Question What do you take the second derivative of (1 point) Since tạ sinh(at) + 3t2 sinh(at) = {2 (sinh(at) + 3 sinh(at)), to find the Laplace transform of tạ sinh(at) + 3t sinh(at) you take the second derivative of sinh(at) Therefore L{t- sinh(at) + 3t2 sinh(at)}(s) ((-8a)(-38^2-a^2))/((s^2-a^2)^3) =

\begin{array}{l}\text { Since; } L\left\{t^{n} F(t)\right\}=(-1)^{n} \frac{d^{n}}{d s^{n}}[L\{F(t)\}] \\\left.\therefore L\left\{t^{2} \sinh (a t)+3 t^{2} \sinh (a t)\right\}\right] \\=L\left[t^{2}\{\sinh (a t)+3 \sinh (a t)\}\right] \\=(-1)^{2} \cdot \frac{d^{2}}{d s^{2}}[L\{\sinh (a t)+3 \sinh (a t)\}] \\=\frac{d^{2}}{d s^{2}}[L\{4 \sinh (a t)\}] \\=\frac{d^{2}}{d s^{2}}\left[\frac{4 \cdot a}{s^{2}-a^{2}}\right] \\\end{array}To find the Leplace tramform of t^{2} \sinh (a t)+3 t^{2} \sinh (a t) you take the seand derivafive of \frac{4 a}{5^{2}-a^{2}} Answen\begin{array}{l}\therefore L\left\{t^{2} \sinh (a t)+3 t^{2} \sinh (a t)\right\} \\=\frac{d^{2}}{d s^{2}}\left[\frac{4 a}{s^{2}-a^{2}}\right] \\=4 a \cdot \frac{d}{d s}\left[\frac{\left(s^{2}-a^{2}\right) \cdot 0-1 \cdot(2 s)}{\left(s^{2}-a^{2}\right)^{2}}\right] \\=-4 a \frac{d}{d s}\left[\frac{2 s}{\left(s^{2}-a^{2}\right)^{2}}\right] \\=-8 a \frac{d}{d s}\left[\frac{s}{\left(s^{2}-a^{2}\right)^{2}}\right] \\=-8 a \cdot\left[\frac{\left.\left(s^{2}-a^{2}\right)^{2} \cdot 1-s \cdot 2\left(s^{2}-a^{2}\right) \cdot 2 s\right]}{\left(s^{2}-a^{2}\right)^{4}}\right] \\\begin{array}{l}=-8 a\left[\frac{\left(s^{2}-a^{2}\right)^{2}-4 s^{2}\left(s^{2}-a^{2}\right)}{\left(s^{2}-a^{2}\right)^{4}}\right] \\=-8 a\left[\left(s^{2}-a^{2}\right)\left(s^{2}-a^{2}-4 s^{2}\right)\right]\end{array} \\=-8 a\left[\frac{\left(s^{2}-a^{2}\right)\left(s^{2}-a^{2}-4 s^{2}\right)}{\left(s^{2}-a^{2}\right)^{4}}\right] \\=\frac{-8 a\left(-3 s^{2}-a^{2}\right)}{\left(s^{2}-a^{2}\right)^{3}} \\=\frac{8 a\left(3 s^{2}+a^{2}\right)}{\left(s^{2}-a^{2}\right)^{3}} \text { (Awwery) } \\\end{array} ...