Trying to solve for the general form of the linear differential equation power series solution...

Question

kainui · Answer

First off, the differential equation I want to solve is:

$$\sum_{i=0}^\infty f_i(x)\frac{d^i y}{dx^i} = 0$$
with:
$$f_i(x) = \sum_{k_i=0}^\infty b_{k_i} x^{k_i}$$
and use the solution:

$$y=\sum_{n=0}^\infty a_nx^n$$

I'll show my progress so far haha...

kainui · Answer

So first off I write the derivatives of y,

$$\frac{d^i y}{dx^i} = \sum_{n=0}^\infty a_{n+i} \frac{(n+i)!}{i!} x^n$$

Then I go ahead and note that this might be useful too:

$$x^n f_i(x) = \sum_{k_i=n}^\infty b_{k_i-n} x^{k_i}$$ 

and then toss this into the differential equation one by one...

$$\sum_{i=0}^\infty \sum_{n=0}^\infty f_i(x) a_{n+i} \frac{(n+i)!}{i!} x^n =0$$

$$\sum_{i=0}^\infty \sum_{n=0}^\infty \sum_{k_i=n}^\infty a_{n+i} \frac{(n+i)!}{i!}b_{k_i-n} x^{k_i}=0$$
And here I think I am a little stuck since I need to now factor out all the terms with the same coefficient on $x^n$ and set those equal to zero to get the "general recurrence relations".

kainui · Answer

It's sorta tough to see what's going on here, since what I'm doing is so scary lol

kainui · Answer

I'm pretty terse too, but I didn't want to confuse what I'm saying too much with way too many symbols unneeded flying around, cause how I derive the general form of the derivatives and shift the terms around is sorta details. #_#

kainui · Answer

Aha I think I found it, just pulling out those terms yields these recurrence relations:

$$\sum_{i=0}^O \sum_{n=0}^{k_i} a_{n+i} b_{k_i-n} \frac{(n+i)!}{i!} = 0$$

Solved!

I put the O instead of infinity because it represents the Order of the differntial equation. First order O=1, second order O=2, etc...

I just gotta test that it works... wew....

math_genius12345 · Answer

I can't believe you solved that whole thing by yourself! Kudos

legomyego180 · Answer

This is going to give me nightmares.
-A Calc II student

reemii · Answer

In the definition of $f_i$, the $k_i$ ranges through what set of values?

kainui · Answer

$k_i$ ranges through 0 to infinity, it's basically because I have this differential equation:

$$0=f_0(x)y(x)+f_1(x)y'(x)+f_2(x)y''(x)+\cdots$$
thought it's intended to be truncated at some smaller value so that it converges, the form is just here to carry through so it can just be plugged into to solve I guess. I kind of forget already since I just made this and forgot about it.