Solve the differential equation $t y''-y'+4t^3y=0$ Solve the differential equation $t y''-y'+4t^3y=0$  using method reduction where $y_1=\sin (t^2)$
my attempt:
$y_2=vy_1=v\sin (t^2)$
$y_2'=v'\sin(t^2)+2t\cos(t^2)\\
y_2''=v''\sin (t^2)+4tv\cos(t^2)+2v\cos(t62)4t^2v\sin(t^2)$
Hence $t y_2''-y_2'+4t^3y_2=0\\
tv''\sin (t^2)+4t^2v\cos(t^2)+2vt\cos(t^2)+4t^3v\sin(t^2)-v'\sin(t^2)-2t\cos(t^2)-4t^3v\sin(t^2)=0\\
v''t\sin(t^2)+4t^2v'\cos(t^2)-v'\sin(t^2)=0$
how to proceed from here?
 A: Let $w = v^\prime.$ What you have is a first order ODE
$$w^\prime t \sin(t^2)  + (4 t^2 \cos(t^2) - \sin(t^2)) w = 0,$$ so 
$$\log w(x) = \int_1^x \frac{(4 t^2 \cos(t^2) - \sin(t^2))}{t \sin(t^2)} d t = 
\int_1^x 4 t \cot t^2 d t - \int_1^t \frac1t d t.$$
I leave both integrals on the RHS to you.
A: Susbstitute $y=v'$ thats a first order equation
$$v''t\sin(t^2)+4t^2v'\cos(t^2)-v'\sin(t^2)=0$$
$$y't\sin(t^2)+y(4t^2\cos(t^2)-\sin(t^2))=0$$
$$y't\sin(t^2)=-y(4t^2\cos(t^2)-\sin(t^2))$$
It's seperable.
$$ \ln|y|=-\int \frac {(4t^2\cos(t^2)-\sin(t^2))}{t\sin(t^2)}dt$$
$$ \ln|y|=-\int \frac  {(4t\cos(t^2)}{\sin(t^2)}dt+\ln|t|$$
Substitute $u =\sin(t^2) \implies du=2t\cos(t^2)dt$
$$ \ln|y|=-2\int \frac  {du}{u}+\ln|t|$$
$$ \ln|y|=-2\ln|\sin(t^2)|+\ln|t|+K$$
$$ y=(\sin(t^2))^{-2} K t$$
$$ v'=((\sin(t^2))^{-2} K t$$
$$ v=K\int \frac {tdt}  {\sin^2(t^2)}$$
Substitute $u=t^2 \implies du=2tdt$
$$ v=K\int \frac {du}  {\sin^2(u)}$$
You should get 
$$\boxed {v=\frac {K_1} {\tan(t^2)}+K_2}$$
And finaly:
$$\boxed {y(t)=K_1{\cos(t^2)}+ {K_2} {\sin(t^2)}}$$

Hint
I used this in the final step...
$$ I=\int \frac {du}  {\sin^2(u)}=\int \frac {\cos^2(u)du}  {\cos^2(u)\sin^2(u)}=\int \frac {d(\tan(u))}  {\tan^2(u)}=-\frac 1 {\tan(u)}$$
