# Thread: Solve differential equation using "Power Series"

1. ## Solve differential equation using "Power Series"

How can I solve the following differential equation using power series?

(1 – x^2)y'' – 6xy' – 4y = 0

I know there are other ways to solve this, but I specifically need to know how to do these using power series.

2. This is just a guess, but maybe rewriting the DEQ as
$\displaystyle y'' - \frac{6x}{1-x^2} y' - \frac{4}{1-x^2} y = 0$
Rewrite the fractions as a power series, and then solve?

3. Well, maybe. Most likely what is intended is to assume a solution of the form

$\displaystyle \displaystyle{y=\sum_{j=0}^{\infty}a_{j}x^{j}},$ plug that into the DE, and turn the crank. Am I right, EmilyL?

4. That sound more like it.

5. Originally Posted by Ackbeet
Well, maybe. Most likely what is intended is to assume a solution of the form

$\displaystyle \displaystyle{y=\sum_{j=0}^{\infty}a_{j}x^{j}},$ plug that into the DE, and turn the crank. Am I right, EmilyL?
Yes, exactly! Instead of "a" and "j", she used "c" and "n" respectively, but I don't think those variables really matter. Can you show me how to go about doing this?

6. Sure. What happens when you plug this expression into the DE?

7. What are you using to type out the equations so neatly? Do you need to know the code, or is there something that does it for you?

For plugging into the DE, I'm not sure what exactly to plug in. For the y, I think its what you showed in the previous post. For the y', it is ncx^(n-1) and for y'' it is n(n-1)cx^(n-1), right? Other than that, is there any more plugging in / substituting to do?

8. Right. In order to type up equations in LaTeX, which is what all those nice-looking equations are in, you have to double-click the Reply to Thread option, or click the Go Advanced button after single-clicking Reply to Thread. That brings you to a new screen, where you can click the button that looks like TeX, close to the far right. That will produce a math environment starting and ending brackets. What you type in-between those brackets gets interpreted as LaTeX code. The best way to learn LaTeX is by doing and observing. On this forum, you can double-click formulas to see how somebody typed them. Another little trick: after double-clicking to see LaTeX source code, you can copy and paste the code for yourself. Saves oodles of time!

So, you've got your original DE thus:

$\displaystyle (1-x^{2})\,y''(x)-6\,x\,y'(x)-4\,y(x)=0,$

as well as your series ansatz (ansatz is a terrific German word often used in the context of DE's. It means "your original guess" or "working hypothesis" for the purposes of computation) as follows:

$\displaystyle \displaystyle{y(x)=\sum_{n=0}^{\infty}c_{n}x^{n}},$ to use your teacher's notation.

Plugging the ansatz into the DE is close to what you said, but not quite. You have

$\displaystyle \displaystyle{y'(x)=\sum_{n=0}^{\infty}nc_{n}x^{n-1}},$ as you said, but

$\displaystyle \displaystyle{y''(x)=\sum_{n=0}^{\infty}n(n-1)c_{n}x^{n-2}}.$

Now these series don't really start at n=0 for the derivatives, do they? The $\displaystyle n$ and $\displaystyle n-1$ multiplying stuff changes it to the following:

$\displaystyle \displaystyle{y'(x)=\sum_{n=1}^{\infty}nc_{n}x^{n-1}},$ and

$\displaystyle \displaystyle{y''(x)=\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2}}.$

So plugging that into the DE produces the following:

$\displaystyle \displaystyle{(1-x^{2})\,\sum_{n=2}^{\infty}n(n-1)c_{n}x^{n-2}-6\,x\,\sum_{n=1}^{\infty}nc_{n}x^{n-1}-4\,\sum_{n=0}^{\infty}c_{n}x^{n}=0}.$

What happens next?

9. I'm not sure. The example in my book indicates that I would substitute k=n-2, k=n-1, and k=n for each series respectively. However, I'm not sure why that is done and what that achieves

10. The reason that's the next step is that what you want to do is add all the series together, gather like terms, etc. You can't do that if the series don't all start at the same beginning value. Each series, currently, isn't on the same page with everyone else. So, what do you get when you make those substitutions?

11. $\displaystyle \displaystyle{(1-x^{2})\,\sum_{k=0}^{\infty}(k+2)(K+1)c_{k+2}x^{k}-6\,x\,\sum_{k=0}^{\infty}nc_{k+1}x^{k}-4\,\sum_{k=0}^{\infty}c_{k}x^{k}=0}.$

right?

I'm not sure how you are allowed to substitute different values for n into the same equation. How can one variable have different values within the same equation?

12. I think you're confused about the shifting. Take the first series: if k = n - 2, then n = k + 2. Just plug that in. What do you get?

13. you're right, I edited my previous post, now getting k=0 for the starting point of each sum

14. Close, very close. I'd clean it up just a bit more:

$\displaystyle \displaystyle{(1-x^{2})\,\sum_{k=0}^{\infty}(k+2)(k+1)c_{k+2}x^{k}-6\,x\,\sum_{k=0}^{\infty}(k+1)c_{k+1}x^{k}-4\,\sum_{k=0}^{\infty}c_{k}x^{k}=0}.$

Now what?

15. Now you can put them all together since the boundaries are the same, right?

$\displaystyle \displaystyle{(1-x^{2})(-6X)(-4)\,\sum_{k=0}^{\infty}(k+2)(k+1)c_{k+2}x^{k}\,+(k +1)c_{k+1}x^{k}\,+c_{k}x^{k}=0}.$

Not sure if I dealt with the coefficients right.

Also, can I get rid of the "sum" sign now?

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# solve the equation near x=0 (1-x^2)y''-6xy'-4y=0

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