Find the indicial equation How do i find the indicial equation to $x^2y''+4xy'+4y=12-12x^2$
I used the frobenius method but got stuck:
$x^2y''+4xy'+4y-12+12x^2=0$
$x^2(y''+12)+4xy'+4y-12=0$
$y''+12+\frac{4y'}{x}+\frac{4y}{x^2}-\frac{12}{x^2}=0$
$y''+\frac{4y'}{x}+\frac{4y}{x^2}+\frac{12x^2-12}{x^2}=0$
let
$y=\sum_{k=0}^{\infty}a_kx^{k+r}$
$y'=\sum_{k=0}^{\infty}(k+r)a_kx^{k+r-1}$
$y''=\sum_{k=0}^{\infty}(k+r)(k+r-1)a_kx^{k+r-2}$
then
$\sum_{k=0}^{\infty}(k+r)(k+r-1)a_kx^{k+r-2}+\sum_{k=0}^{\infty}4(k+r)a_kx^{k+r-2}+$
$\sum_{k=0}^{\infty}4a_kx^{k+r-2}+\frac{12x^2-12}{x^2}=0$
how do i find the co-efficients if they don't depend on each other?
 A: This is an inhomogeneous Euler differential equation. One possible textbook procedure is to solve first the associated homogeneous equation
$$x^2y''+4xy'+4y=0\ .\tag{1}$$
The "Ansatz" $y(x)=x^\alpha$ $\>(x>0)$ with a complex parameter $\alpha$ to be determined is successful if
$$x^2\alpha(\alpha-1)x^{\alpha-2}+4x\alpha x^{\alpha-1}+4x^\alpha\equiv0\ ,$$
and leads to the indicial equation
$$\alpha^2+3\alpha+4=0\ .$$
From its solutions
$$\alpha_{1,2}=-{3\over2}\pm{i\over2}\sqrt{7}$$
we conclude that the general real solution of $(1)$ is given by 
$$y_h(x)=x^{-3/2}\biggl(A\cos\bigl({\sqrt{7}\over2}\log x\bigr)+B\sin\bigl({\sqrt{7}\over2}\log x\bigr)\biggr)\qquad(x>0)\ .$$
In order to obtain a particular solution of the given inhomogeneous equation $(*)$ we introduce the "Ansatz"
$$y_p(x):=ax^2+bx+c$$
with undetermined coefficients $a$, $b$, $c$ into $(*)$ and compare coefficients. I found that
$$y_p(x)=-{6\over7}x^2+3$$
does the job. The general real solution of $(*)$ is then given by $$y(x):=y_h(x)+y_p(x)\ .$$
