Question

In Problems 3 - 24 fill in the blanks or answer true or false. $\mathbf{20}\mathbf{.}{\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{1}{L^2{s}^2+n^2{\pi }^2}\right\}\mathbf{=}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

Short answer

${\mathcal{L}}^{-1}\left\{\frac{1}{L^2{s}^2+n^2{\pi }^2}\right\}=\frac{1}{n\pi L}\sin \frac{n\pi }{L}t$

Step-by-step solution

The Laplace equations

Inverse Laplace Transform of the given expression

${\mathcal{L}}^{-1}\left\{\frac{1}{L^2{s}^2+n^2{\pi }^2}\right\}$

By rearranging the above expression

$\frac{1}{L^2}{\mathcal{L}}^{-1}\left\{\frac{1}{s^2+{\left(\frac{n\pi }{L}\right)}^2}\right\}$

Since we know $\mathcal{L}\left\{\sin kt\right\}=\frac{k}{s^2+k^2}$${\mathcal{L}}^{-1}\left\{\frac{1}{L^2{s}^2+n^2{\pi }^2}\right\}=\frac{1}{n\pi L}{\mathcal{L}}^{-1}\left\{\frac{\frac{n\pi }{L}}{s^2+{\left(\frac{n\pi }{L}\right)}^2}\right\}=\frac{1}{n\pi L}\sin \frac{n\pi }{L}t$

Final proof

${\mathcal{L}}^{-1}\left\{\frac{1}{L^2{s}^2+n^2{\pi }^2}\right\}=\frac{1}{n\pi L}\sin \frac{n\pi }{L}t$