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Mathematics 16 Online
OpenStudy (anonymous):

Use power series to solve (1+x^2)y''+xy'+(3x+2)y=0

OpenStudy (tkhunny):

Well, are you expecting someone else to drag through that for you? Let's see where you started, at least.

OpenStudy (ikram002p):

well maybe he/she dont know the steps :O

OpenStudy (ikram002p):

@keynote start by dividing term on leading coefficient

OpenStudy (anonymous):

I got into the substitution and am stuck there

OpenStudy (ikram002p):

oh then type what u got so far

OpenStudy (tkhunny):

See, this is what I am talking about. Let's see your work so a wonderful, kind, and giving volunteer doesn't have to do it for you. Very good.

OpenStudy (anonymous):

\[\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}+\sum_{n=2}^{\infty} n(n-1)a_nx^{n}+\sum_{n=1}^{\infty} na_nx^{n}+\sum_{n=0}^{\infty} 3a_nx^{n+1}+\sum_{n=0}^{\infty} 2a_nx^{n}+\]

OpenStudy (anonymous):

That's what i have so far

OpenStudy (tkhunny):

That's excellent. Now, you just have to change your indeces so they match up. You can't have one start at 0, one at 1 and the other at 2. Mind you, I can't really tell if that is right, since it is cut off by the screen. It does appear to me that you have the right idea.

OpenStudy (anonymous):

So how do i change the indeces. Thats where my problem is

OpenStudy (tkhunny):

Very carefully... \(\sum\limits_{n=1}^{\infty}na_{n}x^{n} = \sum\limits_{n=0}^{\infty}(n+1)a_{n+1}x^{n+1}\) That's about it.

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