Solve differential equation $x \frac{d^2y}{dx^2} - \frac{dy}{dx}- 4x^3y = 8x^3 \sin(x^2)$? 
$$x \frac{d^2y}{dx^2} - \frac{dy}{dx}- 4x^3y = 8x^3 \sin(x^2)$$

Please solve I think its solve using Cauchy Euler equation but I am stuck help how to proceed.
 A: Symmetry and substitution
The equation is symmetric under the change from $x$ to $-x$, both sides change the sign equally. If a solution can be extended continuously to $x=0$, then $y'(0)=0$ and one can extend this solution as even function that can be written as $y(x)=u(x^2)$.
Or even simpler, as the ODE has a singularity at $x=0$, a regular one, the domain of continuity has to exclude the $y$ axis. Thus the domain of any solution is a subset of $(0,\infty)$ or $(-\infty,0)$. On $(0,\infty)$ one can parametrize $x=\sqrt{s}$, $u(s)=y(\sqrt{s})$ or $y(x)=u(x^2)$.

Solution of the transformed equation
With $y(x)=u(x^2)$ we also get $y'(x)=2xu'(x^2)$, $y''(x)=4x^2u''(x^2)+2u'(x)$ and inserting this the equation becomes
$$
8x^3\sin(x^2)=[4x^3u''(x^2)+2xu'(x)]-[2xu'(x^2)]-4x^3u(x^2)=4x^3(u''(x^2)-u(x^2))
$$
so that the equation reduces to the simpler one
$$
u''(s)-u(s)=2\sin(s)
$$
which can be solved using standard means. As the resulting solutions $y(x)=-\sin(x^2)+Ae^{x^2}+Be^{-x^2}$ are not singular at $x=0$, one can extend the domain to the whole of $\Bbb R$.

Microscopic zoom to $x\approx 0$
One can look for the possibility of solutions with odd terms by considering the power series expansion around $x=0$ where one gets the indicial equation $r(r-2)=0$, thus $y(x)=\sum a_nx^n$ and
$$
x^n:~~ (n^2-1)a_{n+1}-4a_{n-3}=
\begin{cases}0,& n<5~\text{ or }~ 4\nmid(n-5)\\ 8\frac{(-1)^kx^{4k+5}}{(2k+1)!},&n=4k+5,k\in\Bbb N_0\end{cases}
$$
The first of these relations are
$$
-a_1=4a_{-3}=0,~~ 0\cdot a_2=4a_{-2}=0,~~ 3a_3=4a_{-1}=0,~~ 8a_4=4a_0,~~ 15a_5=4a_1=0,~~24a_6=4a_2+8,...
$$
so that also from this side there are no odd terms, the free parameters are $a_0$ and $a_2$.
A: Avoiding the series solution of @Lutzl ODE (where $s=x^2$ and the prime demotes d.w.r.t $s$), the particular solution can be obtained as $$y''-y = 2 \sin s \Rightarrow (D^2-1)y(s)=2 \sin s \Rightarrow y(s)=\frac{2\sin s}{-1-1} =-\sin s.$$ So $y_P(s)=-\sin s.$ The general solution of homogeneous ODE $y''-y=0$ is $y_G(s)=A \sinh s+ B \cosh s, ~or ~ y= P e^s + Q e^{-s}$ 
So the total solution of OP's ODE is:
$$y(x)= A \sinh x^2 + B \cosh x^2 +]sin x^2$$
OR
$$y(x)= P e^{x^2} + Q e^{-x^2} + \sin x^2 $$
Also note that theparticular  solution of $(D^2-1)z= \sin as$ is $z_p(x)= \frac{\sin as}{-a^2-1}$
