# Solve Diff. Eq. using power series

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1. Mar 19, 2017

### JamesonS

1. The problem statement, all variables and given/known data

(1-x)y^{"}+y = 0

I am here but do not understand how to combine the two summations:
Mod note: Fixed LaTeX in following equation.
$$(1-x)\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n+\sum_{n=0}^{\infty}a_nx^n = 0$$

Last edited by a moderator: Apr 1, 2017
2. Mar 19, 2017

### Ray Vickson

Here is your equation using PF-compatible TeX:
$$(1-x)\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2} x^n+\sum_{n=0}^{\infty}a_nx^n = 0$$
Just replace the "\ begin {equation} ... \ end {equation} " by " ...  " (no spaces between the initial and final \$ signs). Also: write "\infty", not "\infinity".

As for your question: write out the first 3 or 4 terms, to see what you get. That will give you insight into what you should do next.

Last edited by a moderator: Apr 1, 2017
3. Mar 19, 2017

### JamesonS

Thanks for the response and Latex help. Writing out the first few terms of each sum:
$$(1-x)[2a_2+6a_3x+12a_4x^2+...]+[a_0+a_1x+a_2x^2+...]$$
I am not sure what to do with the (1-x) term outside the first sum...

Last edited: Mar 19, 2017
4. Mar 19, 2017

### Ray Vickson

What is preventing you from "distributing out" the product? That is, $(1-x) P(x) = P(x) - x P(x).$

Last edited: Mar 20, 2017
5. Mar 31, 2017

### MidgetDwarf

Been a while since I did DE, but doesn't the OP have to be aware of the singular point in this problem? Hence he has to use the method of Frobenius?

6. Apr 1, 2017

### LCKurtz

But the singular point isn't at $x=0$.

7. Apr 2, 2017

### MidgetDwarf

Thanks for the correction. It's been a while. I remembered that there is no singular point if we take the Taylor expansion about x=0? Correct?

8. Apr 2, 2017

### epenguin

Practically after your first equation, or at any later stage, just multiply it out. You have shown that you know how to write a sum of different powers of x in terms of xn by changing the subscript appropriately.

9. Apr 2, 2017

### vela

Staff Emeritus
That's backwards. If you expand about a regular point, then the solution can be written as a Taylor series. If there's a singular point, then you can use the method of Frobenius and end up with a Laurent series.

10. Apr 2, 2017

### MidgetDwarf

Thanks Vela!

It has been a while. I may pop open a differentials to bring the memory back.