inverse laplace transform of $\dfrac{5s^2 - 15s - 11}{(s + 1)(s - 3)^3}$

Submitted by fas on

Find the inverse Laplace transform if
(5s^2-15s-11)/((s+1)(s-3)^3 )

can anyone please help me solve this question...??

Tags
inverse laplace

Jhun Vert Fri, 04/28/2017 - 11:44

$\dfrac{5s^2 - 15s - 11}{(s + 1)(s - 3)^3} = \dfrac{A}{s + 1} + \dfrac{B}{s - 3} + \dfrac{C}{(s - 3)^2} + \dfrac{D}{(s - 3)^3}$

$5s^2 - 15s - 11 = A(s - 3)^3 + B(s + 1)(s - 3)^2 + C(s + 1)(s - 3) + D(s + 1)$

$\begin{align}
5s^2 - 15s - 11 = & \, A(s^3 - 9s^2 + 27s - 27) + B(s^3 - 5s^2 + 3s + 9) \\
& + C(s^2 - 2s - 3) + D(s + 1)
\end{align}$
 

Set s = 3
$5(3^2) - 15(3) - 11 = D(3 + 1)$

$-11 = 4D$

$D = -11/4$
 

Set s = -1
$5(-1)^2 - 15(-1) - 11 = A(-1 - 3)^3$

$9 = -64A$

$A = -9/64$
 

Equate the coefficients of s3
$0 = A + B$

$0 = -\frac{9}{64} + B$

$B = 5/64$
 

Equate the coefficients of s2
$5 = -9A - 5B + C$

$5 = -9(-\frac{9}{64}) - 5(\frac{5}{64}) + C$

$C = 33/8$

 

$\begin{align}
\mathcal{L}^{-1} \left[ \dfrac{5s^2- 15s - 11}{(s + 1)(s - 3)^3} \right] & = \mathcal{L}^{-1} \left[ \dfrac{-9/64}{s + 1} + \dfrac{5/64}{s - 3} + \dfrac{33/8}{(s - 3)^2} + \dfrac{-11/4}{(s - 3)^3} \right] \\
& = -\dfrac{9}{64} e^{-t} + \dfrac{5}{64} e^{3t} + \dfrac{33}{8} \dfrac{e^{3t}t^{2 - 1}}{(2 - 1)!} - \dfrac{11}{4} \dfrac{e^{3t}t^{3 - 1}}{(3 - 1)!} \\
& = -\dfrac{9e^{-t}}{64} + \dfrac{5e^{3t}}{64} + \dfrac{33e^{3t}t}{8} - \dfrac{11e^{3t}t^2}{8}
\end{align}$