6.2 Partial Fractions and Convolutions 369
a.^1
(s^2 +ω^2 )^2
;
c. s
2
(s^2 +ω^2 )^2
;
b. s
(s^2 +ω^2 )^2
;
d. s
3
(s^2 +ω^2 )^2
6.2 Partial Fractions and Convolutions
Because of the formula for the transform of derivatives, the Laplace transform
finds important application to linear differential equations with constant co-
efficients, subject to initial conditions. In order to solve the simple problem
u′+au= 0 , u( 0 )= 1 ,
we transform the entire equation, obtaining
L(u′)+aL(u)= 0
or
sU− 1 +aU= 0 ,
whereU=L(u). The derivative has been “transformed out,” andUis deter-
mined by simple algebra to be
U(s)=
1
s+a.
By consulting Table 2 we find thatu(t)=e−at.
Equationsofhigherordercanbesolvedinthesameway.Whentrans-
formed, the problem
u′′+ω^2 u= 0 , u( 0 )= 1 , u′( 0 )= 0
becomes
s^2 U−s· 1 − 0 +ω^2 U= 0.
Note how both initial conditions have been incorporated into this one equa-
tion. Now we solve the transformed equation algebraically to find
U(s)=
s
s^2 +ω^2 ,
the transform of cos(ωt)=u(t).