How to solve a homogeneous 2nd order linear DE? I want to solve this ODE:
Given $y=x^2$ is a solution to $x^2y''+2xy'-6y=0$ find the general solution:
The answer for the general solution is: $y=Ax^2+B/x^3$
What method do I need to employ to solve this? If it had constant coefficients instead of variables of x I could just treat it as a quadratic, find the roots and then write the appropriate general solution.
What am I supposed to do in this case and what is the name of the method used to solve such an ODE (if any)? e.g. reduction of order etc.
Also, can a full solution be found by hand if it was nonhomogeneous? 
For $x^2y''+2xy'-6y=10x$ can I still use variation of parameters or method of undetermined coefficients to solve for the particular solution?
Please help me, I'm melting. :(
 A: There's a general procedure called reduction of order; it supposes that we can write a second solution as a multiple of the first, so that 
$$y_2 = x^2 v$$
for a particular function $v$. The reason we do this is that
$$y_2' = x^2 v' + 2x v \quad y_2'' = x^2 v'' + 4x v' + 2v$$
Substituting this into the original equation yields
\begin{align*}
0 &= x^2(x^2 v'' + 4xv' + 2v) + 2x (x^2 v' + 2x v) -6 x^2 v \\
&= x^4 v'' + 4x^3 v' + 2x^2 v + 2x^3 v' + 4x^2 v - 6x^2 v \\
&= x^4 v'' + 6x^3 v'
\end{align*}
Now this leads to the essentially first order equation
$$v'' = -\frac{6}{x} v'$$
which can be solved by any number of techniques, e.g. separation of variables. In general, the technique of writing $y_2 = v y_1$ for a suitable function $v$ can be used rather often to carry this procedure out.

If the equation is non-homogeneous, then yes: You can find the equations to the homogeneous equation via these techniques, and then solve the non-homogeneous equation via undetermined coefficients, variation of parameters, or whatever method you please.
A: For $x = e^z$, $$\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}=e^{-z}\frac{dy}{dz}$$
Now $$\frac{d^2y}{dx^2}=\frac{d}{dz}(\frac{dy}{dx}).e^{-z}=e^{-z}\frac{d}{dz}(e^{-z}\frac{dy}{dz})=e^{-z}(e^{-z}\frac{d^2y}{dz^2}-e^{-z}\frac{dy}{dz})=e^{-2z}(\frac{d^2y}{dz^2}-\frac{dy}{dz})$$
Substituting this in your original equation, we will get :$$\frac{d^2y}{dz^2}-\frac{dy}{dz}+2\frac{dy}{dz}-6y=0$$.
Now this is a equation with constant coefficients and you know how to solve it. 
NOw for $$x^2y''+2xy'-6y=10x$$, the new equation becomes $$\frac{d^2y}{dz^2}+\frac{dy}{dz}-6y=10e^z$$ which is also solvable by method of undetermined coefficients or variation of parameters
