I am trying to solve this nonhomogeneous ODE:
$$y''-2x^2y'+4xy=x^2+2x+2$$
I know it's a power series, but when I get down to the very end, I end up with a $C_0$ term, a $C_1$ term, and a $C_2$ term. I am supposed to only have a $C_0$ and $C_1$ term. I cannot find any other ways to solve this, and Wolfram is not helping either.
Any ideas or suggestions?
So my final solution is $$y=c_0+Xc_1+X^2-\frac23X^3c_0-\frac16x^4c_1+0-\frac{2}{45}X^6c_0$$ ....etc
The $X^2$ term is messing me up, because i need this solution for $y_1(x)$ in terms of $c_0$ and $y_2(x)$ in terms of $c_1$..... so that extra $x^2$ is because I got $c_2=1$.
The power series $y(x) = \sum_{n=0}^\infty c_n x^n$ is determined by $c_0$ and $c_1$. Plugging it into the equation and comparing the free terms on both sides yields $$ 2c_2=2 $$ Hence $c_2 = 1$, as you got. You probably found it strange that this coefficient does not depend on $c_0, c_1$. But such is the nature of this nonhomogeneous equation: it forces every solution to have $y''(0)=2$, hence $c_2=1$.
The process continues: comparing the coefficients of $x^1$ on both sides yields $$ 6c_3+4c_0 = 2 $$ which gives $c_3$. Similarly, $c_4$ is obtained from the coefficients of $x^2$. If you just need a few terms of the series, it is reasonable to continue by hand as I did so far. If you need the whole series, there is a recurrence relation to be solved: you get $c_n$ from the coefficients of $x^{n-2}$, $$ n(n-1)c_n-2(n-3)c_{n-3}+4c_{n-3} = 0 \quad \text{(for $n>4$)} $$