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Differential Equations with Polynomial Coefficients
.


4
3
2
s − 6s + 9s − 4s + 8 
( 1 )
1 +  s (s + 1)(s − 2)(s − 3) 
2
× s−2 =  4 3 2  .
2
−2 − 2  −2s + 8s − 8s + 6s − 4 
s
2
s (s + 1)(s − 2)(s − 3)
Using the method of partial fractions we can write
4 1 8 1 7 1 1 1 1 1
Y (s) = · − · + · − · − ·
1
3 s 2 9 s 3 s + 1 3 s − 2 9 s − 3
2 1 10 1 7 1 2 1 1 1
Y (s) = − · + · − · − · − · .
2
3 s 2 9 s 3 s + 1 3 s − 2 9 s − 3
Therefore

4 8 7 1 1
−1 −t 2t 3t
y (t) = L [Y (s)] = t − + e − e − e
1
1
3 9 3 3 9
2 10 7 2 1
3t
2t
−1
y (t) = L [Y (s)] = − t + − e −t − e − e , t ≥ 0.
2
2
3 9 3 3 9
Hence, for t ≥ 0
( ) ( ) ( ) ( ) ( )
4 − 8 7 − 1 − 1
y(t) = t 3 2 + 10 9 + e −t 3 + e 2t 3 + e 3t 9 .
2
1
7
− − −
3 9 3 3 9
. . . .
The vector equation (5.6) is a linear time invariant system whose Laplace input
is given by y + G(s) and the Laplace output Y(s). According to (5.7) the system
0
−1
transformmatrix is given by (sI−A) . We will show that this matrix is the Laplace
tA
tA
transform of the exponential matrix function e . Indeed, e is the solution to the
initial value problem
′
Φ (t) = AΦ(t), Φ(0) = I,
whereI is the n×n identitymatrix and A isaconstantn×n matrix. Taking Laplace
of both sides yields

sL[Φ(t)] − I = AL[Φ(t)].

Solving for L[Φ(t)] we find

tA
L[Φ(t)] = (sI − A) −1 = L[e ].

Differential Equations with Polynomial Coefficients

Recall Theorem 2 that for F(s) = L(y(t)),

d n
n
n
F(s) = (−1) L(t y(t)) (s > α)
ds n
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