# Thread: Finding Recurrence relation for Series Solution of Differential Equation

1. ## Finding Recurrence relation for Series Solution of Differential Equation

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.

2. 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 $k=\tfrac{1}{2}$.