Question (1 point) Since – 5t2 sinh(at) – 9t2 cos(at) = ť(–5 sinh(at) – 9 cos(at)), to find the Laplace transform of –5t2 sinh(at) – 9t² cos(at) you take the second derivative of -(5a/(s^2-a^2))-(95/(s^2+a^2)) Therefore L{-5t- sinh(at) – 9t2 cos(at)}(s) = -

Given: -5 t^{2} \sinh (a t)-9 t^{2} \cos (a t)=t^{2}\{-5 \sinh (a t)-9 \cos (a t)\}on comparing with t^{n}\{f(x)\}, we have\begin{array}{l}n=2 \\f(t)=-5 \sinh (9 t)-9 \cos (a t)\end{array}Wr know that:\begin{aligned}& L\left\{t^{n} f(t)\right\}=(-1)^{n} \frac{d^{n}}{d s^{n}}[L\{f(t)\}] \\\Rightarrow & L\left\{t^{2}(-5 \sinh (a t)-9 \cos (a t))\right\} \\& =(-1)^{2} \frac{d^{2}}{d s^{2}}[L\{-5 \sinh (a t)-9 \cos (a t)] \\& =\frac{d^{2}}{d s^{2}}\left[\frac{-5 a}{\left(s^{2}-a^{2}\right)}-\frac{9 s}{\left(s^{2}+a^{2}\right)}\right]\end{aligned}sina, \left.L\{\sinh (a t)\}=\frac{a}{s^{2}-a^{2}}\right\} (from laplace\left.L\left\{\cos \left(a^{\prime \prime}\right)\right\}=\frac{s}{s^{2}+a^{2}}\right\}table)Therefore, we should take second derivative\text { of } \frac{-5 a}{\left(s^{2}-a^{2}\right)}-\frac{g s}{\left(s^{2}+a^{2}\right)}Now,\begin{aligned}\frac{d^{2}}{d s^{2}}\left[\frac{-5 a}{\left(s^{2}-a^{2}\right)}\right] & =\frac{d}{d s}\left[\frac{5 a \times 2 s}{\left(s^{2}-a^{2}\right)^{2}}\right] \\& =\frac{\left(s^{2}-a^{2}\right)^{2}(10 a)-10 a s\left\{2\left(s^{2}-a^{2}\right) \times 2 s\right\}}{\left\{\left(s^{2}-a^{2}\right)^{2}\right\}^{2}}\end{aligned}( \because using fractionrale)\begin{array}{l}=\frac{10 a\left(s^{2}-a^{2}\right)^{2}-40 a s^{2}\left(s^{2}-a^{2}\right)}{\left(s^{2}-a^{2}\right)^{4}} \\=\frac{\left(s^{2}-a^{2}\right)\left[10 a\left(s^{2}-a^{2}\right)-40 a s^{2}\right]}{\left(s^{2}-a^{2}\right)^{4}}\end{array}\begin{array}{l}=\frac{10 a\left(s^{2}-a^{2}\right)-40 a s^{2}}{\left(s^{2}-a^{2}\right)^{3}} \\=\frac{10 a\left(s^{2}-a^{2}-4 s^{2}\right)}{\left(s^{2}-a^{2}\right)^{3}} \\=\frac{-10 a\left(3 s^{2}+a^{2}\right)}{\left(s^{2}-a^{2}\right)^{3}}\end{array}\frac{d^{2}}{d s^{2}}\left[\frac{9 s}{\left(s^{2}+a^{2}\right)}\right]=\frac{d}{d s}\left[\frac{\left(s^{2}+a^{2}\right) \times 9-9 s(2 s)}{\left(s^{2}+a^{2}\right)^{2}}\right]( \because using fraction rule)\begin{array}{l}=\frac{d}{d s}\left[\frac{9 a^{2}-9 s^{2}}{\left(s^{2}+a^{2}\right)^{2}}\right] \\=\frac{\left(s^{2}+a^{2}\right)^{2}(-18 s)-\left(9 a^{2}-9 s^{2}\right)\left\{2\left(s^{2}+a^{2}\right) \times 2 s\right\}}{\left\{\left(s^{2}+a^{2}\right)^{2}\right\}^{2}}\end{array}(again using fraction rule)\begin{array}{l}=\frac{-18\left(s^{2}+a^{2}\right)^{2} s+4 s\left(9 s^{2}-9 q^{2}\right)\left(s^{2}+a^{2}\right)}{\left(s^{2}+q^{2}\right)^{4}} \\=\frac{\left(s^{2}+q^{2}\right)\left[-18 s\left(s^{2}+a^{2}\right)+36 s\left(s^{2}-q^{2}\right)\right]}{\left(s^{2}+q^{2}\right)^{4}} \\=\frac{-18 s\left(s^{2}+q^{2}\right)+36 s\left(s^{2}-a^{2}\right)}{\left(s^{2}+q^{2}\right)^{3}} \\=\frac{-18 s\left(s^{2}+q^{2}+2 s^{2}-2 q^{2}\right)}{\left(s^{2}+q^{2}\right)^{3}}\end{array}=\frac{-18 s\left(3 s^{2}-9^{2}\right)}{\left(s^{2}+9^{2}\right)^{3}}Therefors,\begin{aligned}& \left\{-5 t^{2} \sinh (a t)-9 t^{2} \cos (a t)\right\} \\= & \frac{d^{2}}{d s^{2}}\left[\frac{-5 a}{\left(s^{2}-a^{2}\right)}-\frac{9 s}{\left(s^{2}+a^{2}\right)}\right] \\= & \frac{d^{2}}{d s^{2}}\left[\frac{-5 a}{\left(s^{2}-a^{2}\right)}\right]-\frac{d^{2}}{d s^{2}}\left[\frac{9 s}{\left(s^{2}+a^{2}\right)}\right] \\= & {\left[\frac{-10 a\left(3 s^{2}+a^{2}\right)}{\left(s^{2}-a^{2}\right)^{3}}\right]-\left[\frac{18 s\left(3 s^{2}-a^{2}\right)}{\left(s^{2}+a^{2}\right)^{3}}\right] } \\= & \left.\frac{-10 a\left(3 s^{2}+a^{2}\right)}{\left(s^{2}-a^{2}\right)^{3}}+\frac{18 s\left(3 s^{2}-a^{2}\right)}{\left(s^{2}+a^{2}\right)^{3}}\right]\end{aligned}An ...