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.


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*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: $y_p=ax^2+b, y_p=\frac{a}{x^2}+b, y_p=ax^6+b$

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 $a$.
