Question

Find l^-1 {1/(s+1)(s+2)^2 by using convolution theorem

Answer 1

By Convolution Theorem :

$\\ L^{-1}\{F_{1}(s).F_{2}(s)\}=\int_0^tF_{1}(u).F_{2}(t-u) \;du$

Here,we need to evaluate $L^{-1}\{\dfrac{1}{(s+1)(s+2)^2}\}=L^{-1}\{\dfrac{1}{s+1}.\dfrac{1}{(s+2)^2}\}$ :

Step-1 : Take $F_1(s)=\dfrac{1}{(s+1)} \;and\; F_2(s)=\dfrac{1}{(s+2)^2}$

Step-2 : $L^{-1}\{F_1(s)\}=L^{-1}\{\dfrac{1}{(s+1)}\}$

Using the properties of inverse laplace transform : $L^{-1}\{\dfrac{1}{s-a}\}=e^{at}\;\;and\;\;L^{-1}\{\dfrac{1}{s^n}\}=\dfrac{t^{n-1}}{(n-1)!}$

$\therefore L^{-1}\{F_1(s)\}= L^{-1}\{\dfrac{1}{(s+1)}\}=e^{-1.t}=e^{-t}=\mathbf{f_1(t)}\;and$

$L^{-1}\{F_2(s)\}=L^{-1}\{\dfrac{1}{(s+2)^2}\}=e^{-2t}L^{-1}\{\dfrac{1}{s^2}\}$

$\implies L^{-1}\{F_2(s)\}=e^{-2t}.\dfrac{t^{2-1}}{(2-1)!}=e^{-2t}.\dfrac{t}{1!}=te^{-2t}=\mathbf{f_2(t)}$

By using the convolution formula,we have -

$L^{-1}\{\dfrac{1}{(s+1)(s+2)^2}\}=L^{-1}\{\dfrac{1}{s+1}.\dfrac{1}{(s+2)^2}\}$

$=\int_0^t\mathbf{f_1(u).f_2{(t-u)du}}$

$=\int_0^te^{-u}.(t-u).e^{-2(t-u)}du$

$=\int_0^t(t-u).e^{-u-2t+2u}du$

$=\int_0^t(t-u).e^{u-2t}du=e^{-2t}\int_0^t(t-u)e^udu$

$=e^{-2t}[t.\int_0^te^udu-\int_0^tue^udu]$

$=e^{-2t}\bigg[t.\bigg[e^u\bigg]_0^t-\bigg[u.\int e^udu-\int \bigg(\dfrac{d}{du}(u).\int e^udu\bigg)du\bigg]_0^t\bigg]$

$=e^{-2t}\bigg[t.[e^t-e^0]-\bigg[u.e^u-\int\bigg(1.e^u\bigg)du\bigg]_0^t\bigg]$

$=e^{-2t}\bigg[t.[e^t-1]-\bigg[ue^u-e^u\bigg]_0^t\bigg]$

$=e^{-2t}\bigg[te^t-t-\bigg[(te^t-e^t)-(0-e^0)\bigg]\bigg]$

$=e^{-2t}\bigg[te^t-t-\bigg[(te^t-e^t)+1)\bigg]\bigg]$

$=e^{-2t}\bigg[te^t-t-te^t+e^t-1\bigg]=e^{-2t}(e^t-t-1)$ .............Ans.