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Application of Laplace Transform

′′
′
′
′
We have L[y (t)] = sL[y (t)] − y (0) = s(sL[y(t)] − y(0)) − y (0) =
2
′
= s L[y(t)] − sy(0) − y (0).
(a)(b) We have
{∫ }
t
L[y(t)] = sL y(s)ds
0
so that
t L[y(t)]
{∫ }
L y(s)ds = .
0 s
. . . .

The above two theorems can be used for solving the following initial value
problem
′
y (t) = Ay + g(t), y(0) = y , t > 0 (5.6)
0
where A is a constant matrix and the components of g(t) are members of PE.
Using the above theorems we can write

sY(s) − y = AY(s) + G(s)
0
or
(sI − A)Y(s) = y + G(s)
0
where L[g(t)] = G(s). If s is not an eigenvalue of A then the matrix sI − A is
invertible and in this case we have

−1
Y(s) = (sI − A) [y + G(s)]. (5.7)
0
To compute y(t) = L 1[Y(s)] we compute the inverse Laplace transform of each
−
.
component of Y(s). We illustrate the above discussion in the next example.

Example 5.13, Solve the initial value problem

( ) ( 2t ) ( )
1 2 e 1
′
y = y + , y(0) = .
2 1 −2t −2
. . . . .
.
Solution. We have

( )
1 s − 1 2
(sI − A) −1 =
(s + 1)(s − 3) 2 s − 1

and
( )
1
G(s) = s−2
2
− 2
s
Thus,
( )
1 s − 1 2
−1
Y(s) = (sI − A) [y + G(s)] = ×
0
. . . . (s + 1)(s − 3) 2 s − 1

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