Series solution for $x^2y''-x(x+6)y'+10y=0$ I have to solve this differential  equation:
$$x^2y''-x(x+6)y'+10y=0$$
by using this method and I am stuck at this step.
Please help me to solve it.
Here is my attempt:
https://i.stack.imgur.com/XglDm.jpg
 A: The given differential equation is a secondorder linear ordinary differential equation. We are required to find a series solution to this problem.
Step 1: Assume that an infinite series solution exists such that
$$
y = \sum_{n = 0}^{\infty}c_{n}x^{n}
$$
Step 2: Calculate $y''$, $y'$:
$$
y' = \sum_{n=1}^{\infty}nc_{n}x^{n-1}\\
y'' = \sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2}
$$
Step 3: Substitute $y''$, $y'$, and $y$ in the given differential equation:
$$
\begin{align}
&x^2y''-x(x+6)y'+10y&=0\\
\implies &x^2(\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2}) - x(x+6)\sum_{n=1}^{\infty}nc_{n}x^{n-1} + 10\sum_{n = 0}^{\infty}c_{n}x^{n} &= 0\\
\implies &\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n} - (x+6)\sum_{n=1}^{\infty}nc_{n}x^{n} + 10\sum_{n = 0}^{\infty}c_{n}x^{n} &= 0\\
\implies &\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n} - x\sum_{n=1}^{\infty}nc_{n}x^{n} -6\sum_{n=1}^{\infty}nc_{n}x^{n} + 10\sum_{n = 0}^{\infty}c_{n}x^{n} &= 0\\
\implies &\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n} - \sum_{n=1}^{\infty}nc_{n}x^{n+1} -\sum_{n=1}^{\infty}6nc_{n}x^{n} + \sum_{n = 0}^{\infty}10c_{n}x^{n} &= 0
\end{align}
$$
All terms, except one, are in terms of $x^n$:
$$
\sum_{n=0}^{\infty}nc_{n}x^{n+1} = \sum_{n=1}^{\infty}(n-1)c_{n-1}x^{n} 
$$
It would be nice if we could take the same limits for all the sums. We are in luck because we can. Upon changing all lower limits to $n=1$:
$$
\begin{align}
\implies &\sum_{n=1}^{\infty}n(n-1)c_{n}x^{n} - \sum_{n=1}^{\infty}nc_{n}x^{n+1} -\sum_{n=1}^{\infty}6nc_{n}x^{n} + \sum_{n=1}^{\infty}10c_{n}x^{n} +10c_{0} &= 0\\
\implies &\sum_{n=1}^{\infty}\left[ n(n-1)c_{n}x^{n} - (n-1)c_{n-1}x^{n} -6nc_{n}x^{n} + 10c_{n}x^{n}\right] +10c_{0}&= 0\\
\implies &\sum_{n=1}^{\infty}\left[ n(n-1)c_{n} - (n-1)c_{n-1} -6nc_{n} + 10c_{n}\right]x^{n} +10c_{0}&= 0
\end{align}
$$
We are required to make an assumption that $c_{0}=0$:
$$
\begin{align}
\implies &\sum_{n=1}^{\infty}\left[ n(n-1)c_{n} - (n-1)c_{n-1} -6nc_{n} + 10c_{n}\right]x^{n} &= 0\\
\implies & n(n-1)c_{n} - (n-1)c_{n-1} -6nc_{n} + 10c_{n} &= 0\\
\implies & (n(n-1)  -6n + 10)c_{n} = (n-1)c_{n-1}&\\
\implies & c_{n} = \dfrac{n-1}{n(n-1)  -6n + 10}c_{n-1}&n>1\\
\end{align}
$$
