Series Solutions to Linear ODEs for Refreshing Your Skills

[arenaninja]

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?

 

[l'Hôpital]

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
<br /> a_k = \frac{a_1}{(k-1)!} , k = 2,3,4,5...<br />

You might be wondering, "What about a_1?" I'll let you figure that part out.

 

[arenaninja]

Ahh ok. Your first sentence explains it.

Many thanks!