Question

\[\mathcal{L}\{t^2sinh(3t)\}=\]

Answer 1

\[\sinh{3t}=\frac{e^{3t}-e^{-3t}}{2}\]\[\mathcal{L}\{t^2\sinh{3t}\}=\int\limits_0^{+\infty}t^2\sinh{3t}\cdot e^{-st}dt=\]\[=\int\limits_0^{+\infty}t^2\frac{e^{3t}-e^{-3t}}{2}\cdot e^{-st}dt=\]\[=\frac{1}{2}\left(\int\limits_0^{+\infty}t^2e^{-(s-3)t}dt-\int\limits_0^{+\infty}t^2e^{-(s+3)t}dt\right)\]\[\mathcal{L}\{t^2\}=\int\limits_0^{+\infty}t^2e^{-st}dt=\frac{2}{s}\int\limits_0^{+\infty}te^{-st}dt=\frac{2}{s^2}\int\limits_0^{+\infty}e^{-st}dt=\frac{2}{s^3}\]\[\mathcal{L}\{t^2\sinh{3t}\}=\frac{1}{2}\left(\frac{2}{(s-3)^3}-\frac{2}{(s+3)^3}\right)=\]\[=\frac{1}{(s-3)^3}-\frac{1}{(s+3)^3}=\frac{18(s^2+3)}{(s^2-9)^3}\]

Answer 2

Thank you so much