Solving the differential equation $x^3y''+2x^2y-6xy = 0$ First question on here, so I hope I'm doing this right.
I've been reading up on differential equations lately and have now stumbled upon one that I have no idea how to solve.
$x^3y''+2x^2y-6xy = 0$ 
$y(1)=2$ 
$y'(1)=-1$
I've gotten fairly confident with differential equations of the type $ay'' + by' + cy = f(x)$, but I can't figure out how to work with this expression.
If someone could point me into the right direction I would be very thankful! I don't have much of an attempted solution to post as I'm completely lost on this one. I would normally start by looking for roots to create the homogenous solution, but I imagine that I have to separate all the occurrences of x to the other side before I can do that. Without those, I would start by finding the roots with $p(r)=r^2 + 2 - 6$ leading to $r= \sqrt{4}$. I imagine I need to do something entirely different for this one though.
 A: $\require{cancel}$
Probably the OP did a typo and the ODE is 
$$x^3y''+2x^2y'-6xy=0$$
First we look for a solution of the ODE on the form $x^n$ so we find
$$n(n-1)x^{n+1}+2nx^{n+1}-6x^{n+1}=0\iff n^2+n-6=0$$
so $n=2$ and then $y:x\mapsto \lambda x^2$ is a solution. Now let the function $z$ such that $y(x)=z(x)x^2$ so $y'=z'x^2+2xz$ and $y''=z''x^2+4xz'+2z$ hence we find
$$x^5z''+4x^4z'+\cancel{2x^3z}+2x^4z'+\cancel{4x^3z}-\cancel{6x^3z}=0\iff z''+6xz'=0$$
Now let $u=z'$ so the last ODE becomes $u'+6xu=0$ and then $ u=Ce^{-3x^2}$ so $z=C_1\int e^{-3x^2}dx+C_2$ and finally
$$y(x)=C_1x^2\int e^{-3x^2}dx+C_2x^2$$
Edited If the ODE is
$$x^3y''+(2x^2-6x)y=0$$
so let's find a solution expanded on power series
$$y(x)=\sum_{n=0}^\infty a_n x^n$$
hence we find
$$\sum_{n=2}^\infty n(n-1)a_nx^{n+1}+2\sum_{n=0}^\infty a_n x^{n+2}-6\sum_{n=0}^\infty a_n x^{n+1}=0\tag 1$$
and by changing the index
$$\sum_{n=3}^\infty (n-2)(n-1)a_{n-1}x^{n}+2\sum_{n=2}^\infty a_{n-2} x^{n}-6\sum_{n=1}^\infty a_{n-1} x^{n}=0$$
so$$(n^2-3n-4)a_{n-1}+2a_{n-2}=0\iff a_{n-1}=-\frac{2}{(n+1)(n-4)}a_{n-2}$$
hence
$$a_{n}=-\frac{2}{(n+2)(n-3)}a_{n-1}\tag2$$
we can find from $(1)$ the value of $a_0,a_1,a_2$ and $a_3$ and we express by induction from $(2)$ $a_n$ as function of $n$ for $n\ge4$.
A: Your equation is $$x^3y''+(2x^2-6x)y = 0, \quad \text{or}$$ $$y''+\left(\frac{2}{x}-\frac{6}{x^2}\right)y = 0 \tag{1}$$
This is a type of Bessel Differential Equation.
We will derive a general result by assuming a solution of the form: $$y = x^\gamma[C_1J_n(\alpha x^{\eta})]$$
and then create a second order differential equation that this $y$ solves. This technique is called a transformation of Bessel's equation. After some computation, we find that $y$ solves:
$$y'' - \dfrac{2\gamma-1}{x}y' + \left(\alpha^2\eta^2x^{2\eta-2}+\dfrac{\gamma^2-n^2\eta^2}{x^2}\right)y = 0 \tag{2}$$
The solution to this (for integer $n$) is $$y = x^{\gamma}[C_1J_n(\alpha x^{\eta})+C_2Y_n(\alpha x^{\eta})]. \tag{3}$$
The next step is to make $(2)$ look like your differential equation. To find your equation simply set $\gamma = \frac{1}{2}$, $\eta = \frac{1}{2}$, $\alpha = 2\sqrt{2}$ and $n = 5$.
Using this, the solution to your equation is 
$$y = \sqrt{x}[C_1J_{5}(2\sqrt{2x}) + C_2Y_{5}(2\sqrt{2x})].$$
All that is left is to find the constants.
