solve differential equation by series

**szpengchao** · May 29th 2008, 10:17 AM

4zy''+2y'+y=0

find series solution about z=0 and identify these solution in closed form

**ThePerfectHacker** · May 29th 2008, 11:39 AM

Originally Posted by szpengchao

4xy''+2y'+y=0

find series solution about x=0 and identify these solution in closed form

If you divide through by x you get $4y'' + \frac{2}{x}y'+\frac{1}{x}y = 0$ . The coefficient function for $y'$ is Laurent series of order $1$ and same for $y$ . This means the equation is solvable through Frobenius method. The Frobenius equation would be $4k(k-1)+2k = 0$ so $k=0,2/3$ . These do not differ by integers so you can solve this equation on $(0,\infty)$ by letting $y = x^{k}\sum_{n=0}^{\infty} a_n x^n$ where $k=0,2/3$ . For each value you will get a new linearly independent solution.

**Moo** · May 29th 2008, 11:40 AM

Hello,

Originally Posted by szpengchao

4zy''+2y'+y=0

find series solution about z=0 and identify these solution in closed form

Let $y(z)=\sum_{n \ge 0} a_n z^n$

You will have to determine $a_n$ ~

$y'(z)=\sum_{n \ge 1} n a_n z^{n-1}$

$y''(z)=\sum_{n \ge 2} n(n-1) a_n z^{n-2}$

$\implies 4zy''(z)+2y'(z)+y(z)=0$

$4 \sum_{n \ge 2} n(n-1) a_n z^{n-1}+2 \sum_{n \ge 1} n a_n z^{n-1}+\sum_{n \ge 0} a_n z^n=0$

Putting all the powers to $z^n$ , by changing the indices ~

In the first and second ones, n-1 becomes n. Then the series will start at 1 and 0..

$4 \sum_{n \ge 1} n(n+1) a_{n+1} z^n+2 \sum_{n \ge 0} (n+1) a_{n +1} z^n+\sum_{n \ge 0} a_n z^n=0$

Now, starting the series at the same point :

$4 \sum_{n \ge 1} n(n+1) a_{n+1} z^n+2 \sum_{n \ge {\color{red}1}} (n+1) a_{n+1} z^n+{\color{red}a_1}+\sum_{n \ge {\color{red}1}} a_n z^n+{\color{red}a_0}=0$

Therefore :

$a_0+a_1+\sum_{n \ge 1} \left[4n(n+1)a_{n+1}+2(n+1)a_{n+1}+a_n\right] z^n=0$

--> $a_0+a_1=0$

And :

$4n(n+1)a_{n+1}+2(n+1)a_{n+1}+a_n=0$

Find the induction formula

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