Finding Recurrence relation for Series Solution of Differential Equation

**Jason Bourne** · October 24th 2008, 08:57 AM

I'm trying to solve the following differential equation with non constant coefficients:

$2xy'' + y' + xy = 0$

I'm totally confused about finding the recurrence relation when solving this when using the method of Frobenius about $x_0 = 0$ . I know it's a regular singular point and all that but I'm just rubbish at finding the recurrence relation, apparently the right answer is

$a_n = - \frac{a_{n-2}}{(n+r)(2(n+r) -1)}$ where r is one of the roots of the indicial equation

Can anyone help me get unconfused about finding the recurrence relation as I'm rubbish at it.

**ThePerfectHacker** · October 24th 2008, 10:27 AM

Originally Posted by Jason Bourne

I'm trying to solve the following differential equation with non constant coefficients:

$2xy'' + y' + xy = 0$ for $x>0$ .

The point $0$ is a regular singular point.
Therefore, solutions shall have the form $y= x^k \sum_{n=0}^{\infty} a_n x^n$ .

The equation associated with finding $k$ is: $2k(k-1)+k=0\implies k = 0 \text{ or }\tfrac{1}{2}$ .

First we do the case when $k=0$ .

We looking for a solution $y=\sum_{n=0}^{\infty} a_n x^n$ .
Substituting that into the equation we get,
$2x\sum_{n=2}^{\infty}n(n-1) a_n x^{n-2} + \sum_{n=1}^{\infty}na_nx^{n-1} + x\sum_{n=0}^{\infty} a_nx^n = 0$
This becomes,
$2\sum_{n=2}^{\infty} n(n-1)a_n x^{n-1} + \sum_{n=1}^{\infty} na_n x^{n-1} + \sum_{n=0}^{\infty}a_nx^{n+1} = 0$
We can rewrite this as,
$2\sum_{n=1}^{\infty} (n+1)na_{n+1}x^n + \sum_{n=0}^{\infty}(n+1)a_{n+1} x^n + \sum_{n=1}^{\infty} a_{n-1} x^n = 0$
Evaluate the middle summation at $n=0$ and combine,
$a_1 + \sum_{n=1}^{\infty} [2(n+1)na_{n+1} + (n+1)a_{n+1} + a_{n-1}]x^n = 0$

This tells us that $a_1=0$ .

The condition in the middle says,
$2(n+1)na_{n+1} + (n+1)a_{n+1}+a_{n-1} = 0 \text{ for }n\geq 1$

Thus, $a_{n+1} = - \frac{a_{n-1}}{(2n+1)(n+1)}$

As a consequence $a_1 = a_3 = a_5 = a_7 = ... = 0$

Taking $a_0$ to be arbitrary (like $a_0=1$ ) shall produce coefficients for $a_2,a_4,a_6,...$ .

That would be one solution to the differencial equation.
Of course the other linearly independent solution is found with $k=\tfrac{1}{2}$ .

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