If y1 and y2 are two linearly independent solutions of $ty'' + 3y' = -te^t$ and if $W (y1, y2) (1) = 2$, find $W (y1, y2) (2)$.
What I have tried:
Bashing my head on the wall, I have absolutely no idea where to start. I thought it has something to do with Abel's theorem but the EDO is not homogeneous.
$$ty'' + 3y' = - te^t$$ $$y'' + \dfrac 3ty' = - e^t$$ You can use Abel's Identity: $$y''+py'+qy=0$$ $$\implies W'+pW=0$$ So that we have : $$ \implies W'+\dfrac 3tW=0$$ Solve the ODE to get the Wronskian $W$: $$(\ln W)'=-\dfrac 3t$$ $$W(t)=Ct^{-3}$$ $$W(1)=2 \implies C=2$$ $$W(t)=2t^{-3}$$ Now you can deduce $W(2)$.