# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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Frequently, a Laplace transform F(s) is expressible as a sum of several terms,

F(s) = F() + F2(s) + ¦¦¦ + Fn (s). (17)

Suppose that f1(t) = L~l{F1(s)},..., fn(t) = L~l{Fn(s)}. Then the function

f (t) = f() + ¦¦¦+ fn (t)

has the Laplace transform F(s). By the uniqueness property stated previously there is no other continuous function f having the same transform. Thus

L1{ F(s)} = L1{ F^s)} + ¦¦¦ + L1{ Fn (s)};(18)

that is, the inverse Laplace transform is also a linear operator.

In many problems it is convenient to make use of this property by decomposing a given transform into a sum of functions whose inverse transforms are already known or can be found in the table. Partial fraction expansions are particularly useful in this

304

Chapter 6. The Laplace Transform

TABLE 6.2.1 Elementary Laplace Transforms

f(t) = L-1{ F(s)} F(s) = L{ f (t)} Notes

1. 1 1 Sec. 6.1; Ex. 4

s > 0

s

2. eat 1 Sec. 6.1; Ex. 5

-, s > a

s- a

3. tn; n = positive integer n! Sec. 6.1; Prob. 27

---γγ, s > 0

s^1

4. tp, p > -1 T(p + 1) 0 Sec. 6.1; Prob. 27

-λ---, s > 0

sp+1

5. sin at a Sec. 6.1; Ex. 6

2 2 s > 0

s2 + a2

6. cos at s Sec. 6.1; Prob. 6

2 2, s > 0

s2 + a2

7. sinh at a Sec. 6.1; Prob. 8

2 2 s > lal

s2 - a2

8. cosh at s Sec. 6.1; Prob. 7

2 2, s > lal

s2 - a2

9. 5 b Sec. 6.1; Prob. 13

^ ^, s > a

(s - a)2 + b2

10. eat cos bt s- a Sec. 6.1; Prob. 14

s > a

(s - a)2 + b2

11. tneat, n = positive integer n! Sec. 6.1; Prob. 18

. ³ , s > a

(s - a)n+1

12. 4'( e-cs Sec. 6.3

a -, s > 0

s

13. u e Sec. 6.3

(5 i

) s

f (F

't- )s

1

)c

14. f F(s - c) Sec. 6.3

ct

e

15. f(ct) 1F ( C ) c > 0 Sec. 6.3; Prob. 19

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