Page 71 - LAPLACE TRANSFORM

Example 2

By using Laplace transform method, elaborate y(t) for the given differential equation.

3t
y’’− 3y’ + 2y = 4e ; given y (0) =0 and y’ (0) = 5

Solution

3t
y’’− 3y’ + 2y = 4e
2
{s Y(s) – s y (0) – y’ (0)} – 3 {s Y(s) – y (0)} + 2Y(s) = 4
(S−3)
2
{s Y(s) – s (0) – (5)} – 3 {s Y(s) – (0)} + 2Y(s) = 4
(S−3)
2
{s Y(s) – (5)} – 3 {s Y(s)} + 2Y(s) = 4
(S−3)
2
s Y(s) – 5 – 3sY(s) + 2Y(s) = 4
(S−3)
2
s Y(s) – 3sY(s) + 2Y(s) = 4 + 5
(S−3)
2
Y(s) (s – 3s + 2) = 4 + 5. (S−3)
(S−3) (S−3)
2
Y(s) (s – 3s + 2) = 4 + 5S − 15
(S−3) (S−3)
Y(s) {(s – 1) (s – 2)} = 4 + 5S − 15
(S−3)
4 + 5S − 15 A B C
Y(s) = = + +
(s – 1)(s – 2)(S−3) (s−1) (s−2) (s−3)
+ − = A ( − ) ( − ) + B ( − ) ( − ) + C ( − ) ( − )

when (s − 1) = 0

s= 1; 4 + 5S − 15 = A ( − 2) ( − 3) + B ( − 1) ( − 3) + C ( − 1) ( − 2)
4 + 5S − 15 = A ( − 2) ( − 3)

4 + 5(1) − 15 = A (1 − 2) (1 − 3)
−6 = 2A

−
A = = −

59