# differential equation using series expansion

Trying to solve xy'= xy + y using the series y(x) = $\sum\limits_{i=0}^\infty a_nx^n$ This is what i have so far.

y'(x)= $\sum\limits_{i=0}^\infty na_nx^{n-1}$

xy' - xy - y = 0

x $\sum\limits_{i=0}^\infty na_nx^{n-1}$ - x $\sum\limits_{i=0}^\infty a_nx^n$ - $\sum\limits_{i=0}^\infty a_nx^n$ = 0

$\sum\limits_{i=0}^\infty na_nx^{n}$ - $\sum\limits_{i=0}^\infty a_nx^{n+1}$ - $\sum\limits_{i=0}^\infty a_nx^n$ = 0

then i re-index the second term to get an $x^n$ term

$\sum\limits_{i=0}^\infty na_nx^{n}$ - $\sum\limits_{i=1}^\infty a_{n-1}x^{n}$ - $\sum\limits_{i=0}^\infty a_nx^n$ = 0

Now i'm trying to get a recursion relation but seem to be a little stuck here, I get from n=0 that

$a_0$ = 0

then for n = 1

$na_1$ - $a_0$ - $a_1$ = 0

I think I'm close, just a little confuse with the indexing and how to actually create the recursion relation to solve this. Any help would be greatly appreciated. Thanks!

## 1 Answer

I would say:

It is necessary to use the series $y=\sum\limits_{i=0}^\infty a_nx^{n+r},\,\,y'=\sum\limits_{i=0}^\infty (n+r)a_nx^{n+r-1}$ and determine the appropriate number r:

xy' - xy - y = 0

$\sum\limits_{i=0}^\infty (n+r)a_nx^{n+r}-\sum\limits_{i=0}^\infty a_nx^{n+r+1}-\sum\limits_{i=0}^\infty a_nx^{n+r}=0$

Then:

$(0+r)a_0-a_0=0 \Rightarrow a_0(r-1)=0 \Rightarrow r = 1$

$(n+r)a_n-a_{n-1}-a_n = 0 \Rightarrow a_n=\frac{a_{n-1}}{(n+r-1)_{r=1}}\Rightarrow a_n=\frac{a_{n-1}}{n},\,\,\, n \ge 1$

So:

$a_n=\frac{a_0}{n!}\Rightarrow y=\sum\limits_{i=0}^\infty \frac{a_0}{n!}x^{n+1}=a_0xe^x$