- Thread starter EmilyL
- Start date

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?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?

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?

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?