Power Series Solution

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July 9th 2008, 09:09 PM
Kyeong

Power Series Solution

The given equation is $2x^2y'' + (x^2 + 3x)y' - y = 0$

1. Find two linearly independent solutions of the form $y = (x^r)\sum_{n=0}^{\infty} a_nx^{n}$ . Using the recurrence relation.

I changed the given y to $y = \sum_{n=0}^{\infty} a_nx^{n+r}$

Then I found the first and second derivatives.

$y' = \sum_{n=0}^{\infty} (n+r)a_nx^{n+r-1}$

$y'' = \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r-2}$

When I plug it in back into the original equation, I get:

$2 \sum_{n=0}^{\infty} (n+r)(n+r-1)a_nx^{n+r}$ + $\sum_{n=0}^{\infty} (n+r)a_nx^{n+r+1}$ + $3\sum_{n=0}^{\infty} (n+r)a_nx^{n+r}$ - $\sum_{n=0}^{\infty} a_nx^{n+r}$

Ok, I understand up to this part. Now I'm just confused after this, but this is what I got. I took out $x^r$

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

Then I tried to get $a_0$ and changed some of the n values to $j$ and $j+1$ similar to what my professor did.

$x^r [(2r(r-1)+3r-1)a_0$ + $\sum_{n=1}^{\infty}$ $2(j+r)(j+r-1)a_j + 3(j+r)a_j - a_j - (j-1+r)a_{j-1}]x^j$

Then I tried to find the r by:

$2r^2 +r - 1 = 0$
$(2r-1)(r+1)$

I got r = 1/2 and r = -1.

Is this correct so far? When my professor went over a homework question similar to this problem, he said that the r's would be -1 and -1/2. I just don't see where the -1/2 came from and even so. When I plug -1 , I get zero in the third term which ruins it unless that's ok?

I know what to do pass this part, but it's very confusing. I spent more than 5 hours trying to do this and I know I'm doing something wrong or thinking of something else. Please help me lead to the right direction.
July 9th 2008, 09:17 PM
ThePerfectHacker

What equation?
July 9th 2008, 09:25 PM
Kyeong

Sorry, just added it.
July 9th 2008, 09:33 PM
ThePerfectHacker

Quote:

Originally Posted by Kyeong

The given equation is $2x^2y'' + (x^2 + 3x)y' - y = 0$

The point $0$ is a regular singular point of the differencial equation. Note, $\lim_{x\to 0}x\cdot \frac{x^2+3x}{x^2} =3$ and $\lim_{x\to 0}x^2\cdot \frac{-1}{x^2} = -1$ . This means the indicial equation is $2k(k-1) + 3k - 1 = 0$ .
July 9th 2008, 09:52 PM
Kyeong

Well, I believe I already got the indicial equation. It's near the end of the post, just in terms of r. First part of that long function.

$x^r [(2r(r-1)+3r-1)a_0$ ..........

I got $r_1 = \frac {1} {2}$ and $r_2 = -1$ .

Then I plug in the r values and set up all the functions $\frac {a_{n-1}} {a_n}$ . I did this for each r value and when r = -1, the third term is zero and ruins it and can't get the fourth non-zero term.

Maybe I'm looking at this a different way?

when $r = \frac {1} {2}$
$a_j = \frac {-(j- \frac {1} {2})a_{j-1}} {2j^2 + 3j}$

when $r = 1$
$a_j = \frac {-(j- 2)a_{j-1}} {2j^2 - 3j}$

Did I make a sign error or something?