Question

find the inverse Laplace transform for 4s+5/(s-1)^2(s+2)​

Answer 1

Answer:

refer the attachment :)

Answer 2

We need to recall the concept of inverse Laplace transformation.

This transformation transforms the function back to a function in a domain.

Given:

$\frac{4s+5}{(s+1)^{2}(s+2) }$

We can rewrite the function as,

$\frac{4s+5}{(s+1)^{2}(s+2)} =\frac{P}{s-1} + \frac{Q}{(s-1)^2} + \frac{R}{s+2}$

Solving both sides, we get

$4s+5=P(s-1)(s+2)+Q(s+2)+R(s-1)^{2}$

Substitute $s=-2$

$4(-2)+5=P(-2-1)(-2+2)+Q(-2+2)+R(-2-1)^{2}$

$-3=9R$

$R=\frac{-1}{3}$

Substitute $s=1$

$4(1)+5=P(1-1)(1+2)+Q(1+2)+R(1-1)^{2}$

$9=3Q$

$Q=3$

Substitute $s=0$

$4(0)+5=P(0-1)(0+2)+Q(0+2)+R(0-1)^{2}$

$5=-2P+2Q+R$

$5=-2P+2(3)+\frac{-1}{3}$

$P=\frac{1}{3}$

Hence, we get

$\frac{4s+5}{(s+1)^{2}(s+2)} =\frac{\frac{1}{3} }{s-1} + \frac{3}{(s-1)^2} + \frac{\frac{-1}{3} }{s+2}$

So, the inverse Laplace transform is,

$L^{-1} (\frac{4s+5}{(s+1)^{2}(s+2)} )=L^{-1}(\frac{1}{3} \frac{ 1}{s-1} + 3\frac{1}{(s-1)^2} + \frac{-1}{3}\frac{ }{s+2})$

$L^{-1} (\frac{4s+5}{(s+1)^{2}(s+2)} )=\frac{1}{3}L^{-1} (\frac{ 1}{s-1} )+ 3L^{-1} (\frac{1}{(s-1)^2} )+ \frac{-1}{3}L^{-1} (\frac{ }{s+2})$

$L^{-1} (\frac{4s+5}{(s+1)^{2}(s+2)} )=\frac{1}{3}e^{x} + 3e^{x}L^{-1} (\frac{ 1}{s^2})-\frac{ 1}{3}e^{-2x}$

$L^{-1} (\frac{4s+5}{(s+1)^{2}(s+2)} )=\frac{1}{3}e^{x} + 3e^{x}x-\frac{ 1}{3}e^{-2x}$

$L^{-1} (\frac{4s+5}{(s+1)^{2}(s+2)} )=\frac{1}{3}e^{x} + 3xe^{x}-\frac{ 1}{3}e^{-2x}$