Laplace transform and IVP at $\infty$

Question

Solving the following differential equation $$ty^{''}\left ( t \right )+\left ( t-1 \right )y^{'}\left ( t \right )-y\left ( t \right )=0$$ with initial values $$y\left ( 0 \right )=5, y\left ( \infty \right )=0$$ I suspect the identity $$\lim _{t\rightarrow \infty}y\left ( t \right )=\lim _{s\rightarrow 0}sF\left ( s \right ) \left ( * \right )$$ where $s$ is a complex varaible would be of help here. But for the Laplace transform of the second derivative we need the initial value of the first derivative. Can we somehow obtain that value using equation $\left ( * \right )$?

Answer 1

$ty''(t)+(t-1)y'(t)-y(t)=0$

$t(y'(t)+y(t))'-(y'(t)+y(t))=0$

$\dfrac{(y'(t)+y(t))'}{y'(t)+y(t)}=\dfrac{1}{t}$

$\int\dfrac{(y'(t)+y(t))'}{y'(t)+y(t)}dt=\int\dfrac{1}{t}dt$

$\ln(y'(t)+y(t))=\ln t+c$

$y'(t)+y(t)=C_1t$

$(e^ty(t))'=C_1te^t$

$e^ty(t)=C_1(t-1)e^t+C_2$

$y(t)=C_1(t-1)+C_2e^{-t}$

$y(\infty)=0$ :

$C_1=0$

$\therefore y(t)=C_2e^{-t}$

$y(0)=5$ :

$C_2=5$

$\therefore y(t)=5e^{-t}$