Solving second order differential equation - Important for thesis Thank you so much in advance for any help that you might give me. I'm stuck and can't solve the following equation
$y(x)+C_1y'(x) x+C_2 y''(x)x²=0$
Where $C_1$ and $C_2$ are constants. 
More importantly, 
$-r y(x)+\theta (\mu - x)y'(x)+\frac{\sigma ²}{2}y''(x)x²=0$
Any help?

Looking at the first equation:
$0
=y(x)+cy'(x) x+d y''(x)x^2
$.
Let
$y(x)
=\sum_{n=0}^{\infty} a_nx^n
$.
$y'(x)
=\sum_{n=1}^{\infty} na_nx^{n-1}
$
so
$cxy'(x)
=\sum_{n=1}^{\infty} cna_nx^{n}
$.
$y''(x)
=\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}
$
so
$dx^2y''(x)
=\sum_{n=2}^{\infty} dn(n-1)a_nx^{n}
$.
Therefore
$\begin{array}\\
0
&=y(x)+cy'(x) x+d y''(x)x^2\\
&=a_0+a_1x+ca_1x +\sum_{n=2}^{\infty}a_nx^n(1+cn+dn(n-1))\\
\end{array}
$
Therefore, no solution of this type.
If $y = ax^b$,
$cxy' = abcx^{b}$
and
$dx^2y'' = ab(b-1)dx^{b}$,
so
$0
=x^b(a+abc+ab(b-1)d)
=ax^b(1+bc+b(b-1)d)
$
so
$\begin{array}\\
0
&=1+bc+b(b-1)d\\
&=1+bc-bd+b^2d\\
&=1+b(c-d)+b^2d\\
\end{array}
$
Therefore,
if $d \ne 0$,
$\begin{array}\\
b
&=\frac{d-c\pm\sqrt{(c-d)^2-4d}}{2d}\\
&=\frac{d-c\pm\sqrt{c^2-6cd+d^2}}{2d}\\
&=\frac{1-r\pm\sqrt{r^2-6r+1}}{2}
\qquad\text{where }r = \frac{c}{d}\\
\end{array}
$
If $d = 0$,
$b = -\frac1{c}
$.
