Solving the ODE $y''+ty'-y=0$ by Laplace transform

Question

Question:

Solve the ODE $y''+ty'-y=0$ when $y(0)=0$ and $y'(0)=5$ by Laplace transform

My try:

$Y:=L\{y\}$

$s(sY-y(0))-5+L\{ty'\}-Y=0 \implies s^2 Y-5-(L\{y'\})'-Y=0 $ $\implies Y(s^2-1)-5-(Y+sY')=0 \implies Y(s^2-2)-5-sY'=0 $ $ \implies Y'+\frac{2-s^2}{s}Y=\frac{-5}{s} \implies Y=ce^{\frac{s^2}{2}}(\frac{1}{s^2})(1+5e^{-\frac{s^2}{2}})$

Now, the problem is that I don't know how to find the inverse Laplace of $Y=ce^{\frac{s^2}{2}}(\frac{1}{s^2})(1+5e^{-\frac{s^2}{2}})$
Any idea?

Answer 1

You have made a mistake continue where you have stayed: $$ Y'+\frac{2-s^2}{s}Y=\frac{-5}{s}.$$ The integrating factor for this equation is $s^2e^{-s^2/2}$. Multiplying both sides we get $$\bigg(s^2e^{-s^2/2}Y(s)\bigg)'=-5se^{-s^2/2}.$$ Integrating we get $$s^2e^{-s^2/2}Y(s)=5e^{-s^2/2}\Longrightarrow Y(s)=\frac{5}{s^2}\Longrightarrow y(t)=5t.$$