Question

Find inverse laplace transform of f(t) f1t) - 5+2 5² (5+1) 2

Answer 1

F(S) s+2 $2 (s +1) 2 S 2

take partial fraction

$\frac{s+2}{s^2\left(s+1\right)^2}=\frac{a_0}{s}+\frac{a_1}{s^2}+\frac{a_2}{s+1}+\frac{a_3}{\left(s+1\right)^2}$...................(1)

$s+2=a_0s\left(s+1\right)^2+a_1\left(s+1\right)^2+a_2s^2\left(s+1\right)+a_3s^2$...................(2)

take s=0

$0+2=a_0\cdot \:0\cdot \left(0+1\right)^2+a_1\left(0+1\right)^2+a_2\cdot \:0^2\left(0+1\right)+a_3\cdot \:0^2$

2= 01

${\color{Blue} a_1=2}$

.

$s+2=a_0s\left(s+1\right)^2+a_1\left(s+1\right)^2+a_2s^2\left(s+1\right)+a_3s^2$

take s=-1

${\color{black} \left(-1\right)+2=a_0\left(-1\right)\left(\left(-1\right)+1\right)^2+a_1\left(\left(-1\right)+1\right)^2+a_2\left(-1\right)^2\left(\left(-1\right)+1\right)+a_3\left(-1\right)^2}$

$1=a_3$

${\color{Blue} a_3=1}$

put it back in equation 2

.

$s+2=a_0s\left(s+1\right)^2+2\left(s+1\right)^2+a_2s^2\left(s+1\right)+1\cdot \:s^2$

$s+2=a_0s^3+a_2s^3+2a_0s^2+a_2s^2+3s^2+a_0s+4s+2$

$1\cdot \:s+2=s^3\left(a_0+a_2\right)+s^2\left(2a_0+a_2+3\right)+s\left(a_0+4\right)+2$

compare coefficients both sides

$\begin{bmatrix}a_0+4=1\\ 2a_0+a_2+3=0\\ a_0+a_2=0\end{bmatrix}$

${\color{Blue} a_0=-3,\:a_2=3}$

put all constants in equation 1

$F(s)=\frac{\left(-3\right)}{s}+\frac{2}{s^2}+\frac{3}{s+1}+\frac{1}{\left(s+1\right)^2}$

$F(s)=-\frac{3}{s}+\frac{2}{s^2}+\frac{3}{s+1}+\frac{1}{\left(s+1\right)^2}$

take inverse laplace

$f(t)=L^{-1}\left\{-\frac{3}{s}+\frac{2}{s^2}+\frac{3}{s+1}+\frac{1}{\left(s+1\right)^2}\right\}$

$f(t)=-L^{-1}\left\{\frac{3}{s}\right\}+L^{-1}\left\{\frac{2}{s^2}\right\}+L^{-1}\left\{\frac{3}{s+1}\right\}+L^{-1}\left\{\frac{1}{\left(s+1\right)^2}\right\}$

${\color{Red} f(t)=-3+2t+3e^{-t}+te^{-t}}$