Answer to Question #277192 in Calculus for nitish

Question

find inverse laplace transorm : se^-2s/(s^2 + pi^2)

Answer 1

By property of Laplace Transform,

$\begin{aligned} &\text { if } \mathcal{L}^{-1}(G(s))=g(t), \text { then } \\ &\mathcal{L}^{-1}\left(e^{-a s} G(s)\right)=u(t-a) \times g(t-a) \\ &\text { Now, } \mathcal{L}(f(t))=\frac{s e^{-2 s}}{s^{2}+\pi^{2}}=F(s) \\ &\Rightarrow f(t)=\mathcal{L}^{-1}(F(s))=\mathcal{L}^{-1}\left(\frac{s e^{-2 s}}{s^{2}+\pi^{2}}\right)=\mathcal{L}^{-1}\left(e^{-2 s} \times \frac{s}{s^{2}+\pi^{2}}\right) \\ &\qquad \begin{array}{l} \text { But, } \mathcal{L}^{-1}(G(s))=\mathcal{L}^{-1}\left(e^{-2 s} \times G(s)\right), \text { where } G(s)=\frac{s}{s^{2}+\pi^{2}} \\ \Rightarrow f(t)=\mathcal{L}^{-1}\left(e^{-2 s} \times G(s)\right) \\ \quad=u(t-2) \times g(t-2), \text { from the above stated property where } \mathrm{a}=2 \\ \quad=u(t-2) \times \cos (\pi(t-2)), \text { since } g(t)=\cos (\pi t) \end{array} \\ &\text { Thus }, \mathcal{L}^{-1}\left(\frac{s e^{-2 s}}{s^{2}+\pi^{2}}\right)=u(t-2) \times \cos (\pi(t-2)) . \end{aligned}$