Is there a solution to this differential equation? I am trying to find a function $y(x)$ that is a solution to
$$
\left(a_3 x^3+a_1 x\right) y''(x)-\left(3 a_3 x^2+2 a_1\right) y'(x)+3 a_3\, x \,y(x)=a_0 x^4+a_2
$$
I tried using mathematica but it ran for hours without giving a solution.
Thank you.
 A: Notice the LHS of the ODE can be rewritten in operator form:
$$\begin{align}
LHS 
&= \left[x\frac{d}{dx} - 2\right]\left[(a_1 + a_3 x^2)\frac{d}{dx} - 3a_3 x\right]y\\
&= \left[ x^3 \frac{d}{dx} x^{-2}\right]\left[ (a_1 + a_3 x^2)^{5/2} \frac{d}{dx}  
\frac{1}{(a_1 + a_3 x^2)^{3/2}} \right] y
\end{align}
$$
Unwinding the leftmost differential operator, we find for suitable chosen integration constant $A$, we have
$$\left[ (a_1 + a_3 x^2)^{5/2} \frac{d}{dx}  
\frac{1}{(a_1 + a_3 x^2)^{3/2}} \right] y
= x^2 \int^x \frac{a_0 t^4 + a_2}{t^3} dt
= \frac{a_0 x^4 + A x^2 - a_2}{2}
$$
Unwinding the remaining differential operator, we can express $y(x)$ as an integral
$$y(x) = (a_1+a_3 x^2)^{3/2}\left[\frac{y(0)}{a_1^{3/2}} + \frac12 \int_0^x 
\frac{a_0 t^4 + A t^2 - a_2}{(a_1 + a_3 t^2)^{5/2}} dt\right]
$$
With help of an CAS, we can evaluate the integral and get
$$\begin{align}
y(x) &= 
(a_1+a_3 x^2)^{3/2}\left[\frac{y(0)}{a_1^{3/2}} + \frac{a_0}{2a_3^{5/2}}\sinh^{-1}\left(\sqrt{\frac{a_3}{a_1}}x\right)\right]\\
& \quad - \frac16\left[
\frac{a_0 x (3a_1 + 4 a_3 x^2)}{a_3^2}
-\frac{A x^3}{a_1}
+\frac{a_2 x (3a_1 + 2 a_3 x^2)}{a_1^2}
\right]
\end{align}
$$
A: If y is a polynomial of degree 3,
then all the terms
are polynomials of degree 4.
Put in the polynomial
and equate coefficients
of each degree.
