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Application of Laplace Transform
.

s − 1 s − 1
Y (s) = = .
2
2
(s − 2)(s − s − 2) (s − 2) (s + 1)
Use partial fraction:

s − 1 A B C
= + + .
2
(s − 2) (s + 1) s + 1 s − 2 (s − 2) 2
Compare the numerators:

2
s − 1 = A(s − 2) + B(s + 1)(s − 2) + C(s + 1).

2 1
Set s = −1, we get A = − . Set s = 2, we get C = . Set s = 0 (any
9 3
2
convenient values of s can be used in this step), we get B = . So
9
2 1 2 1 1 1
Y (s) = − + + ,
9 s + 1 9 s − 2 3 (s − 2) 2

and
2 2 1
2t
−1
2t
y(t) = L [Y (s)] = − e −t + e + te .
9 9 3
Compare this to the method of undetermined coefficients: general solution
of the equation should be y = y +y , where y is the general solution to the
h
p
h
homogeneous equation and y is a particular solution. The characteristic
p
2
equation is k − k − 2 = (k + 1)(k − 2) = 0, so k = −1, k = 2, and
1
2
−t
2t
y = C e +C e . Since 2 is a root, so the form of the particular solution
h
2
1
2t
is y = Ate . This discussion concludes that the solution should be of the
p
form
2t
2t
y(t) = C e −t + C e + Ate ,
2
1
. . . . for some constants C , C , A. This fits well with our result.
1
2
.
Example 5.8, (Pure imaginary roots)
′
y + y = cos 2t, y(0) = 2, y (0) = 1.
′′
. . . . .
.
Solution. Again, let’s first predict the terms in the solution:

2
k + 1 = 0; k 1,2 = ±i; y = C cos t + C sin t; y = A cos 2t.
1
p
2
h
so
p
h
. . . . y = y + y = C cos t + C sin t + A cos 2t,
2
1
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