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