Answer:
[tex]\frac{s^2-25}{(s^2+25)^2} [/tex]
Step-by-step explanation:
Let's use the definition of the Laplace transform and the identity given:[tex]\mathcal{L}[t \cos 5t]=(-1)F'(s)[/tex] with [tex]F(s)=\mathcal{L}[\cos 5t][/tex].
Now, [tex]F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt[/tex]. Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that [tex]F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt[/tex].
Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that
[tex]F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)[/tex].
Solving for F(s) on the last equation, [tex]F(s)=\frac{s}{s^2+25}[/tex], then the Laplace transform we were searching is [tex]-F'(s)=\frac{s^2-25}{(s^2+25)^2} [/tex]
