Series solution for $x^2y''-x(x+6)y'+10y=0$

Question

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

Answer 1

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 $c_{0}=0$: $$ \begin{align} \implies &\sum_{n=1}^{\infty}\left[ n(n-1)c_{n} - (n-1)c_{n-1} -6nc_{n} + 10c_{n}\right]x^{n} &= 0\\ \implies & n(n-1)c_{n} - (n-1)c_{n-1} -6nc_{n} + 10c_{n} &= 0\\ \implies & (n(n-1) -6n + 10)c_{n} = (n-1)c_{n-1}&\\ \implies & c_{n} = \dfrac{n-1}{n(n-1) -6n + 10}c_{n-1}&n>1\\ \end{align} $$