Question

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

19.

Short answer

$L^{-1}\left\{\frac{s+\pi }{s^2+{\pi }^2}{e}^{-s}\right\}=\cos \pi (t-1)U(t-1)+\sin \pi (t-1)U(t-1)$

Step-by-step solution

The Laplace equations

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

$f{\left(t\right)}^{*}=\cos \pi (t-1)U(t-1)+\sin \pi (t-1)U(t-1)$$\left\{\mathrm{If}{\mathcal{L}}^{-1}\left\{F\right(s\left)\right\}=f\left(t\right)\mathrm{and}a>0,\mathrm{then}f(t-a)U(t-a)={\mathcal{L}}^{-1}\left\{e^{-as}F\left(s\right)\right\}\right.$$\sin kt={\mathcal{L}}^{-1}\left\{\frac{k}{s^2+k^2}\right\}$$\cos kt={\mathcal{L}}^{-1}\left\{\frac{s}{s^2+k^2}\right\}$

Final proof

$L^{-1}\left\{\frac{s+\pi }{s^2+{\pi }^2}{e}^{-s}\right\}=\cos \pi (t-1)U(t-1)+\sin \pi (t-1)U(t-1)$