# 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:

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

I'm totally confused about finding the recurrence relation when solving this when using the method of Frobenius about $\displaystyle x_0 = 0$. 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

$\displaystyle a_n = - \frac{a_{n-2}}{(n+r)(2(n+r) -1)}$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:

$\displaystyle 2xy'' + y' + xy = 0$ for $\displaystyle x>0$.
The point $\displaystyle 0$ is a regular singular point.
Therefore, solutions shall have the form $\displaystyle y= x^k \sum_{n=0}^{\infty} a_n x^n$.

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

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

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

This tells us that $\displaystyle a_1=0$.

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

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

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

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

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