Difficult Reduction of Order with $y_1 = \sin(2t^2)$ I found this problem in the extra questions section of my text, and I can't figure it out.  It is:
Use reduction of order to find a second solution for
$ty''-y'+16t^3y=0, \ \ \ \ y_1=\sin(2t^2)$.
I tried going about it by reducing a general second order differential to 
$v''y_1+v'(2y_1'+p(t)y_1)v'=0$ 
$v''\sin(2t^2)+v'(8t\cos(2t^2)-t^{-1})=0$
But from here I am stuck.  Any and all help is appreciated!  Thanks!
 A: We have the DEQ:
$$\tag 1 ty''-y'+16t^3y=0$$
If we take $y_2(t) = v(t) y_1(t) = v(t) \sin(2 t^2)$, we have:


*

*$y_2''(t) = \sin \left(2 t^2\right) v''(t)+8 t \cos \left(2 t^2\right) v'(t)-16 t^2 v(t) \sin \left(2 t^2\right)+4 v(t) \cos \left(2 t^2\right)$

*$y_2'(t) = \sin \left(2 t^2\right) v'(t)+4 t v(t) \cos \left(2 t^2\right)$

*$y_2(t) = v(t) \sin(2t^2)$


If we substitute these three into $(1)$, we have:
$-\sin \left(2 t^2\right) v'(t)+t \left(\sin \left(2 t^2\right) v''(t)+8 t \cos \left(2 t^2\right) v'(t)-16 t^2 v(t) \sin \left(2 t^2\right)+4 v(t) \cos \left(2 t^2\right)\right)-4 t v(t) \cos \left(2 t^2\right)+16 t^3 v(t) \sin \left(2 t^2\right)=0$
This simplifies to:
$$\tag 2 8 t^2 \cos \left(2 t^2\right) v'(t)+\sin \left(2 t^2\right) \left(t v''(t)-v'(t)\right) = 0$$
Now, we let $w(t) = v'(t) \implies w'(t) = v''(t)$ and substitute these into $(2)$, yielding:
$$\sin \left(2 t^2\right) \left(t w'(t)-w(t)\right)+8 t^2 w(t) \cos \left(2 t^2\right)=0$$
You can solve this using an Integrating Factor, yielding:
$$w(t) = c_1 t \csc^2(2 t^2)$$
We have $w(t) = v'(t)$, hence:
$$v'(t) = c_1 t \csc^2(2 t^2)$$
Solve by integrating, yielding:
$$v(t) = c_1-\frac{1}{4} c_2 \cot \left(2 t^2\right)$$
Lastly, we have $y_2(t) = v(t) y_1(t) = v(t) \sin(2 t^2)$, which reduces to:
$$y_2(t) = \frac{1}{4} \left(4 c_1 \sin \left(2 t^2\right)-c_2 \cos \left(2 t^2\right)\right) = c_1 \sin(2 t^2) + c_2 \cos(2 t^2)$$
