1. ## Power Series Solution

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

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

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

Then I found the first and second derivatives.

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

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

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

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

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

$\displaystyle x^r$$\displaystyle [\sum_{n=0}^{\infty} 2(n+r)(n+r-1)a_nx^{n} + \displaystyle \sum_{n=0}^{\infty} (n+r)a_nx^{n+1} + \displaystyle \sum_{n=0}^{\infty} 3(n+r)a_nx^{n} -\displaystyle \sum_{n=0}^{\infty} a_nx^{n}] Then I tried to get \displaystyle a_0 and changed some of the n values to \displaystyle j and \displaystyle j+1 similar to what my professor did. \displaystyle x^r [(2r(r-1)+3r-1)a_0 + \displaystyle \sum_{n=1}^{\infty}$$\displaystyle 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:

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

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.

2. What equation?

4. Originally Posted by Kyeong
The given equation is $\displaystyle 2x^2y'' + (x^2 + 3x)y' - y = 0$
The point $\displaystyle 0$ is a regular singular point of the differencial equation. Note, $\displaystyle \lim_{x\to 0}x\cdot \frac{x^2+3x}{x^2} =3$ and $\displaystyle \lim_{x\to 0}x^2\cdot \frac{-1}{x^2} = -1$. This means the indicial equation is $\displaystyle 2k(k-1) + 3k - 1 = 0$.

5. 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.

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

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

Then I plug in the r values and set up all the functions $\displaystyle \frac {a_{n-1}} {a_n}$. 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 $\displaystyle r = \frac {1} {2}$
$\displaystyle a_j = \frac {-(j- \frac {1} {2})a_{j-1}} {2j^2 + 3j}$

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

Did I make a sign error or something?