## Series solutions to ODEs

Hey everyone. I'm trying to refresh myself of solving linear ODEs. For simplicity's sake, I began by trying to solve
$$xy'=xy+y$$
This is actually a separable ODE, and the solution is $$y = c_{1}xe^{x}$$. I am attempting to derive the same result from a series solution.
First, rewrite this as a homogeneous equation:
$$xy' - xy - y = 0$$
Then, substitute for y and y':
$$x \sum_{n=1}^{\infty}na_{n}x^{n-1} - x \sum_{n=0}^{\infty}a_{n}x^{n} - \sum_{n=0}^{\infty}a_{n}x^{n}=0$$
Distribute the x terms into the series:
$$\sum_{n=1}^{\infty}na_{n}x^{n} - \sum_{n=0}^{\infty}a_{n}x^{n+1} - \sum_{n=0}^{\infty}a_{n}x^{n}=0$$
Powercounting, we see that all but the last term begin at $$x^{1}$$. So we take out one term from the last one to obtain
$$\sum_{n=1}^{\infty}na_{n}x^{n} - \sum_{n=0}^{\infty}a_{n}x^{n+1} - \sum_{n=1}^{\infty}a_{n}x^{n} - a_{0}=0$$
And I believe this effectively implies that $$a_{0}=0$$. Moving on, the powers are now correct. Now for a change of index. For the first and last series terms, $$n = k$$. For the second, $$k = n + 1$$.
$$\sum_{k=1}^{\infty}\left(ka_{k} - a_{k-1} - a_{k}\right)x^{k} = 0$$
So this implies that
$$ka_{k} - a_{k-1} - a_{k}= 0$$
And from here we obtain the recurring relation
$$a_{k} = \frac{a_{k-1}}{k-1}, k = 1, 2, 3,...$$
My issue with this answer is that it fails at k = 1 (because the denominator would be zero). And even if it weren't, all subsequent elements in the series would be zero.

Where did I go wrong?
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 The problem is you are dividing by zero, silly! By dividing by (k-1) you are implicitly assuming that's not = 0. Thus, handle that case separately, so you obtain a_0 = 0. Now, with induction, it shouldn't be hard to show $$a_k = \frac{a_1}{(k-1)!} , k = 2,3,4,5...$$ You might be wondering, "What about a_1?" I'll let you figure that part out.
 Ahh ok. Your first sentence explains it. Many thanks!

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