MHB Dharshan's question via email about a Laplace Transform

[Prove It]

Evaluate $\displaystyle \mathcal{L}\left\{ 5\sin{ \left( 11 \, t \right) } \sinh{ \left( 11\,t \right) } \right\} $.

$\displaystyle \begin{align*}
\mathcal{L} \left\{ 5\sin{ \left( 11\,t \right) } \sinh{ \left( 11\,t \right) } \right\} &= \mathcal{L} \left\{ 5\sin{ \left( 11\,t \right) } \cdot \frac{1}{2} \left( \mathrm{e}^{11\,t} - \mathrm{e}^{-11\,t} \right) \right\} \\
&= \frac{5}{2} \,\mathcal{L} \left\{ \mathrm{e}^{11\,t} \sin{ \left( 11\,t \right) } - \mathrm{e}^{-11\,t} \sin{ \left( 11\,t \right) } \right\} \\
&= \frac{5}{2} \left[ \mathcal{L}\left\{ \mathrm{e}^{11\,t}\sin{ \left( 11\,t \right) } \right\} - \mathcal{L}\left\{ \mathrm{e}^{-11\,t} \sin{ \left( 11\,t \right) } \right\} \right] \\
&= \frac{5}{2} \left\{ \left[ \frac{11}{s^2 + 11^2} \right]_{s \to s - 11} - \left[ \frac{11}{s^2 + 11^2} \right]_{s \to s + 11} \right\} \\
&= \frac{55}{2} \left[ \frac{1}{\left( s - 11 \right) ^2 + 121} - \frac{1}{\left( s + 11 \right) ^2 + 121} \right]
\end{align*} $

It would be fine to leave your answer in this form, but if you get a common denominator and simplify, you could write the answer as $\displaystyle \frac{1210\,s}{s^4 + 58564}$.

 

[benorin]

the transforms are correct, I didn’t check the algebra on your last simplification.