t) and g(t) are both integrable functions for all t t0 suchthat |f (t)| |g(t ) for t t0. If

t 0 g(t )dt is convergent, then

t 0 f (t )dt isalso convergent. If, on the other hand,

t0 f (t)dt is divergent then

t0 f (t)dtis also divergent.

Theorem 41.2 (Existence)Suppose that f (t) is piecewise continuous on t 0 and has an exponentialorder at innity with |f (t)| Me at for t C. Then the Laplace transform

F (s) =

0f (t)e st dt

exists as long as s > a. Note that the two conditions above are sufficient, butnot necessary, for F (s) to exist.

Proof.The integral in the denition of F (s) can be splitted into two integrals asfollows

0f (t)e st dt = C 0 f (t)e st dt +

C f (t)e st dt.

Since f (t) is piecewise continuous in 0 t C, it is bounded there. Byletting A = max {|f (t)| : 0 t C }we have

C 0 f (t)e st dt A C 0 e st dt = A 1s e sC

s< .

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Now, by Example 41.1(a), the integral

C f (t )e st dt is convergent for s > a.

By Theorem 41.1 the integral on the left is also convergent. That is, f (t)possesses a Laplace transform

In what follows, we will denote the class of all piecewise continuous func-tions with exponential order at innity by PE . The next theorem shows thatany linear combination of functions in PE is also in PE . The same is true forthe product of two functions in PE .Theorem 41.3Suppose that f (t) and g(t ) are two elements of PE with

|f (t)

| M 1ea 1 t , t

C 1 and

|g(t)

| M 2ea 1 t , t

C 2.

(i) For any constants and the function f (t) + g (t) is also a member of

PE . MoreoverL[f (t) + g (t)] = L[f (t )] + L[g(t)].

(ii) The function h(t) = f (t)g(t) is an element of PE .Proof.(i) It is easy to see that f (t) + g (t ) is a piecewise continuous function.Now, let C = C 1 + C 2, a = max {a1, a 2}, and M = | |M 1 + | |M 2. Then fort

C we have

|f (t) + g (t )| | ||f (t )|+ | ||g(t)| | |M 1ea 1 t + | |M 2ea 2 t Me at .This shows that f (t) + g (t) is of exponential order at innity. On theother hand,

L[f (t) + g (t)] = limT T 0 [f (t ) + g (t)]dt= lim

T T 0 f (t)dt + limT T 0 g(t)dt=

L[f (t)] +

L[g(t )].

(ii) It is clear that h(t) = f (t)g(t ) is a piecewise continuous function. Now,letting C = C 1 + C 2, M = M 1M 2, and a = a1 + a2 then we see that for t C we have

|h(t )| = |f (t)||g(t)| M 1M 2e(a 1 + a 2 )t = Me at .8


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Hence, h(t) is of exponential order at innity. By Theorem 41.2 , L[h(t)]exists for s > a

We next discuss the problem of how to determine the function f (t) if F (s)is given. That is, how do we invert the transform. The following result onuniqueness provides a possible answer. This result establishes a one-to-onecorrespondence between the set PE and its Laplace transforms. Alterna-tively, the following theorem asserts that the Laplace transform of a memberin PE is unique.Theorem 41.4Let f (t) and g(t) be two elements in PE with Laplace transforms F (s) andG(s) such that F (s) = G(s) for some s > a. Then f (t) = g(t) for all t 0where both functions are continuous.The standard techniques used to prove this theorem( i.e., complex analysis,residue computations, and/or Fouriers integral inversion theorem) are gen-erally beyond the scope of an introductory differential equations course. Theinterested reader can nd a proof in the book Operational Mathematicsby Ruel Vance Churchill or in D.V. Widder The Laplace Transform.With the above theorem, we can now officially dene the inverse Laplacetransform as follows: For a piecewise continuous function f of exponentialorder at innity whose Laplace transform is F, we call f the inverse Laplacetransform of F and write f = L 1[F (s)]. Symbolically

f (t) = L 1[F (s)]F (s) = L[f (t)].

Example 41.5Find L

1 1s 1 , s > 1.

Solution.From Example 41.1(a), we have that L[eat ] = 1s a , s > a. In particular, fora = 1 we nd that L[et ] = 1s 1 , s > 1. Hence, L

1 1s 1 = e

t , t 0 .

The above theorem states that if f (t ) is continuous and has a Laplace trans-form F (s), then there is no other function that has the same Laplace trans-form. To nd L

1[F (s)], we can inspect tables of Laplace transforms of known functions to nd a particular f (t) that yields the given F (s).When the function f (t) is not continuous, the uniqueness of the inverse

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Laplace transform is not assured. The following example addresses the

uniqueness issue.Example 41.6Consider the two functions f (t) = h(t)h(3 t) and g(t) = h(t) h(t 3).(a) Are the two functions identical?(b) Show that L[f (t)] = L[g(t ).Solution.(a) We have

f (t) = 1, 0 t 30, t > 3and

g(t) = 1, 0 t < 30, t 3So the two functions are equal for all t = 3 and so they are not identical.(b) We have

L[f (t)] = L[g(t )] = 30 e st dt = 1 e 3s

s, s > 0.

Thus, both functions f (t) and g(t) have the same Laplace transform even

though they are not identical. However, they are equal on the interval(s)where they are both continuous

The inverse Laplace transform possesses a linear property as indicated inthe following result.

Theorem 41.5Given two Laplace transforms F (s) and G(s) then

L 1[aF (s) + bG(s)] = aL

1[F (s)] + bL 1[G(s)]

for any constants a and b.

Proof.Suppose that L[f (t)] = F (s) and L[g(t)] = G(s). Since L[af (t) + bg(t)] =aL[f (t)]+ bL[g(t)] = aF (s)+ bG(s) then L 1[aF (s)+ bG(s)] = af (t)+ bg(t ) =aL

1[F (s)] + bL 1[G(s)]

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Practice Problems

Problem 41.1Determine whether the integral

01

1+ t2 dt converges. If the integral con-verges, give its value.


0t



0 e t cos(e t )dt converges. If the integral

converges, give its value.Problem 41.4Using the denition, nd L[e3t ], if it exists. If the Laplace transform existsthen nd the domain of F (s).Problem 41.5Using the denition, nd L[t 5], if it exists. If the Laplace transform existsthen nd the domain of F (s).Problem 41.6

Using the denition, nd L[e(t

1)2

], if it exists. If the Laplace transformexists then nd the domain of F (s).

Problem 41.7Using the denition, nd L[(t 2)2], if it exists. If the Laplace transformexists then nd the domain of F (s).Problem 41.8Using the denition, nd L[f (t)], if it exists. If the Laplace transform existsthen nd the domain of F (s).

f (t) = 0, 0 t < 1t 1, t 1

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Problem 41.9

Using the denition, nd L[f (t)], if it exists. If the Laplace transform existsthen nd the domain of F (s).

f (t) =0, 0 t < 1t 1, 1 t < 20, t 2.

Problem 41.10Let n be a positive integer. Using integration by parts establish the reductionformula

tn e st dt =

tn e st

s+

ns

tn 1e st dt, s > 0.

Problem 41.11For s > 0 and n a positive integer evaluate the limits

limt 0 tn e st (b) lim t t n e st

Problem 41.12(a) Use the previous two problems to derive the reduction formula for theLaplace transform of f (t ) = tn ,

L[tn ] =ns L[tn

1], s > 0.

(b) Calculate L[tk], for k = 1 , 2, 3, 4, 5.(c) Formulate a conjecture as to the Laplace transform of f (t ), t n with n apositive integer.

From a table of integrals,

eu sin udu = eu sin u sin u 2 + 2 eu cos udu = eu cos u + sin u 2 + 2Problem 41.13Use the above integrals to nd the Laplace transform of f (t) = cos t, if itexists. If the Laplace transform exists, give the domain of F (s).

Problem 41.14Use the above integrals to nd the Laplace transform of f (t) = sin t, if itexists. If the Laplace transform exists, give the domain of F (s).

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Problem 41.15

Use the above integrals to nd the Laplace transform of f (t) = cos (t 2),if it exists. If the Laplace transform exists, give the domain of F (s).Problem 41.16Use the above integrals to nd the Laplace transform of f (t) = e3t sin t, if itexists. If the Laplace transform exists, give the domain of F (s).

Problem 41.17Use the linearity property of Laplace transform to nd L[5e

7t + t + 2 e2t ].Find the domain of F (s).

Problem 41.18

Consider the function f (t) = tan t.

(a) Is f (t) continuous on 0 t < , discontinuous but piecewise contin-uous on 0 t < , or neither?(b) Are there xed numbers a and M such that |f (t)| Me at for 0 t < ?Problem 41.19Consider the function f (t) = t2e t .

(a) Is f (t) continuous on 0 t < , discontinuous but piecewise contin-uous on 0

t 0.

A Heaviside function at 0 is the shifted function h(t ) ( units to theright). For this function, the Laplace transform is

L[h(t )] =

0h(t )e

st dt =

e st dt =

e st

s

=

e s

s, s > 0.

Laplace Tranform of eatThe Laplace transform for the function f (t) = eat is

L[eat ] =

0e (s a)t dt =

e (s a)t

s a

0=

1s a

, s > a.

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Laplace Tranforms of sin at and cos at

Using integration by parts twice we nd

L[sinat ] =

0e st sin atdt

= e st sin at

s ae st cos at

s2

0 a 2

s2

0e st sin atdt

= as2

a 2

s2 L[sin at ]s2 + a 2

s2 L[sinat ] =as2

L[sinat ] =a

s2 + a2 , s > 0. (2)A similar argument shows that

L[cosat ] =s

s2 + a2, s > 0.

Laplace Transforms of cosh at and sinh atUsing the linear property of Lwe can write

L[coshat ] =12 L[e

at ] + L[e at ]

=

1

2

1s a +

1s + a

, s >

|a

|=

ss2 a 2

, s > |a|.A similar argument shows that

L[sin at ] =a

s2 a2, s > |a|.

Laplace Transform of a PolynomialLet n be a positive integer. Using integration by parts we can write

0tn e st dt =

tn e st

s

0+

n

s

0tn 1e st dt.

By repeated use of LH opitals rule we nd lim t tn e st = lim t n !s n est = 0for s > 0. Thus,

L[tn ] =ns L[tn

1], s > 0.

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Using induction on n = 0 , 1, 2, one can easily eastablish that

L[tn ] =n !

sn +1, s > 0.

Using the above result together with the linearity property of Lone can ndthe Laplace transform of any polynomial.The next two results are referred to as the rst and second shift theorems.As with the linearity property, the shift theorems increase the number of functions for which we can easily nd Laplace transforms.

Theorem 42.1 (First Shifting Theorem)If f (t) is a piecewise continuous function for t 0 and has exponential orderat innity with |f (t )| Me

at, t C, then for any real number we have

L[et f (t)] = F (s ), s > a + where L[f (t)] = F (s).Proof.From the denition of the Laplace transform we have

L[eat f (t)] =

0e st eat f (t)dt =

0e (s a)t f (t)dt.

Using the change of variable = s a the previous equation reduces toL[eat f (t)] =

0e st eat f (t)dt =

0e t f (t)dt = F ( ) = F (sa), s > a +

Theorem 42.2 (Second Shifting Theorem)If f (t) is a piecewise continuous function for t 0 and has exponential orderat innity with |f (t)| Me at , t C, then for any real number 0 wehave

L[f (t )h(t )] = e s F (s), s > a

where L[f (t)] = F (s) and h(t) is the Heaviside step function.Proof.From the denition of the Laplace transform we have

L[f (t )h(t )] =

0f (t )h(s )e

st dt =

f (t )e

st dt.

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Using the change of variable = t the previous equation reduces to

L[f (t )h(t )] =

0f ( )e s( + )d

= e s

0f ( )e s d = e s F (s), s > a

Example 42.1Find

(a) L[e2t t2] (b) L[e3t cos2t] (c) L 1[e 2t s2].

Solution.

(a) By Theorem 42.1, we have L[e2t

t2] = F (s 2) where L[t

2] =

2!s3 =F (s), s > 0. Thus, L[e2t t2] = 2(s 2)3 , s > 2.

(b) As in part (a), we have L[e3t cos2t] = F (s 3) where L[cos2t] = F (s 3).But L[cos2t] = ss2 +4 , s > 0. Thus,

L[e3t cos2t ] =s 3

(s 3)2 + 4, s > 3

(c) Since L[t] = 1s2 , by Theorem 42.2, we havee 2t

s2= L[(t 2)h(t 2)].

Therefore,

L 1 e

2t

s2= ( t 2)h(t 2) =

0, 0 t < 2t 2, t 2The following result relates the Laplace transform of derivatives and integralsto the Laplace transform of the function itself.

Theorem 42.3Suppose that f (t) is continuous for t 0 and f (t) is piecewise continuousof exponential order at innity with |f (t)| Me at , t C Then(a) f (t) is of exponential order at innity.(b) L[f (t)] = sL[f (t)] f (0) = sF (s) f (0), s > max{a, 0}+ 1 .(c) L[f (t)] = s2L[f (t )] sf (0) f (0) = s2F (s) sf (0) f (0), s >max{a, 0}+ 1 .(d) L t0 f (u)du =

L [f (t )]s =

F (s)s , s > max{a, 0}+ 1 .

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Proof.

(a) By the Fundamental Theorem of Calculus we have f (t ) = f (0) t

0 f (u)du.Also, since f is piecewise continuous then |f (t)| T for some T > 0 andall 0 t C. Thus,

|f (t)| = f (0) t0 f (u)du = |f (0) C 0 f (u)du tC f (u)du ||f (0)|+ T C + M tC eau du.

Note that if a > 0 then

t

C eau

du =1a (e

at

eaC

) eat

a

and so

|f (t)| [|f (0)|+ T C +M a

]eat .

If a = 0 then

tC eau du = t C and therefore

|f (t)| |f (0)|+ T C + M (t C ) (|f (0)|+ T C + M )et .Now, if a < 0 then

tC eau du = 1a (eat eaC ) 1|a |so that

|f (t)| (|f (0)|+ T C +M

|a|)et

It follows that

|f (t)| Ne bt , t 0where b = max

{a, 0

}+ 1 .

(b) From the denition of Laplace transform we can write

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Since f (t ) may have jump discontinuities at t1, t 2, , t N in the interval0 t A, we can write

A0 f (t)e st dt = t10 f (t)e st dt + t2t1 f (t)e st dt + + AtN f (t)e st dt.Integrating each term on the RHS by parts and using the continuity of f (t)to obtain

t 10 f (t)e st dt = f (t1)e st 1 f (0) + s t 10 f (t )e st dt

t 2

t 1f (t)e st dt = f (t2)e

st 2

f (t1)e

st 1 + s

t2

t 1f (t )e st dt

...

tN tN 1 f (t)e st dt = f (tN )e st N f (tN 1)e st N 1 + s tN t N 1 f (t)e st dt AtN f (t)e st dt = f (A)e sA f (t N )e st N + s AtN f (t )e st dt.

Also, by the continuity of f (t ) we can write

A

0 f (t)e st

dt = t 1

0 f (t)e st

dt + t2

t1 f (t)e st

dt + + A

t N f (t)e st

dt.

Hence,

A0 f (t)e st dt = f (A)e sA f (0) + s A0 f (t)e st dt.Since f (t) has exponential order at innity, lim A f (A)e sA = 0 . Hence,

L[f (t )] = sL[f (t)] f (0).(c) Using part (b) we nd

L[f (t)] = sL[f (t)] f (0)= s(sF (s) f (0)) f (0)= s2F (s) sf (0) f (0), s > max{a, 0}+ 1 .

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(d) Since ddt

t

0 f (u)du = f (t), by part (b) we have

F (s) = L[f (t)] = sL t0 f (u)duand therefore

L t0 f (u)du = L[f (t)]s = F (s)s , s > max{a, 0}+ 1The argument establishing part (b) of the previous theorem can be extendedto higher order derivatives.

Theorem 42.4Let f (t), f (t), , f (n 1) (t) be continuous and f (n )(t ) be piecewise continu-ous of exponential order at innity with |f (n )(t)| Me at , t C. Then

L[f (n )(t)] = snL[f (t)]sn 1f (0)sn

2f (0)f (n 1) (0), s > max{a, 0}+1 .

We next illustrate the use of the previous theorem in solving initial valueproblems.

Example 42.2

Solve the initial value problemy 4y + 9 y = t, y (0) = 0 , y (0) = 1 .

Solution.We apply Theorem 42.4 that gives the Laplace transform of a derivative. Bythe linearity property of the Laplace transform we can write

L[y ]4L[y ] + 9L[y] = L[t].Now since

L[y ] =s2L[y]sy (0) y (0) = s2Y (s) 1L[y ] =sY (s) y(0) = sY (s)L[t] =

1s2

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where L[y] = Y (s), we obtains2Y (s) 1 4sY (s) + 9 Y (s) =

1s2

.

Rearranging gives

(s2 4s + 9) Y (s) =s2 + 1

s2.

Thus,

Y (s) =s2 + 1

s2(s2 4s + 9)and

y(t) = L

1s2 + 1

s2(s2 4s + 9)In the next section we will discuss a method for nding the inverse Laplacetransform of the above expression.

Example 42.3Consider the mass-spring oscillator without friction: y + y = 0 . Supposewe add a force which corresponds to a push (to the left) of the mass as itoscillates. We will suppose the push is described by the function

f (t) = h(t 2 ) + u(t (2 + a))for some a > 2 which we are allowed to vary. (A small a will correspondto a short duration push and a large a to a long duration push.) We areinterested in solving the initial value problem

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

Solution.To begin, determine the Laplace transform of both sides of the DE:

L[y + y] = L[f (t)]or

s2Y sy (0) y (0) + Y (s) = 1s

e 2s +1s

e (2 + a)s .

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f(t) F(s)

h(t) = 1, t 00, t < 0 1s , s > 0t n , n = 1 , 2, n !s n +1 , s > 0et 1s , s >

sin(t ) s2 + 2 , s > 0

cos(t ) ss2 + 2 , s > 0

sinh(t ) s2 2 , s > ||cosh(t ) ss2 2 , s > ||et f (t), with |f (t)| Me at F (s ), s > + aet h(t) 1s , s >

et tn , n = 1 , 2, n !(s )n +1 , s >

et

sin(t )

(s )2 + 2 , s >

et cos(t ) s (s )2 + 2 , s >

f (t )h(t ), 0 e s F (s), s > a

with |f (t )| Me at

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f(t) F(s) (continued)

h(t ), 0e s

s , s > 0tf (t) -F (s)

t2 sin t

s(s2 + 2 )2 , s > 0

123 [sin t t cos t ] 1(s2 + 2 )2 , s > 0f (t), with f (t) continuous sF (s) f (0)and |f (t)| Me at s > max{a, 0}+ 1f (t), with f (t) continuous s 2F (s) sf (0) f (0)and |f (t)| Me at s > max{a, 0}+ 1f (n )(t), with f (n 1) (t) continuous s n F (s) sn

1f (0) and |f (n )(t)| Me at -sf (n 2) (0) f (n

1) (0)s > max{a, 0}+ 1

t0 f (u)du, with |f (t)| Me at F (s)s , s > max{a, 0}+ 1Table L

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Practice Problems

Problem 42.1Use Table Lto nd L[2et + 5] .Problem 42.2Use Table Lto nd L[e3t

3h(t 1)].Problem 42.3Use Table Lto nd L[sin2 t ].Problem 42.4

Use Table Lto nd L[sin3t cos3t].Problem 42.5Use Table Lto nd L[e2t cos3t].Problem 42.6Use Table Lto nd L[e4t (t2 + 3 t + 5)] .Problem 42.7Use Table Lto nd L

1[ 10s2 +25 +4

s 3 ].

Problem 42.8Use Table Lto nd L 1[ 5(s 3)4 ].

Problem 42.9Use Table Lto nd L 1[e

2 s

s 9 ].

Problem 42.10Use Table Lto nd L

1[e 3 s (2s+7)

s2 +16 ].

Problem 42.11Graph the function f (t ) = h(t 1) + h(t 3) for t 0, where h(t) is theHeaviside step function, and use Table Lto nd L[f (t)].Problem 42.12Graph the function f (t) = t [h(t 1) h(t 3)] for t 0, where h(t) is theHeaviside step function, and use Table Lto nd L[f (t)].

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Problem 42.13

Graph the function f (t) = 3[h(t 1) h(t 4)] for t 0, where h(t) is theHeaviside step function, and use Table Lto nd L[f (t)].Problem 42.14Graph the function f (t ) = |2 t|[h(t 1) h(t 3)] for t 0, where h(t) isthe Heaviside step function, and use Table Lto nd L[f (t )].Problem 42.15Graph the function f (t) = h(2t) for t 0, where h(t) is the Heaviside stepfunction, and use Table Lto nd L[f (t)].

Problem 42.16Graph the function f (t ) = h(t 1) + h(4 t) for t 0, where h(t) is theHeaviside step function, and use Table Lto nd L[f (t)].Problem 42.17The graph of f (t) is given below. Represent f (t) as a combination of Heav-iside step functions, and use Table Lto calculate the Laplace transform of f (t).

Problem 42.18The graph of f (t) is given below. Represent f (t) as a combination of Heav-iside step functions, and use Table Lto calculate the Laplace transform of f (t).

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Problem 42.19

Using the partial fraction decomposition nd L 1 12

(s 3)( s+1) .

Problem 42.20Using the partial fraction decomposition nd L

1 24e 5 s

s2 9 .

Problem 42.21Use Laplace transform technique to solve the initial value problem

y + 4 y = g(t), y(0) = 2

where

g(t) = 0, 0 t < 112, 1 t < 30, t 3Problem 42.22Use Laplace transform technique to solve the initial value problem

y 4y = e3t , y(0) = 0 , y (0) = 0 .Problem 42.23Obtain the Laplace transform of the function

2 tf ( )d in terms of L[f (t )] =

F (s) given that 2

0 f ( )d = 3 .

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43 The Laplace Transform and the Method

of Partial FractionsIn the last example of the previous section we encountered the equation

y(t) = L 1 s

2 + 1s2(s2 4s + 9)

.

We would like to nd an explicit expression for y(t ). This can be done usingthe method of partial fractions which is the topic of this section. Accordingto this method, nding L

1 N (s)D (s) , where N (s) and D (s) are polynomials,

require decomposing the rational function into a sum of simpler expressions

whose inverse Laplace transform can be recognized from a table of Laplacetransform pairs.The method of integration by partial fractions is a technique for integratingrational functions, i.e. functions of the form

R (s) =N (s)D (s)

where N (s) and D (s) are polynomials.The idea consists of writing the rational function as a sum of simpler frac-tions called partial fractions . This can be done in the following way:

Step 1. Use long division to nd two polynomials r (s) and q (s) such thatN (s)D (s)

= q (s) +r (s)D (s)

.

Note that if the degree of N (s) is smaller than that of D (s) then q (s) = 0and r (s) = N (s).

Step 2. Write D (s) as a product of factors of the form ( as + b)n or (as 2 +bs + c)n where as 2 + bs + c is irreducible, i.e. as 2 + bs + c = 0 has no real zeros.

Step 3. Decomposer (s)

D (s) into a sum of partial fractions in the followingway:(1) For each factor of the form ( s )k write

A1s

+A2

(s )2+ +

Ak(s )k

,

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where the numbers A1, A2, , Ak are to be determined.(2) For each factor of the form ( as

2

+ bs + c)k

writeB 1s + C 1

as 2 + bs + c+

B 2s + C 2(as 2 + bs + c)2

+ +B ks + C k

(as 2 + bs + c)k,

where the numbers B 1, B 2, , B k and C 1, C 2, , C k are to be determined.Step 4. Multiply both sides by D (s) and simplify. This leads to an ex-pression of the form

r (s) = a polynomial whose coefficients are combinations of A i, Bi, and C i.

Finally, we nd the constants, Ai , B i , and C i by equating the coefficients of like powers of s on both sides of the last equation.

Example 43.1Decompose into partial fractions R (s) = s

3 + s2 +2s2 1 .

Solution.Step 1. s

3 + s2 +2s2 1 = s + 1 +

s+3s2 1 .

Step 2. s2 1 = ( s 1)(s + 1) .Step 3. s+3(s+1)( s 1) = As+1 + Bs 1 .Step 4. Multiply both sides of the last equation by ( s

1)(s + 1) to obtain

s + 3 = A(s 1) + B (s + 1) .Expand the right hand side, collect terms with the same power of s , andidentify coefficients of the polynomials obtained on both sides:

s + 3 = ( A + B )s + ( B A).Hence, A + B = 1 and B A = 3 . Adding these two equations gives B = 2 .Thus, A = 1 and so

s3 + s2 + 2s2 1 = s + 1

1s + 1 +

2s 1 .

Now, after decomposing the rational function into a sum of partial fractionsall we need to do is to nd the Laplace transform of expressions of the form

A(s )n or

Bs + C (as 2 + bs+ c)n .

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Example 43.2

Find L 1 1

s(s 3) .

Solution.We write

1s(s 3)

=As

+B

s 3.

Multiply both sides by s(s 3) and simplify to obtain1 = A(s 3) + Bs

or1 = ( A + B )s

3A.

Now equating the coefficients of like powers of s to obtain 3A = 1 andA + B = 0 . Solving for A and B we nd A = 13 and B = 13 . Thus,

L 1 1

s(s 3)=

13L

1 1s

+13L

1 1s 3

= 13

h(t) +13

e3t , t 0where h(t) is the Heaviside unit step function

Example 43.3

Find L

1 3s+6s2 +3 s .Solution.We factor the denominator and split the integrand into partial fractions:

3s + 6s(s + 3)

=As

+B

s + 3.

Multiplying both sides by s(s + 3) to obtain

3s + 6 = A(s + 3) + Bs= ( A + B )s + 3 A

Equating the coefficients of like powers of x to obtain 3 A = 6 and A + B = 3.Thus, A = 2 and B = 1 . Finally,

L 1 3s + 6

s2 + 3 s= 2 L

1 1s

+ L 1 1

s + 3= 2 h(t) + e 3t , t 0

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Example 43.4

Find L 1 s2 +1

s(s+1) 2 .

Solution.We factor the denominator and split the rational function into partial frac-tions:

s2 + 1s(s + 1) 2

=As

+B

s + 1+

C (s + 1) 2

.

Multiplying both sides by s(s + 1) 2 and simplifying to obtain

s2 + 1 = A(s + 1) 2 + Bs (s + 1) + Cs

= ( A + B )s2 + (2 A + B + C )s + A.Equating coefficients of like powers of s we nd A = 1 , 2A + B + C = 0and A + B = 1 . Thus, B = 0 and C = 2. Now nding the inverse Laplacetransform to obtain

L 1 s

2 + 1s(s + 1) 2

= L 1 1

s 2L 1 1

(s + 1) 2= h(t) 2te

t , t 0Example 43.5Use Laplace transform to solve the initial value problem

y + 3 y + 2 y = e t

, y(0) = y (0) = 0 .Solution.By the linearity property of the Laplace transform we can write

L[y ] + 3L[y ] + 2L[y] = L[e t ].

Now since

L[y ] =s2L[y]sy (0) y (0) = s2Y (s)L[y ] =sY (s) y(0) = sY (s)

L[e t

] =1

s + 1

where L[y] = Y (s), we obtains2Y (s) + 3 sY (s) + 2 Y (s) =

1s + 1

.

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Rearranging gives

(s2 + 3 s + 2) Y (s) =1

s + 1 .

Thus,Y (s) =

1(s + 1)( s2 + 3 s + 2)

.

andy(t ) = L

1 1(s + 1)( s2 + 3 s + 2)

.

Using the method of partial fractions we can write

1(s + 1)( s2 + 3 s + 2) =

1s + 2

1s + 1 +

1(s + 1) 2 .

Thus,

y(t) = L 1 1

s + 2 L 1 1

s + 1+ L

1 1(s + 1) 2

= e 2t e t + te t , t 0

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Practice Problems

In Problems 43.1 - 43.4, give the form of the partial fraction expansion forF (s). You need not evaluate the constants in the expansion. However, if thedenominator has an irreducible quadratic expression then use the completingthe square process to write it as the sum/difference of two squares.

Problem 43.1

F (s) =s3 + 3 s + 1

(s 1)3(s 2)2.

Problem 43.2

F (s) =s2 + 5 s 3

(s2 + 16)( s 2).

Problem 43.3

F (s) =s3 1

(s2 + 1) 2(s + 4) 2.

Problem 43.4

F (s) = s4 + 5 s2 + 2 s 9(s2 + 8 s + 17)( s 2)2.

Problem 43.5Find L 1 1(s+1) 3 .Problem 43.6Find L

1 2s 3s2 3s+2 .

Problem 43.7Find

L 1 4s2 + s+1

s3 + s .

Problem 43.8Find L

1 s2 +6 s+8s4 +8 s2 +16 .

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Problem 43.9

Use Laplace transform to solve the initial value problemy + 2 y = 26sin3 t, y (0) = 3 .

Problem 43.10Use Laplace transform to solve the initial value problem

y + 2 y = 4 t, y (0) = 3 .


y + 3 y + 2 y = 6 e t , y(0) = 1 , y (0) = 2 .


y + 4 y = cos2 t, y (0) = 1 , y (0) = 1 .


y 2y + y = e2t , y(0) = 0 , y (0) = 0 .Problem 43.14Use Laplace transform to solve the initial value problem

y + 9 y = g(t), y(0) = 1 , y (0) = 0

whereg(t) = 6, 0 t < 0, t <

Problem 43.15Determine the constants , ,y 0, and y0 so that Y (s) = 2s

1s2 + s+2 is the Laplace

transform of the solution to the initial value problem

y + y + y = 0 , y(0) = y0, y (0) = y0.

Problem 43.16Determine the constants , ,y 0, and y0 so that Y (s) = s(s+1) 2 is the Laplacetransform of the solution to the initial value problem

y + y + y = 0 , y(0) = y0, y (0) = y0.

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44 Laplace Transforms of Periodic Functions

In many applications, the nonhomogeneous term in a linear differential equa-tion is a periodic function. In this section, we derive a formula for the Laplacetransform of such periodic functions.Recall that a function f (t) is said to be T periodic if we have f (t+ T ) = f (t)whenever t and t + T are in the domain of f (t ). For example, the sine andcosine functions are 2periodic whereas the tangent and cotangent func-tions are periodic.If f (t) is T periodic for t 0 then we dene the function

f T (t) =f (t), 0 t T

0, t > T The Laplace transform of this function is then

L[f T (t)] =

0f T (t)e

st dt = T 0 f (t )e st dt.The Laplace transform of a T periodic function is given next.Theorem 44.1If f (t) is a T periodic, piecewise continuous fucntion for t 0 then

L[f (t)] = L[f T (t)]

1 e sT , s > 0.

Proof.Since f (t) is piecewise continuous, it is bounded on the interval 0 t T.By periodicity, f (t) is bounded for t 0. Hence, it has an exponential orderat innity. By Theorem 41.2, L[f (t)] exists for s > 0. Thus,

L[f (t)] =

0f (t)e st dt =

n =0 T

0f T (t nT )h(t nT )e

st dt,

where the last sum is the result of decomposing the improper integral into asum of integrals over the constituent periods.By the Second Shifting Theorem (i.e. Theorem 42.2) we have

L[f T (t nT )h(t nT )] = e nT s L[f T (t)], s > 0

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Figure 44.2

Solution.The given function is periodic of period b. For the rst period the functionis dened by

f b(t) =ab

t[h(t) h(t b)].So we have

L[f b(t )] =L[ab

t(h(t) h(t b))]= ab dds L[h(t) h(t b)].

But

L[h(t) h(t b)] =L[h(t )] L[h(t b)]=

1s

e bs

s, s > 0.

Hence,

L[f b(t)] =

a

b

1

s2

bse bs + e bs

s2 .

Finally,

L[f (t)] = L[f b(t)]

1 e bs=

ab

1 e bs bse

bs

s2(1 e bs)

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Example 44.3

Find L 1 1

s2 e s

s(1 e s ) .

Solution.Note rst that

1s2

e s

s(1 e s )=

1 e s se ss2(1 e s )

.

According to the previous example with a = 1 and b = 1 we nd that

L 1 1s2 e s

s(1 e s ) is the sawtooth function shown in Figure 44.2

Linear Time Invariant Systems and the Transfer Function

The Laplace transform is a powerful technique for analyzing linear time-invariant systems such as electrical circuits, harmonic oscillators, optical de-vices, and mechanical systems, to name just a few. A mathematical modeldescribed by a linear differential equation with constant coefficients of theform

a n y(n ) + an 1y(n 1) + + a1y + a 0y = bm u (m ) + bm 1u (m

1) + + b1u + b0uis called a linear time invariant system. The function y(t) denotes thesystem output and the function u(t) denotes the system input. The system iscalled time-invariant because the parameters of the system are not changing

over time and an input now will give the same result as the same input later.Applying the Laplace transform on the linear differential equation with nullinitial conditions we obtain

a n sn Y (s)+ a n 1sn 1Y (s)+ + a0Y (s) = bm sm U (s)+ bm 1sm

1U (s)+ + b0U (s).The function

(s) =Y (s)U (s)

=bm sm + bm 1sm 1 + + b1s + b0a n sn + an 1sn 1 + + a 1s + a 0

is called the system transfer function. That is, the transfer function of a linear time-invariant system is the ratio of the Laplace transform of itsoutput to the Laplace transform of its input.

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Example 44.4

Consider the mathematical model described by the initial value problemmy + y + ky = f (t), y(0) = 0 , y (0) = 0 .

The coefficients m,, and k describe the properties of some physical system,and f (t) is the input to the system. The solution y is the output at time t.Find the system transfer function.

Solution.By taking the Laplace transform and using the initial conditions we obtain

(ms 2 + s + k)Y (s) = F (s).

Thus,

(s) =Y (s)F (s)

=1

ms 2 + s + k(3)

Parameter IdenticationOne of the most useful applications of system transfer functions is for systemor parameter identication.

Example 44.5

Consider a spring-mass system governed bymy + y + ky = f (t), y(0) = 0 , y (0) = 0 . (4)

Suppose we apply a unit step force f (t ) = h(t) to the mass, initially atequilibrium, and you observe the system respond as

y(t) = 12

e t cos t 12

e t sin t +12

.

What are the physical parameters m,, and k?

Solution.Start with the model (4)) with f (t) = h(t ) and take the Laplace transform of both sides, then solve to nd Y (s) = 1s(ms 2 + s + k) . Since f (t) = h(t), F (s) =

1s .

Hence(s) =

Y (s)F (s)

=1

ms 2 + s + k.

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On the other hand, for the input f (t) = h(t) the corresponding observed

output isy(t) =

12

e t cos t 12

e t sin t +12

.

Hence,

Y (s) = L[12

e t cos t 12

e t sin t +12

]

= 12

s + 1(s + 1) 2 + 1

12

1(s + 1) 2 + 1

+12s

=1

s(s2 + 2 s + 2).

Thus,

(s) =Y (s)F (s)

=1

s2 + 2 s + 2.

By comparison we conclude that m = 1 , = 2 , and k = 2

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Practice Problems

Problem 44.1Find the Laplace transform of the periodic function whose graph is shown.



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Problem 44.4

Find the Laplace transform of the periodic function whose graph is shown.

Problem 44.5State the period of the function f (t) and nd its Laplace transform where

f (t) =sin t, 0 t < f (t + 2 ) = f (t), t 0.0, t < 2

Problem 44.6State the period of the function f (t) = 1 e t , 0 t < 2, f (t + 2) = f (t),and nd its Laplace transform.Problem 44.7Using Example 44.3 nd

L 1 s

2 ss3

+e s

s(1 e s ).

Problem 44.8An object having mass m is initially at rest on a frictionless horizontal surface.At time t = 0 , a periodic force is applied horizontally to the object, causingit to move in the positive x-direction. The force, in newtons, is given by

f (t) =

f 0, 0 t T 2f (t + T ) = f (t), t 0.0, T 2 < t < T

The initial value problem for the horizontal position, x(t), of the object is

mx (t) = f (t ), x(0) = x (0) = 0 .

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(a) Use Laplace transforms to determine the velocity, v(t) = x (t ), and the

position, x(t), of the object.(b) Let m = 1 kg, f 0 = 1 N, and T = 1 sec. What is the velocity, v, andposition, x, of the object at t = 1 .25 sec?

Problem 44.9Consider the initial value problem

ay + by + cy = f (t ), y(0) = y (0) = 0 , t > 0

Suppose that the transfer function of this system is given by ( s) = 12s2 +5 s+2 .(a) What are the constants a, b, and c?(b) If f (t) = e t , determine F (s), Y (s), and y(t).

Problem 44.10Consider the initial value problem

ay + by + cy = f (t ), y(0) = y (0) = 0 , t > 0

Suppose that an input f (t ) = t, when applied to the above system producesthe output y(t) = 2( e t 1) + t(e

t + 1) , t 0.(a) What is the system transfer function?(b) What will be the output if the Heaviside unit step function f (t) = h(t)is applied to the system?

Problem 44.11

Consider the initial value problemy + y + y = f (t), y(0) = y (0) = 0 ,

where

f (t) =1, 0 t 1 f (t + 2) = f (t)1, 1 < t < 2

(a) Determine the system transfer function ( s).(b) Determine Y (s).

Problem 44.12

Consider the initial value problemy 4y = et + t, y (0) = y (0) = y (0) = 0 .

(a) Determine the system transfer function ( s).(b) Determine Y (s).

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Problem 44.13

Consider the initial value problemy + by + cy = h(t), y(0) = y0, y (0) = y0, t > 0.

Suppose that L[y(t )] = Y (s) = s2 +2 s+1

s3 +3 s2 +2 s . Determine the constants b, c, y0,and y0.

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Example 46.2

Find f g where f (t) = e

t

and g(t) = sin t.Solution.Using integration by parts twice we arrive at

(f g)( t ) = t0 e (t s) sin sds=

12

e (t s)(sin s cos s)t0

=e t

2+

12

(sin t cos t)Graphical Interpretation of Convolution OperationFor the convolution

(f g)( t) = t0 f (t s)g(s)dswe perform the following:Step 1. Given the graphs of f (s) and g(s).(Figure 46.1(a) and (b))Step 2. Time reverse f (s). (See Figure 46.1(c))Step 3. Shift f (s) right by an amount t to get f (t s). (See Figure 46.1(d))Step 4. Determine the product f (t

s)g(s). (See Figure 46.1(e))

Step 5. Determine the area under the graph of f (t s)g(s) between 0 and t.(See Figure 46.1(e))

Figure 46.1

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Next, we state several properties of convolution product, which resemble

those of ordinary product.Theorem 46.1Let f (t), g(t), and k(t) be three piecewise continuous scalar functions denedfor t 0 and c1 and c2 are arbitrary constants. Then(i) f g = gf (Commutative Law)(ii) (f g)k = f (gk) (Associative Law)(iii) f (c1g + c2k) = c1f g + c2f k (Distributive Law)Proof.(i) Using the change of variables = t s we nd

(f g)( t) = t0 f (t s)g(s)ds= 0t f ( )g(t )d = t0 g(t )f ( )d = ( gf )( t).

(ii) By denition, we have

[(f g)

k)](t) = t

0 (f g)( t u)k(u)du

= t0 t u

0f (t u w)g(w)k(u)dw du.

For the integral in the bracket, make change of variable w = s u . We have

[(f g)k)](t) = t0 tu f (t s)g(s u)k(u)ds du.This multiple integral is carried over the region

{(s, u ) : 0 u s t}as depicted by shaded region in the following graph.

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Figure 46.2

Changing the order of integration, we have

[(f g)

k)](t) = t

0 s

0f (t s)g(s u)k(u)du ds

= t0 f (t s)(gk)(s)ds=[ f (gk)](t)

(iii) We have

(f (c1g + c2k))( t) = t0 f (t s)(c1g(s) + c2k(s))ds= c1

t

0f (t s)g(s)ds + c2

t

0f (t s)k(s)ds

= c1(f g)( t) + c2(f k)( t)

Example 46.3Express the solution to the initial value problem y + y = g(t), y(0) = y0in terms of a convolution integral.

Solution.Solving this initial value problem by the method of integrating factor we nd

y(t) = e t

y0 + t

0 e (t s)

g(s)ds = e t

y0 + e t

g(t)

The following theorem, known as the Convolution Theorem, provides a wayfor nding the Laplace transform of a convolution integral and also ndingthe inverse Laplace transform of a product.

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We next introduce the change of variables = t . The region of integrationbecomes 0, t 0. In this case, we have

L[(f g)( t )] =

=0

=0e s( + ) f ( )g( )dd

=

=0e s g( )d

=0e s f ( )d

= G(s)F (s) = F (s)G(s)

Example 46.4Use the convolution theorem to nd the inverse Laplace transform of

H (s) = 1(s2 + a 2)2

.

Solution.Note that

H (s) =1

s2 + a21

s2 + a2.

So, in this case we have, F (s) = G(s) = 1s2 + a 2 so that f (t) = g(t ) =1a sin(at ).

Thus,

(f

g)( t) =1

a2 t

0sin(at

as )sin( as )ds =

1

2a 3(sin(at )

at cos(at ))

Convolution integrals are useful in solving initial value problems with forcingfunctions.

Example 46.5Solve the initial value problem

4y + y = g(t ), y(0) = 3 , y (0) = 7Solution.

Take the Laplace transform of all the terms and plug in the initial conditionsto obtain4(s2Y (s) 3s + 7) + Y (s) = G(s)

or(4s2 + 1) Y (s) 12s + 28 = G(s).

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Solving for Y (s) we nd

Y (s) =12s 28

4 s2 + 14+

G(s)4 s2 + 14

=3s

s2 + ( 122 7

12

2

s2 + 122 +

14

G(s)12

2

s2 + 122 .

Hence,

y(t) = 3 cost2 14sin

t2

+12 t0 sin s2 g(t s)ds.

So, once we decide on a g(t) all we need to do is to evaluate the integral andwell have the solution

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Practice Problems

Problem 46.1Consider the functions f (t) = g(t) = h(t), t 0 where h(t) is the Heavisideunit step function. Compute f g in two different ways.(a) By directly evaluating the integral.(b) By computing L

1[F (s)G(s)] where F (s) = L[f (t)] and G(s) = L[g(t)].Problem 46.2Consider the functions f (t ) = et and g(t) = e 2t , t 0. Compute f g intwo different ways.(a) By directly evaluating the integral.

(b) By computing L 1

[F (s)G(s)] where F (s) = L[f (t)] and G(s) = L[g(t)].Problem 46.3Consider the functions f (t) = sin t and g(t) = cos t, t 0. Compute f g intwo different ways.(a) By directly evaluating the integral.(b) By computing L 1[F (s)G(s)] where F (s) = L[f (t)] and G(s) = L[g(t)].Problem 46.4Compute and graph f g where f (t ) = h(t ) and g(t) = t[h(t) h(t 2)].Problem 46.5Compute and graph f g where f (t) = h(t) h(t 1) and g(t ) = h(t 1) 2h(t 2)].Problem 46.6Compute ttt .

Problem 46.7Compute h(t )e

t

e 2t .

Problem 46.8Compute t

e t

et .

Problem 46.9

Suppose it is known that

n functions

h(t)h(t) h(t) = Ct 8. Determine the con-stants C and the poisitive integer n.53


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Problem 46.10

Use Laplace transform to solve for y(t) :

t0 sin(t )y( )d = t2.Problem 46.11Use Laplace transform to solve for y(t) :

y(t ) t0 e(t )y( )d = t.Problem 46.12Use Laplace transform to solve for y(t) :

t

y(t) = t2(1 e t ).

Problem 46.13Solve the following initial value problem.

y y = t0 (t )e d, y (0) = 1.

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47 The Dirac Delta Function and Impulse Re-

sponseIn applications, we are often encountered with linear systems, originally atrest, excited by a sudden large force (such as a large applied voltage to anelectrical network) over a very short time frame. In this case, the outputcorresponding to this sudden force is referred to as the impulse response.Mathematically, an impulse can be modeled by an initial value problem witha special type of function known as the Dirac delta function as the externalforce, i.e., the nonhomogeneous term. To solve such IVP requires nding theLaplace transform of the delta function which is the main topic of this section.

An Example of Impulse ResponseConsider a spring-mass system with a time-dependent force f (t ) applied tothe mass. The situation is modeled by the second-order differential equation

my + y + ky = f (t) (5)

where t is time and y(t) is the displacement of the mass from equilibrium.Now suppose that for t 0 the mass is at rest in its equilibrium position, soy(0) = y (0) = 0 . Hence, the situation is modeled by the initial value problem

my + y + ky = f (t), y(0) = 0 , y (0) = 0 . (6)

Solving this equation by the method of variation of parameters one nds theunique solution

y(t) = t0 (t s)f (s)ds (7)where

(t) =e( / 2m )t sin t km 24m 2

m

k

m 2

4m2

.

Next, we consider the problem of strucking the mass by an instantaneoushammer blow at t = 0 . This situation actually occurs frequently in practice-asystem sustains a forceful, almost-instantaneous input. Our goal is to modelthe situation mathematically and determine how the system will respond.

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In the above situation we might describe f(t) as a large constant force applied

on a very small time interval. Such a model leads to the forcing function

f (t ) =1 , 0 t 0, otherwise

where is a small positive real number. When is close to zero the appliedforce is very large during the time interval 0 t and zero afterwards. Apossible graph of f (t ) is given in Figure 47.1

Figure 47.1

In this case, its easy to see that for any choice of we have

f dt = 1

and

lim 0+

f (t) = 0 , t = 0 , lim 0+

f (0) = . (8)Our ultimate interest is the behavior of the solution to equation (5) withforcing function f (t) in the limit 0+ . That is, what happens to thesystem output as we make the applied force progressively sharper andstronger?.Let y (t) be the solution to equation (5) with f (t ) = f (t). Then the uniquesolution is given by

y (t ) = t0 (t s)f (s)ds.56


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For t the last equation becomesy (t) = 1 0 (t s)ds.

Since (t) is continuous for all t 0, we can apply the mean value theoremfor integrals and writey (t) = (t )

for some 0 . Letting 0+ and using the continuity of we ndy(t) = lim

0+y (t) = (t).

We call y(t) the impulse response of the linear system.

The Dirac Delta FunctionThe problem with the integral

t0 (t s)f (s)dsis that lim 0+ f (0) is undened. So it makes sense to ask the question of whether we can nd a function (t) such that

lim 0+

y (t) = lim 0+

t

0(t s)f (s)ds

= t0 (t s) (s)ds= (t)

where the role of (t) would be to evaluate the integrand at s = 0 . Note thatbecause of Fig 47.1 and (8), we cannot interchange the opeartions of limitand integration in the above limit process. Such a function exist in thetheory of distributions and can be dened as follows:

If f (t) is continuous in a t b then we dene the function (t) by theintegral equation

ba f (t) (t t 0)dt = lim 0+ ba f (t)f (t t 0)dt.57


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The object (t) on the left is called the Dirac Delta function, or just the

delta function for short.Finding the Impulse Function Using Laplace TransformFor > 0 we can solve the initial value problem (6) using Laplace transforms.To do this we need to compute the Laplace transform of f (t), given by theintegral

L[f (t)] =

0f (t)e st dt =

1 0 e st dt = 1 e s

s.

Note that by using LHopitals rule we can write

lim 0+ L[f (t)] = lim 0+ 1 e

s

s= 1 , s > 0.

Now, to nd y (t), we apply the Laplace transform to both sides of equation(5) and using the initial conditions we obtain

ms 2Y (s) + sY (s) + kY (s) =1 e

s

s.


Y (s) =1

ms 2 + s + k1

e s

s .

Letting 0+ we ndY (s) =

1ms 2 + s + k

which is the transfer function of the system. Now inverse transform Y (s) tond the solution to the initial value problem. That is,

y(t) = L 1 1

ms 2 + s + k= (t).

Now, impulse inputs are usually modeled in terms of delta functions. Thus,knowing the Laplace transform of such functions is important when solvingdifferential equations. The next theorem nds the Laplace transform of thedelta function.

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Theorem 47.1

With (t) dened as above, if a t0 < b

ba f (t) (t t 0)dt = f (t 0).Proof.We have

ba f (t) (t t 0) = lim 0+ ba f (t)f (t t0)dt= lim

0+

1

t 0 +

t0f (t)dt

= lim 0+

1f (t0 + ) = f (t 0)

where 0 < < 1 and the mean-value theorem for integrals has been used

Remark 47.1Since p (t t0) = 1 for t0 t t0 + and 0 otherwise we see that ba f (t) (t a )dt = f (a) and ba f (t) (t t0)dt = 0 for t0 b.It follows immediately from the above theorem that

L[ (t t0)] =

0e st (t t0)dt = e

st 0 , t0 0.In particular, if t0 = 0 we nd

L[ (t)] = 1 .The following example illustrates the formal use of the delta function.

Example 47.1A spring-mass system with mass 2, damping 4, and spring constant 10 issubject to a hammer blow at time t = 0 . The blow imparts a total impulse of 1 to the system, which was initially at rest. Find the response of the system.

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Solution.

The situation is modeled by the initial value problem2y + 4 y + 10 y = (t), y(0) = 0 , y (0) = 0 .

Taking Laplace transform of both sides we nd

2s2Y (s) + 4 sY (s) + 10 Y (s) = 1 .


Y (s) =1

2s2 + 4 s + 10.

The impulsive response is

y(t) = L 1 1

21

(s + 1) 2 + 2 2=

14

e t sin2t

Example 47.2A 16 lb weight is attached to a spring with a spring constant equal to 2lb/ft. Neglect damping. The weight is released from rest at 3 ft below theequilibrium position. At t = 2 sec, it is struck with a hammer, providing animpulse of 4 lb-sec. Determine the displacement function y(t) of the weight.

Solution.This situation is modeled by the initial value problem

1632

y + 2 y = 4 (t 2 ), y(0) = 3 , y (0) = 0 .Apply Laplace transform to both sides to obtain

s2Y (s) 3s + 4 Y (s) = 8 e 2s .


Y (s) = 3ss2 + 4 + e 2s

s2 + 4 .

Now take the inverse Laplace transform to get

y(t) = L 1[Y (s)] = 3 cos 2t + 8 h(t 2 )f (t 2 )

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where

f (t) = L

11

s2 + 4 =12 sin2t.

Hence,

y(t) = 3cos 2 t + 4 h(t 2 )sin2( t 2 ) = 3cos 2 t + 4 h(t 2 )sin2tor more explicitly

y(t) = 3cos2t, t < 23cos2t + 4 sin 2 t, t 2

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Practice Problems

Problem 47.1Evaluate

(a) 30 (1 + e t ) (t 2)dt.(b) 1 2(1 + e t ) (t 2)dt.Problem 47.2Let f (t ) be a function dened and continuous on 0 t < . Determine

(f

)( t) =

t

0f (t

s) (s)ds.

Problem 47.3Determine a value of the constant t0 such that 10 sin2 [ (t t 0)] (t12 )dt = 34 .Problem 47.4If 51 tn (t 2)dt = 8, what is the exponent n?Problem 47.5Sketch the graph of the function g(t) which is dened by g(t) = t0 s0 (u 1)duds, 0 t < .Problem 47.6The graph of the function g(t) = t0 et (t t0)dt, 0 t < is shown.Determine the constants and t 0.

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Problem 47.7

(a) Use the method of integarting factor to solve the initial value problemy y = h(t ), y(0) = 0 .(b) Use the Laplace transform to solve the initial value problem = (t ), (0) = 0 .(c) Evaluate the convolution h(t) and compare the resulting function withthe solution obtained in part(a).

Problem 47.8Solve the initial value problem

y + y = 2 + (t 1), y(0) = 0 , 0 t 6.Graph the solution on the indicated interval.


y = (t 1) (t 3), y(0) = 0 , y (0) = 0 , 0 t 6.Graph the solution on the indicated interval.

Problem 47.10

Solve the initial value problemy 2y = (t 1), y(0) = 1 , y (0) = 0 , 0 t 2.

Graph the solution on the indicated interval.


y + 2 y + y = (t 2), y(0) = 0 , y (0) = 1 , 0 t 6.Graph the solution on the indicated interval.

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48 Solutions to Problems

Section 41


01


Solution.We have

0

11 + t2

dt = limA A0 11 + t 2 dt = limA [arctan t]A0

= limA

arctan A =2

So the integral is convergent


0t


Solution.We have

0

t1 + t2

dt =12

limA A0 2t1 + t2 dt = 12 limA ln (1 + t2) A0

=12

limA

ln (1 + A2) = Hence, the integral is divergent


0 e t cos(e t )dt converges. If the integral

converges, give its value.

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Solution.

Using the substitution u = e t

we nd

0e t cos(e t )dt = lim

A e A

1 cos udu= lim

A[sin u]

e A

1 = limA [sin1sin(e A)]

=sin1

Hence, the integral is convergent

Problem 41.4Using the denition, nd L[e3t ], if it exists. If the Laplace transform existsthen nd the domain of F (s).Solution.We have

L[e3t ] = limA A0 e3t e st dt = limA A0 et (3 s)dt= lim

A

et (3 s)

3 sA

0

= limA

eA(3 s)

3 s 1

3 s= 1

s 3, s > 3

Problem 41.5Using the denition, nd L[t 5], if it exists. If the Laplace transform existsthen nd the domain of F (s).Solution.Using integration by parts we nd

L[t 5] = limA A0 (t 5)e st dt = limA (t 5)e st

s

A

0+

1s A0 e st dt

= limA

(A 5)e sA + 5

s e st

s2A

0

=1s2

5s

, s > 0

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Problem 41.6

Using the denition, nd L[e(t 1) 2

], if it exists. If the Laplace transformexists then nd the domain of F (s).

Solution.We have

0e(t 1)

2e st dt =

0e(t 1)

2 st dt.

Since limt (t 1)2 st = lim t t2 1 (2+ s)

t +1t2 = , for any xed s

we can choose a positive C such that ( t 1)2 st 0 for t C. In this case,e(t 1)2 st 1 and this implies that

0 e(t 1) 2 st dt

C dt. The integral on

the right is divergent so that the integral on the left is also divergent by thecomparison theorem of improper integrals. Hence, f (t) = e(t 1) 2 does nothave a Laplace transform

Problem 41.7Using the denition, nd L[(t 2)2], if it exists. If the Laplace transformexists then nd the domain of F (s).Solution.We have

L[(t

2)2] = lim

T

T

0(t

2)2e st dt.

Using integration by parts with u = e st and v = ( t 2)2 we nd

T 0 (t 2)2e st dt = (t 2)2e st

s

T

0+

2s T 0 (t 2)e st dt

=4s

(T 2)2e sT

s+

2s T 0 (t 2)e st dt.

Thus,

limT

T

0

(t

2)2e st dt =

4

s+

2

slim

T

T

0

(t

2)e st dt.

Using by parts with u = e st and v = t 2 we nd

T 0 (t 2)e st dt = (t 2)e st

s 1s2

e stT

0.

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Letting T in the above expression we ndlim

T T

0(t 2)e

st dt = 2s

+1s2

, s > 0.

Hence,

F (s) =4s

+2s

2s

+1s2

=4s

4s2

+2s3

, s > 0

Problem 41.8Using the denition, nd L[f (t)], if it exists. If the Laplace transform existsthen nd the domain of F (s).

f (t) = 0, 0 t < 1t 1, t 1Solution.We have

L[f (t)] = limT T 1 (t 1)e st dt.Using integration by parts with u = e st and v = t 1 we nd

limT

T

1(t 1)e

st dt = limT

(t 1)e sts

1s2

e stT

1=

e s

s2, s > 0

Problem 41.9Using the denition, nd L[f (t)], if it exists. If the Laplace transform existsthen nd the domain of F (s).

f (t) =0, 0 t < 1t 1, 1 t < 20, t 2.

Solution.We have

L[f (t)] = 2

1(t 1)e

st dt = (t 1)e

st

s 1s2

e st2

1

= e 2s

s+

1s2

(e s e 2s ), s = 0

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Problem 41.10

Let n be a positive integer. Using integration by parts establish the reductionformula

tn e st dt = tn e st

s+

ns tn 1e st dt, s > 0.

Solution.Let u = e st and v = tn . Then u = e

st

s and v = ntn 1. Hence,

tn e st dt = tn e st

s+

ns tn 1e st dt, s > 0

Problem 41.11

For s > 0 and n a positive integer evaluate the limits(a) lim t 0 tn e st (b) lim t tn e st

Solution.(a) lim t 0 tn e st = lim t 0 t

n

est =01 = 0 .

(b) Using LHopitals rule repeatedly we nd

limt

tn e st = = limt n !

sn est= 0

Problem 41.12(a) Use the previous two problems to derive the reduction formula for theLaplace transform of f (t ) = tn ,

L[tn ] =ns L[tn

1], s > 0.

(b) Calculate L[tk], for k = 1 , 2, 3, 4, 5.(c) Formulate a conjecture as to the Laplace transform of f (t ), t n with n apositive integer.

Solution.(a) Using the two previous problems we nd

L[tn ] = limT T

0t n e st dt = lim

T tn e st

sT

0+ n

s T

0t n 1e st dt

=ns

limT T 0 t n 1e st dt = ns L[tn 1], s > 0

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(b) We have

L[t] = 1s2L[t 2] =

2s L[t ] =

2s3

L[t 3] =3s L[t 2] =

6s4

L[t 4] =4s L[t 3] =

24s5

L[t 5] =5s L[t 4] =

120s5

(c) By induction, one can easily show that for n = 0 , 1, 2,

L[tn ] =

n !sn +1

, s > 0

From a table of integrals,

eu sin udu = eu sin u cos u 2 + 2 eu cos udu = eu cos u + sin u 2 + 2

Problem 41.13Use the above integrals to nd the Laplace transform of f (t) = cos t, if itexists. If the Laplace transform exists, give the domain of F (s).

Solution.We have

L[cost ] = limT e st s cos t + sin t

s2 + 2T

0=

ss2 + 2

, s > 0

Problem 41.14Use the above integrals to nd the Laplace transform of f (t) = sin t, if itexists. If the Laplace transform exists, give the domain of F (s).

Solution.We have

L[sin t ] = limT e st s sin t + cos t

s2 + 2T

0=

s2 + 2

, s > 0

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Problem 41.15

Use the above integrals to nd the Laplace transform of f (t) = cos (t 2),if it exists. If the Laplace transform exists, give the domain of F (s).Solution.Using a trigonometric identity we can write f (t) = cos (t 2) = cos t cos2+sin t sin2. Thus, using the previous two problems we nd

L[cos(t 2)] =s cos2 + sin2

s2 + 2, s > 0

Problem 41.16Use the above integrals to nd the Laplace transform of f (t) = e3t sin t, if it

exists. If the Laplace transform exists, give the domain of F (s).Solution.We have

L[e3t sin t] = limT T 0 e (s 3)t sin tdt= lim

T e (s 3) t (s 3)sin t + cos t

(s 3)2 + 1T

0

=1

(s

3)2 + 1

, s > 3

Problem 41.17Use the linearity property of Laplace transform to nd L[5e 7t + t + 2 e2t ].Find the domain of F (s).Solution.We have L[e

7t ] = 1s+7 , s > 7, L[t] = 1s2 , s > 0, and L[e2t ] = 1s 2 , s > 2.Hence,

L[5e 7t + t + 2 e2t ] = 5L[e

7t ] + L[t ] + 2L[e2t ] =5

s + 7+

1s2

+2

s 2, s > 2

Problem 41.18Consider the function f (t) = tan t.(a) Is f (t) continuous on 0 t < , discontinuous but piecewise continuouson 0 t < , or neither?(b) Are there xed numbers a and M such that |f (t)| Me at for 0 t < ?

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Problem 41.21

Consider the oor function f (t) = t , where for any integer n we havet = n for all n t < n + 1 .(a) Is f (t) continuous on 0 t < , discontinuous but piecewise continuouson 0 t < , or neither?(b) Are there xed numbers a and M such that |f (t)| Me at for 0 t < ?Solution.(a) The oor function is a piecewise continuous function on 0 t < .(b) Since t t < e t for 0 t < we nd M = 1 and a = 1Problem 41.22

Find L 1 3

s 2 .

Solution.Since L 1s a = 1s a , s > a we nd

L 1 3

s 2= 3 L

1 1s 2

= 3 e2t , t 0Problem 41.23Find L

1 2s2 + 1s+1 .

Solution.Since L[t ] = 1s2 , s > 0 and L 1s a = 1s a , s > a we nd

L 1

2s2

+1

s + 1= 2L

1 1s2

+ L 1 1

s + 1= 2t + e

t , t 0Problem 41.24Find L

1 2s+2 +

2s 2 .

Solution.We have

L 1 2

s + 2+

2s 2

= 2L 1 1

s + 2+2 L

1 1s 2

= 2( e 2t + e2t ), t 0

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Section 42

Problem 42.1Use Table Lto nd L[2et + 5] .Solution.

L[2et + 5] = 2 L[et ] + 5L[1] =2

s 1+

5s

, s > 1

Problem 42.2

Use Table Lto nd L[e3t 3

h(t 1)].Solution.

L[e3t 3h(t 1)] = L[e3(t

1) h(t 1)] = e sL[e3t ] =

e s

s 3, s > 3

Problem 42.3Use Table Lto nd L[sin2 t ].Solution.

L[sin2

t ] = L[1

cos2t2 ] =

12(L[1]L[cos2t ]) =

12

1s

ss2 + 4 2 , s > 0

Problem 42.4Use Table Lto nd L[sin3t cos3t].Solution.

L[sin3t cos3t] = Lsin6t

2=

12L[sin6t] =

3s2 + 36

, s > 0

Problem 42.5Use Table

Lto nd

L[e2t cos3t].

Solution.

L[e2t cos3t] =s 2

(s 2)2 + 9, s > 2

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Problem 42.6

Use Table Lto nd L[e4t

(t2

+ 3 t + 5)] .Solution.

L[e4t (t2+3 t+5)] = L[e4t t 2]+3L[e4t t]+5L[e4t ] =2

(s 4)3+

3(s 4)2

+5

s 4, s > 4

Problem 42.7Use Table Lto nd L 1[ 10s2 +25 + 4s 3 ].Solution.

L 1[ 10s2 + 25 + 4s 3] = 2L 1[ 5s2 + 25 ]+4L 1[ 1s 3] = 2sin 5

t +4 e3t , t 0Problem 42.8Use Table Lto nd L 1[ 5(s 3)4 ].Solution.

L 1[

5(s 3)4

] =56L

1[3!

(s 3)4] =

56

e3t t3, t 0Problem 42.9

Use Table Lto nd L 1

[e 2 s

s 9 ].Solution.

L 1[

e 2s

s 9] = e9( t 2) h(t 2) =

0, 0 t < 2e9( t 2) , t 2Problem 42.10Use Table Lto nd L

1[e 3 s (2s+7)

s2 +16 ].

Solution.We have

L 1[

e 3s (2s + 7)s2 + 16

] =2L 1[

e 3s ss2 + 16

] +74L

1[e 3s4

s2 + 16]

=2cos4( t 3)h(t 3) +74

sin4(t 3)h(t 3), t 0

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Problem 42.13Graph the function f (t) = 3[h(t 1) h(t 4)] for t 0, where h(t) is theHeaviside step function, and use Table Lto nd L[f (t)].Solution.

Note that

f (t ) =0, 0 t < 13, 1 t < 40, t 4

The graph of f (t) is shown below. Using Table Lwe nd

L[f (t)] = 3L[h(t 1)] 3L[h(t 4)] =3e s

s 3e 4s

s, s > 0

Problem 42.14Graph the function f (t ) = |2 t|[h(t 1) h(t 3)] for t 0, where h(t) isthe Heaviside step function, and use Table Lto nd L[f (t )].Solution.

Note thatf (t) =

0, 0 t < 1|2 t|, 1 t < 30, t 3

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L[f (t)] =(2 t)h(t 1) + 2( t 2)h(t 2) (t 2)h(t 3)= L[(t 1)h(t 1) + h(t 1) + 2( t 2)h(t 2) (t 3)h(t 3) h(t 3)]= L[(t 1)h(t 1)] + L[h(t 1)] + 2L[(t 2)h(t 2)]L[(t 3)h(t 3)] L[h(t 3)]=

e s

s2+

e s

s+

2e 2s

s2 e 3s

s2 e 3s

s, s > 0

Problem 42.15Graph the function f (t) = h(2t) for t 0, where h(t) is the Heaviside stepfunction, and use Table Lto nd L[f (t)].Solution.Note that

f (t) = 1, 0 t 20, t > 2The graph of f (t ) is shown below. From this graph we see that f (t) =h(t) h(t 2)h(t 2). Using Table Lwe nd

L[f (t)] = L[h(t)] L[h(t 1)h(t 1)] =1 e 2s

s, s > 0

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Problem 42.16

Graph the function f (t ) = h(t 1) + h(4 t) for t 0, where h(t) is theHeaviside step function, and use Table Lto nd L[f (t)].Solution.Note that

f (t) =1, 0 t < 12, 1 t 41, t > 4


L[f (t)] = L[h(t1)]+L[h(4t)] =e s

s+

4

0e st dt =

1 + e s e 4s

s, s > 0

Problem 42.17The graph of f (t) is given below. Represent f (t) as a combination of Heav-iside step functions, and use Table

Lto calculate the Laplace transform of

f (t).

Solution.From the graph we see that

f (t) = ( t 2)h(t 2) (t 3)h(t 3) h(t 4)Thus,

L[f (t)] = L[(t2)h(t2)]L[(t3)h(t3)]L[h(t4)] =e 2s e

3s

s2 e 4s

s, s > 0

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Problem 42.18

The graph of f (t) is given below. Represent f (t) as a combination of Heav-iside step functions, and use Table Lto calculate the Laplace transform of f (t).

Solution.From the graph we see that

f (t) = h(t 1) + h(t 2) 2h(t 3).Thus,

L[f (t)] = L[h(t 1)]2L[h(t3)]+ L[h(t2)] =e s 2e

3s + e 2s

s, s > 0

Problem 42.19Using the partial fraction decomposition nd

L 1 12

(s 3)( s+1) .

Solution.Write

12(s 3)(s + 1)

=A

s 3+

Bs + 1

.

Multiply both sides of this equation by s 3 and cancel common factors toobtain12

s + 1= A +

B (s 3)s + 1

.

Now, nd A by setting s = 3 to obtain A = 3 . Similarly, by multiplying both

sides bys

+ 1 and then settings

= 1 in the resulting equation leads toB = 3. Hence,12

(s 3)(s + 1)= 3

1s 3

1s + 1

.

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Finally,

L 1 12

(s 3)(s + 1)=3 L

1 1s 3

3L 1 1

s + 1=3 e3t 3e

t , t 0Problem 42.20Using the partial fraction decomposition nd L

1 24e 5 s

s2 9 .

Solution.Write

24

(s 3)(s + 3)=

A

s 3+

B

s + 3.

Multiply both sides of this equation by s 3 and cancel common factors toobtain24

s + 3= A +

B (s 3)s + 3

.

Now, nd A by setting s = 3 to obtain A = 4 . Similarly, by multiplying bothsides by s + 3 and then setting s = 3 in the resulting equation leads toB = 4. Hence,

24(s

3)(s + 3)

= 41

s

3

1s + 3

.

Finally,

L 1 24e

5s

(s 3)(s + 3)=4 L

1 e 5s

s 3 4L

1 e 5s

s + 3=4[e3(t 5) e

3(t 5) ]h(t 5), t 0Problem 42.21Use Laplace transform technique to solve the initial value problem

y + 4 y = g(t), y(0) = 2

where

g(t) =0, 0 t < 112, 1 t < 30, t 3

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Solution.

Note rst that g(t) = 12[h(t 1) h(t 3)] so thatL[g(t)] = 12L[h(t 1)] 12L[h(t 3)] =

12(e s e 3s )

s, s > 0.

Now taking the Laplace transform of the DE and using linearity we nd

L[y ] + 4L[y] = L[g(t)].But L[y ] = sL[y]y(0) = sL[y]2. Letting L[y] = Y (s) we obtain

sY (s)

2 + 4Y (s) = 12

e s e 3s

s.


Y (s) =2

s + 4+ 12

e s e 3s

s(s + 4).

But

L 1 2

s + 4= 2 e 4t

and

L 1 12e

s e 3ss(s + 4) =3 L 1 (e s e

3s ) 1s

1s + 4

=3 L 1 e

s

s 3L 1 e

3s

s 3L 1 e

s

s + 4+ 3L

1 e 3s

s + 4=3 h(t 1) 3h(t 3) 3e

4( t 1) h(t 1) + 3e 4(t 3) h(t 3)

Hence,

y(t) = 2 e 4t +3[ h(t1)h(t3)]3[e 4(t 1) h(t1)e

4( t 3) h(t3)], t 0

Problem 42.22Use Laplace transform technique to solve the initial value problem

y 4y = e3t , y(0) = 0 , y (0) = 0 .

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Solution.

Taking the Laplace transform of the DE and using linearity we nd

L[y ]4L[y] = L[e3t ].But L[y ] = s2L[y]sy (0) y (0) = s2L[y]. Letting L[y] = Y (s) we obtain

s2Y (s) 4Y (s) =1

s 3.


Y (s) =1

(s 3)(s 2)(s + 2).

Using partial fraction decomposition

1(s 3)(s 2)(s + 2)

=A

s 3+

Bs + 2

+C

s 2we nd A = 15 , B =

120 , and C = 14 . Thus,

y(t) = L 1 1

(s 3)(s 2)(s + 2)=

15L

1 1s 3

+120L

1 1s + 2

14L

1 1s 2

=15e

3t

+120e

2t

14e

2t

, t 0Problem 42.23Obtain the Laplace transform of the function t2 f ( )d in terms of L[f (t )] =F (s) given that 20 f ( )d = 3 .Solution.We have

L t2 f ( )d = L t0 f ( )d 20 f ( )d= F (s)

s L[3]=

F (s)s

3s

, s > 0

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In Problems 43.1 - 43.4, give the form of the partial fraction expansion forF (s). You need not evaluate the constants in the expansion. However, if thedenominator has an irreducible quadratic expression then use the completingthe square process to write it as the sum/difference of two squares.

Problem 43.1

F (s) =s3 + 3 s + 1

(s 1)3(s 2)2.

Solution.

F (s) = A1(s 1)3

+ A2(s 1)2

+ A3s 1

+ B 1(s 2)2

+ B 2s 2

Problem 43.2

F (s) =s2 + 5 s 3

(s2 + 16)( s 2).

Solution.

F (s) =A1s + A2

s2 + 16+

B 1

s 2Problem 43.3

F (s) =s3 1

(s2 + 1) 2(s + 4) 2.

Solution.

F (s) =A1s + A2(s2 + 1) 2

+A3s + A4

s2 + 1+

B 1(s + 4) 2

+B 2

s + 4

Problem 43.4

F (s) =s4 + 5 s2 + 2 s 9

(s2 + 8 s + 17)( s 2)2.

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Solution.

F (s) =A1

(s 2)2+

A2s 2

+B 1s + B 2

s2 + 8 s + 17

Problem 43.5Find L

1 1(s+1) 3 .

Solution.Using Table Lwe nd L

1 1(s+1) 3 =

12 e

t t2, t 0Problem 43.6

Find L 1 2s 3

s2 3s+2 .


2s 3(s 1)(s 2)

=A

s 1+

Bs 2

.

Multiplying both sides by ( s 1)(s 2) and simplifying to obtain2s

3 = A(s

2) + B (s

1)

=( A + B )s 2A B.Equating coefficients of like powers of s we obtain the system

A + B = 2

2A B = 3.Solving this system by elimination we nd A = 1 and B = 1 . Now ndingthe inverse Laplace transform to obtain

L 1 2s 3

(s 1)(s 2)=

L 1 1

s 1+

L 1 1

s 2= et + e2t , t

0.

Problem 43.7Find L

1 4s2 + s+1s3 + s .

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Solution.

We factor the denominator and split the rational function into partial frac-tions:

4s2 + s + 1s(s2 + 1)

=As

+Bs + C s2 + 1

.

Multiplying both sides by s(s2 + 1) and simplifying to obtain

4s2 + s + 1 = A(s2 + 1) + ( Bs + C )s=( A + B )s2 + Cs + A.

Equating coefficients of like powers of s we obtain A + B = 4 , C = 1 , A = 1 .Thus, B = 3 . Now nding the inverse Laplace transform to obtain

L 1 4s2 + s + 1

s(s2 + 1)= L

1 1s

+ 3L 1 s

s2 + 1+ L

1 1s2 + 1

=1 + 3cos t + sin t, t 0Problem 43.8Find L

1 s2 +6 s+8s4 +8 s2 +16 .


s2 + 6 s + 8(s2 + 4) 2

= B 1s + C 1s2 + 4

+ B 2s + C 2(s2 + 4) 2

.

Multiplying both sides by ( s2 + 4) 2 and simplifying to obtain

s2 + 6 s + 8 =( B 1s + C 1)(s2 + 4) + B 2s + C 2= B 1s3 + C 1s2 + (4 B 1 + B 2)s + 4 C 1 + C 2.

Equating coefficients of like powers of s we obtain B 1 = 0 , C 1 = 1 , B 2 = 6 ,and C 2 = 4 . Now nding the inverse Laplace transform to obtain

L

1 s2 + 6 s + 8

(s2

+ 4)2 =

L

1 1s2

+ 4

+ 6

L

1 s

(s2

+ 4)2 + 4

L

1 1

(s2

+ 4)2

=12

sin2t + 6t4

sin2t + 4116

[sin2t 2t cos2t]=

32

t sin2t +34

sin2t 12

t cos2t, t 0

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Problem 43.9

Use Laplace transform to solve the initial value problemy + 2 y = 26sin3 t, y (0) = 3 .

Solution.Taking the Laplace of both sides to obtain

L[y ] + 2L[y] = 26L[sin3t].Using Table Lthe last equation reduces to

sY (s) y(0) + 2 Y (s) = 263

s2 + 9 .

Solving this equation for Y (s) we nd

Y (s) =3

s + 2+

78(s + 2)( s2 + 9)

.

Using the partial fraction decomposition we can write

1(s + 2)( s2 + 9)

=A

s + 2+

Bs + C s2 + 9

.

Multipliying both sides by ( s + 2)( s2 + 9) to obtain

1 = A(s2 + 9) + ( Bs + C )(s + 2)=( A + B )s2 + (2 B + C )s + 9 A + 2 C.

Equating coefficients of like powers of s we nd A + B = 0 , 2B + C = 0 , and9A + 2 C = 1 . Solving this system we nd A = 113 , B = 113 , and C = 213 .Thus,

Y (s) =9

s + 2 6s

s2 + 9+ 4

3s2 + 9

.

Finally,

y(t) = L 1[Y (s)] = 9L

1 1s + 2 6L

1 ss2 + 9

+ 4L 1 3

s2 + 9= 9 e 2t 6cos3t + 4sin 3 t, t 0

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Problem 43.10

Use Laplace transform to solve the initial value problemy + 2 y = 4 t, y (0) = 3 .


L[y ] + 2L[y] = 4L[t ].Using Table Lthe last equation reduces to

sY (s) y(0) + 2 Y (s) =4s2 .


Y (s) =3

s + 2+

4(s + 2) s2

.


1(s + 2) s2

=A

s + 2+

Bs + C s2

.

Multipliying both sides by ( s + 2) s2 to obtain

1 = As 2 + ( Bs + C )(s + 2)= ( A + B )s2 + (2 B + C )s + 2 C

Equating coefficients of like powers of s we nd A + B = 0 , 2B + C = 0 , and2C = 1 . Solving this system we nd A = 14 , B = 14 , and C = 12 . Thus,

Y (s) =4

s + 2 1s

+ 21s2

.

Finally,

y(t) = L 1[Y (s)] = 4L

1 1s + 2 L

1 1s

+ 2L 1 1

s2

= 4 e 2t 1 + 2 t, t 0

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Problem 43.11

Use Laplace transform to solve the initial value problemy + 3 y + 2 y = 6 e t , y(0) = 1 , y (0) = 2 .


L[y ] + 3L[y ] + 2L[y] = 6L[e t ].

Using Table Lthe last equation reduces to

s2

Y (s) sy (0) y (0) + 3( sY (s) y(0)) + 2 Y (s) =6

s + 1 .


Y (s) =s + 5

(s + 1)( s + 2)+

6(s + 2)( s + 1) 2

=s2 + 6 s + 11

(s + 1) 2(s + 2).


s2 + 6 s + 11(s + 2)( s + 1) 2

=A

s + 2+

Bs + 1

+C

(s + 1) 2.

Multipliying both sides by ( s + 2)( s + 1) 2 to obtain

s2 + 6 s + 11 = A(s + 1) 2 + B (s + 1)( s + 2) + C (s + 2)= ( A + B )s2 + (2 A + 3 B + C )s + A + 2 B + 2 C

Equating coefficients of like powers of s we nd A + B = 1 , 2A +3 B + C = 6 ,and A + 2 B + 2 C = 11 . Solving this system we nd A = 3 , B = 2, andC = 6 . Thus,

Y (s) =3

s + 2 2

s + 1+

6(s + 1) 2

.

Finally,

y(t) = L 1[Y (s)] = 3L

1 1s + 2 2L

1 1s + 1

+ 6L 1 1

(s + 1) 2

= 3 e 2t 2e t + 6 te t , t 0

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Problem 43.12

Use Laplace transform to solve the initial value problemy + 4 y = cos2 t, y (0) = 1 , y (0) = 1 .


L[y ] + 4L[y] = L[cos2t].Using Table Lthe last equation reduces to

s2Y (s)

sy (0)

y (0) + 4 Y (s) =

s

s2 + 4.


Y (s) =s + 1s2 + 4

+s

(s2 + 4) 2.

Using Table Lagain we ndy(t) = L

1 ss2 + 4

+12L

1 2s2 + 4

+ L 1 s

(s2 + 4) 2

= cos2 t +12 sin2t +

t4 sin2t, t 0


y 2y + y = e2t , y(0) = 0 , y (0) = 0 .Solution.Taking the Laplace of both sides to obtain

L[y ]

2

L[y ] +

L[y] =

L[e2t ].

Using Table Lthe last equation reduces tos2Y (s) sy (0) y (0) 2sY (s) + 2 y(0) + Y (s) =

1s 2

.

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Y (s) = 1(s 1)2(s 2)

.

Using the partial fraction decomposition, we can write

Y (s) =A

s 1+

B(s 1)2

+C

s 2.

Multipliying both sides by ( s 2)(s 1)2 to obtain1 = A(s 1)(s 2) + B (s 2) + C (s 1)2

= ( A + C )s2 + ( 3A + B 2C )s + 2 A 2B + C Equating coefficients of like powers of s we nd A+ C = 0 , 3A+ B 2C = 0 ,and 2A 2B + C = 1 . Solving this system we nd A = 1, B = 1, andC = 1 . Thus,

Y (s) = 1

s 1 1

(s 1)2+

1s 2

.

Finally,

y(t) = L 1[Y (s)] = L

1 1s

1 L

1 1(s

1)2

+ L 1 1

s

2

= et te t + e2t , t 0Problem 43.14Use Laplace transform to solve the initial value problem

y + 9 y = g(t), y(0) = 1 , y (0) = 3

whereg(t) = 6, 0 t < 0, t <


L[y ] + 9L[y] = L[g(t)] = 6L[h(t) h(t )].

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Y (s) =sy0 + ( y0 + y 0)

s2 + s + =

2s 1s2 + s + 2

.

By identication we nd = 1 , = 2 , y0 = 2 , and y0 = 3Problem 43.16Determine the constants , ,y 0, and y0 so that Y (s) = s(s+1) 2 is the Laplacetransform of the solution to the initial value problem

y + y + y = 0 , y(0) = y0, y (0) = y0.

Solution.Taking the Laplace transform of both sides we nd

s2Y (s) sy 0 y0 + sY (s) y 0 + Y (s) = 0 .Solving for Y (s) we nd

Y (s) =sy 0 + ( y0 + y 0)

s2 + s + =

ss2 + 2 s + 1

.

By identication we nd = 2 , = 1 , y0 = 1 , and y0 = 2

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Section 44


Solution.The function is of period T = 2 . Thus,

3 10 e st dt + 21 e st dt = 3s e st1

0 e st

s

2

1=

1s

(3 2e s e

2s ).

Hence,L[f (t)] =

3 2e s e

2s

s(1 e 2s )Problem 44.2Find the Laplace transform of the periodic function whose graph is shown.

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Solution.

The function is of period T = 4 . Thus,

2 10 e st dt + 31 e st dt = 2s e st1

0 e st

s

3

1=

1s

(2 e s e

3s ).

Hence,

L[f (t)] =2 e

s e 3s

s(1 e 4s )Problem 44.3Find the Laplace transform of the periodic function whose graph is shown.

Solution.

The function is of period T = 2 . Thus,

21 (t 1)e st dt = 21 te st dt 21 e st dt=

ts

e st e st

s2+

e st

s

2

1

= e s

s2[(1 + s)e s 1]

Hence,

L[f (t)] =e s

s2(1 e 2s ) [1(s + 1) e s

]


94

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