Solve the equation $ty''-y'+4t^3y=0, t>0$ given that $\sin(t^2)$ is a solution.
(I am aware that this has been asked before but the answers did not help me).
My attempt: I want to use reduction of order, so I assume the second solution is of the form $y_2(t)=u(t)\sin(t^2)$. So: $$y'=u'(t)\sin(t^{2})+2tu(t)\cos(t^{2})$$ $$y''=2t\cos(t^{2})u'(t)+u''(t)\sin(t^{2})+2t(u'(t)\cos(t^{2})-2t\sin(t^{2})u(t))+2u'(t)\cos(t^{2})$$ Plugging into the equation and rearranging, we get the following equation: $$u''(t)(t\sin(t^{2}))+u'(t)(4t^{2}\cos(t^{2})-\sin(t^{2}))=0$$ And by a change of variable $v=u'$, we then have the following seperable equation: $$\frac{v'(t)}{v(t)}=(\frac{1}{t}-4t\frac{\cos(t^2)}{\sin(t^{2})})$$ Integrating both sides and taking exponenets, we have: $$v(t)=-2t\sin(t^2)$$ And therefore: $$u(t)=\cos(t^2)$$ So the solution is $y_2(t)=\cos(t^2)\sin(t^2)$. However this is incorrect, and $y_2(t)$ should be just equal to $\cos (t^2)$ (or some linear combination of $\sin(t^2)$ and $\cos(t^2)$.
What am I missing here? Any help would be appreactied.
You have some typos/transcription errors in the derivatives, that however have no follow-on effects.
The integration of the equation for $v$ results in $$ \ln|v|=c+\ln|t|-2\ln|\sin(t^2)|. $$ A particular solution for $v$ can be selected so that $$ u'=v=\frac{-2t}{\sin^2(t^2)} \implies u=\frac{\cos(t^2)}{\sin(t^2)}. $$