Question

Use the convolution theorem to find the inverse Laplace transform of the given function. 4 s (s2 + 4) **** 14.30-0 $(2+4)

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Answer #1

Convolution theorem:

L^{-1}[f(s).g(s)]=\int^{t}_{0}f(u).g(t-u)du

Split the given function into two terms and apply inverse laplace transform.

L^{-1}[\frac{1}{s^2+b^2}]=\frac{sinbt}{b}

L^{-1}[\frac{1}{s^n}]=\frac{t^{n-1}}{(n-1)!}

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:

Sin^2x+cos^2x=1

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-Now-apply limits; Toto to the 20 te = (2t - 1) Sinat & at cosat 0 과 t 1² los au du = (2t²-1) Sinat + 2t cos at -4 # 24 O nex= at sinat +(1-2t² cos 2t 4. Now, t t sinst fu² cossu du osat Juzsinau du Sinat [let 2_1)s.in2t & at tos at 4 Cosat at sinat22 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-

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