I'm having trouble using the Frobenius Method to solve $x^2 \frac{d^2 y}{dx^2}-2x \frac{dy}{dx}+2y=0$

Question

I've successfully found the first solution ($y_1=c_1 x^2$) using the Frobenius Method, but when it comes to finding the other solution ($y_2=c_2 x$), I get an recurring relation on the summation that reduces to $[n(n-1)]c_n$. Since in this case we do have an integer $n \geq 1$ that can make the relation zero, we can't just say $c_n=0$. How do I get around this block?

Answer 1

$$ y=\sum c_nx^{k+n}\implies \left\{\begin{aligned} 0&=(n+k)(n+k-1)c_{n}-2(n+k)c_{n}+2c_n\\ &=(n+k-2)(n+k-1)c_{n} \end{aligned}\right. $$ The indicial equation for $n=0$ has solutions $k=1$ and $k=2$. The case $k=2$ is contained in the case $k=1$ which then implies $c_n=0$ for $n\ge2$, $c_0,c_1$ are free leading to $y=c_0x^{1+0}+c_1x^{1+1}$.

Answer 2

$$y=\sum_{n=0}^{\infty}c_nx^n$$ $$y'=\sum_{n=0}^{\infty}c_n nx^{n-1}$$ $$y''=\sum_{n=0}^{\infty}c_n n(n-1)x^{n-2}$$ $$x^2y''-2xy'+2y=0=\sum_{n=0}^{\infty}\left(n(n-1)-2n+2 \right)c_nx^n$$ $$\sum_{n=0}^{\infty}(n-2)(n-1)c_nx^n=0$$ This has to be true any $x$. Thus, for any $n\neq 1$ and $n\neq 2$ this implies $c_n=0$.

For $n=1$ and $n=2$ the terms $(n-2)(n-1)c_nx^n$ are nul any $x$. So, $c_1$ and $c_2$ can be any constants. The only two remaining terms are : $$y=c_1x+c_2x^2 \qquad \text{any } c_1 , c_2$$