Find both solutions to $x^2(x^2+1)y''-2x^3y'+2(x^2-1)y=0$, given that one of the solutions is of the form $y_1 = x^n$ or $e^{ax}$.
I can solve $x^2y''+Cxy'+Dy=0$ equations, but dividing by $x^2$ in this case results in an equation which I'm not sure how to solve either.
Basically how to solve equations of the form $Ax^2y''+Bxy'+Cy=0$ where $A, B, C$ are functions of $x$.
This equation has no solutions of the given forms. I suspect you meant $$ x^2 (x^2 + 1) y'' - 2 x^3 y' + 2 (x^2 - 1) y = 0 $$
To find $n$, substitute $y = x^n$ into the differential equation and see what has to be true in order for this to simplify to $0 = 0$.
To find the other solution, once you know what $n$ is, substitute $y = x^n g$ and you'll get a simpler equation for $g$ (the method is called Reduction of Order).