[tex]y_1=\sin(x^2)\implies{y_1}'=2x\cos(x^2)\implies{y_1}''=2\cos(x^2)-4x^2\sin(x^2)[/tex]
Substituting [tex]y_1(x)[/tex] and its derivatives into the ODE gives
[tex]x\left(2\cos(x^2)-4x^2\sin(x^2)\right)-2x\cos(x^2)+4x^3\sin(x^2)=0[/tex]
as required.
Assume a second solution of the form [tex]y_2(x)=u(x)y_1(x)[/tex]. Then
[tex]y_2=uy_1\implies{y_2}'=u'y_1+u{y_1}'\implies{y_2}''=u''y_1+2u'{y_1}'+u{y_1}''[/tex]
Substitute these into the ODE:
[tex]x(u''y_1+2u'{y_1}'+u{y_1}'')-(u'y_1+u{y_1}')+4x^3uy_1=0[/tex]
[tex]x\sin(x^2)u''+\left(4x^2\cos(x^2)-\sin(x^2)\right)u'=0[/tex]
This ODE is separable as
[tex]\implies\dfrac{\mathrm du''}{\mathrm du}=\dfrac{\sin(x^2)-4x^2\cos(x^2)}{x\sin(x^2)}\,\mathrm dx[/tex]
Integrating both sides gives
[tex]\ln|u'|=\displaystyle\int\left(\frac1x-4x\cot(x^2)\right)\,\mathrm dx[/tex]
For the remaining integral, we have
[tex]\displaystyle\int\frac{\mathrm dx}x=\ln|x|+C[/tex]
[tex]\displaystyle\int4x\cot(x^2)\,\mathrm dx=2\int\cot(x^2)(2x\,\mathrm dx)=2\ln|\sin(x^2)|+C[/tex]
So we have
[tex]\ln|u'|=\ln|x|-2\ln|\sin(x^2)|+C[/tex]
[tex]\implies u'=Cx\csc^2(x^2)[/tex]
Integrate both sides to solve for [tex]u(x)[/tex]:
[tex]\displaystyle\int u'\,\mathrm dx=C\int x\csc^2(x^2)\,\mathrm dx[/tex]
[tex]\implies u=\displaystyle\frac C2\int\csc^2(x^2)(2x\,\mathrm dx)[/tex]
[tex]\implies u=C_1\cot^2(x^2)+C_2[/tex]
Then our second solution turns out to be
[tex]y_2=uy_1=C_1\cot^2(x^2)\sin(x^2)+C_2\sin(x^2)[/tex]
[tex]y_1(x)[/tex] already captures the second term here, so the second fundamental solution is
[tex]\boxed{y_2(x)=\cos(x^2)\cot(x^2)}[/tex]
and the general solution is
[tex]\boxed{y(x)=C_1\sin(x^2)+C_2\cos(x^2)\cot(x^2)}[/tex]
