ODE $(2x^6+5x^3-3x)y''(x)+(-14x^5-20x^2+6)y'(x)+30x^4y(x)=0$ $A(x)y''+B(x)y'+C(x)y=0$ represents the equation.


*

*Tried using $t=\phi(x)$, but the $\int{\sqrt{\frac{C(x)}{A(x)}}}$
gives too complicated results.

*$y=UV$ substitution didn't do anything.
Particular solutions are only thing left, combinations I've tried and that didn't work for me: $y_p=ax^2+b, y_p=\frac{a}{x^2}+b, y_p=ax^6+b$
 A: Try $y(x)=\sum_{n=a}^{\infty}b_nx^n$.
Rearrange the equation so you get $$\sum_{n=a}^{\infty}x^nf(b_n,b_{n+1},...)=0$$
That way, $f(b_n,b_{n+1},...)=0$, and you have a recursion for the $b_n$.
These recursions may give a value of $a$ for which $b_{a-1}=b_{a-2}=0\neq b_a$ is possible, so that is the starting point of the series.
The recursions may also give a value of $a$ for which $b_{a+1}=b_{a+2}=0\neq b_a$ is possible, so the series might end at that value of $a$.
A: You can simplify the equation a bit: introduce
\begin{equation}
 p(x) = 2 x^7 + 5 x^4 - 3 x^2,
\end{equation}
then the ODE has the form
\begin{equation}
\frac{p(x)}{x} y'' - \frac{p'(x)}{x} y' + 30 x^4 y = 0.
\end{equation}
You can rewrite this as
\begin{equation}
 p^2\left(\frac{y'}{p}\right)' + 30 x^5 y = 0,
\end{equation}
which is equivalent to
\begin{equation}
 p^3 \left(\frac{1}{p} \frac{\text{d}}{\text{d} x}\right)^2 y + 30 x^5 y = 0.
\end{equation}
Introducing
\begin{equation}
 \xi(x) = \int^x \!p(\bar{x})\,\text{d}\bar{x} = \frac{1}{4} x^8 + x^5 - x^3 \,\,(+ \text{constant}),
\end{equation}
you can write
\begin{equation}
 \frac{1}{p} \frac{\text{d}}{\text{d} x} = \frac{\text{d}}{\text{d} \xi},
\end{equation}
so the original ODE can be 'simplified' to
\begin{equation}
 \frac{\text{d}^2 y}{\text{d} \xi^2} + F(\xi)\, y = 0,
\end{equation}
where
\begin{equation}
 F(\xi) = \frac{30 x(\xi)^5}{p(x(\xi))^3}.
\end{equation}
Admittedly, it might be a bit cumbersome (to say the least) to invert the coordinate transformation $x \to \xi$, but it might be worth a try. 
If all else fails, you can indeed follow @Michael's suggestion and try substituting a series expansion. The singular points of the coefficient multiplying $y''$ are not easy to find however, so that approach might be of limited value -- except for solutions around $x = 0$, where the 'usual' series expansion will give you some information.
A: For large $x$ an approximate ODE is : $2x^6y''-14x^5y'+30x^4y=0$
The solutions on the form $x^r$ are obtained in solving $2r(r-1)-14r+30=0$ which roots are $r=5$ and $r=3$. 
So, it is possible (but not sure) that particular solutions of the complete ODE be  polynomials of degree 3 and/or 5.
Let try $y(x)=$ polynomial of degree 5 with indetermined coefficients. Puting it into the complete ODE allows to identify the coefficients. This is an easy calculus leading to a first solution $y=x^5+1$
An even simpler calculus with $y=$polynomial of degree 3 leads to $y=x^3+1$
So, the general solution of the complete ODE is :
$$y=c_1 (x^5+1) + c_2 (x^3+1)$$
Of course this method "with guess" is not always succesfull. By chance, it was the case this time !
