Convolution theorem:
![L^{-1}[f(s).g(s)]=\int^{t}_{0}f(u).g(t-u)du](http://img.homeworklib.com/questions/adda7ae0-100d-11eb-be88-2fa28de2e7c2.png?x-oss-process=image/resize,w_560)
Split the given function into two terms and apply inverse laplace transform.
![L^{-1}[\frac{1}{s^2+b^2}]=\frac{sinbt}{b}](http://img.homeworklib.com/questions/ae316f00-100d-11eb-8dbf-911826245555.png?x-oss-process=image/resize,w_560)
![L^{-1}[\frac{1}{s^n}]=\frac{t^{n-1}}{(n-1)!}](http://img.homeworklib.com/questions/ae87af30-100d-11eb-bf12-f93d8a426526.png?x-oss-process=image/resize,w_560)
After that apply in the formula, put t=u in f(t) function and put t=t-u in g(t) function.
Now do integration, apply limits, get final equation.
Formula used:


![ul. Sien | 2t - 24) du - tg u² [ Sinat. Coszu los at. Sin 2u] du ts 2 Sinet cos 24 du – fulles at. Sinzudu - Sinat i un cos s](http://img.homeworklib.com/questions/afff3510-100d-11eb-82d6-a72c358cf28a.png?x-oss-process=image/resize,w_560)


![22 sin et + cos² et] et cos at म 4 , 3. - 무 + los at स 4 १ 2 है 2 + cos at 2. स 4 2 + D । [ो 4 53(s+4) Los(26) में न 2-](http://img.homeworklib.com/questions/b33fbe30-100d-11eb-b38a-5701f5276896.png?x-oss-process=image/resize,w_560)
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