Answer
See below
Work Step by Step
Given $y''-x^{-1}y'+4x^2y=0$
We first compute the appropriate derivatives:
$y'=\sin(x^2)(u'+u)\\
y''=\sin(x^2)(u''+u'+u)$
Substitute:
$(u''+u'+u)-x^{-1}(u'+u)+4x^2\sin(x^2)=0$
which simplifies to
$\sin(x^2)u''+4xu'\cos(x^2)-\frac{\sin(x^2)}{x}u'=0$
or equivalently,
$w'x+w=0$
where $w=u'$
Separating the variables yields
$\frac{1}{x}-\frac{4x\cos (x^2)}{\sin(x^2)}=\frac{w'}{w}$
By integrating, we obtain
$\ln w=-2\ln (\sin(x^2))+\ln x +C\\
wx=k$
Therefore,
$u'=\frac{kx}{(\sin(x^2))^2}$
Integration by parts gives
$u=-\frac{k}{2}\cot(x^2)+C$
The general solution is
$y=k\sin (x^2)\cot(x^2)+C\sin (x^2)$
