372 Chapter 6 Laplace Transform
The algebraic determination of theA’s is very tedious, but notice that
(s−r 1 )q(s)
p(s)=A 1 +A 2 s−r^1
s−r 2+···+Aks−r^1
s−rk.
Ifsis set equal tor 1 , the right-hand side is justA 1. The left-hand side becomes
0 /0, but L’Hôpital’s rule gives
slim→r
1(s−r 1 )q(s)
p(s)=slim→r
1(s−r 1 )q′(s)+q(s)
p′(s)=q(r^1 )
p′(r 1 ).
ThereforeA 1 and all the otherA’s are given by
Ai=q(ri)
p′(ri).Consequently, our rational function takes the form
q(s)
p(s)=q(r 1 )
p′(r 1 )1
s−r 1 +···+q(rk)
p′(rk)1
s−rk.From this point we can easily obtain the inverse transform, as expressed in the
conclusion of the following theorem.
Theorem 1.Let p and q be polynomials, q of lower degree than p, and let p have
only simple roots, r 1 ,r 2 ,...,rk.Then
L−^1(q(s)
p(s))
=
q(r 1 )
p′(r 1 )exp(r^1 t)+···+q(rk)
p′(rk)exp(rkt). (2)(Equation(2)is known as Heaviside’s formula.)
Let us apply the theorem to the example in whichq(s)=s+4,p(s)=s^2 +
3 s+2,p′(s)= 2 s+3. Then
L−^1(
s+ 4
s^2 + 3 s+ 2)
= −^1 +^4
2 (− 1 )+ 3
e−t+ −^2 +^4
2 (− 2 )+ 3e−^2 t= 3 e−t− 2 e−^2 t.In nonhomogeneous differential equations also, the Laplace transform is a
useful tool. To solve the problem
u′+au=f(t), u( 0 )=u 0 ,we again transform the entire equation, obtaining
sU−u 0 +aU=F(s),
U(s)= u^0
s+a+^1
s+aF(s).