# Power series solution for ODE

1. Nov 28, 2013

### Rct33

1. The problem statement, all variables and given/known data

I am trying to find the recursion relation for the coefficients of the series around x=0 for the ODE: $y'''+x^2y'+xy=0$

3. The attempt at a solution
Therefore letting:

$y=\sum_{m=0}^\infty y_mx^m$

$\therefore y'=\sum_{m=1}^\infty my_mx^{m-1}$

$\therefore y''=\sum_{m=2}^\infty m(m-1)y_mx^{m-2}$

$\therefore y'''=\sum_{m=3}^\infty m(m-1)(m-2)y_mx^{m-3}$

Subbing this back in gives:

$\sum_{m=3}^\infty m(m-1)(m-2)y_mx^{m-3}+x^2\sum_{m=1}^\infty my_mx^{m-1}+x\sum_{m=0}^\infty y_mx^m=0$

Fixing $y'''$:

$\sum_{m=3}^\infty m(m-1)(m-2)y_mx^{m-3}=\sum_{m=2}^\infty (m+1)(m+2)(m+3)y_{m+3}x^{m}+6y_3+24xy_4$

Fixing $y'$:

$x^2\sum_{m=1}^\infty my_mx^{m-1}=\sum_{m=2}^\infty (m-1)y_{m-1}x^m$

Fixing $y$:

$x\sum_{m=0}^\infty y_mx^m=\sum_{m=2}^\infty y_{m-1}x^m+xy_0$

Therefore combining these terms gives:

$\sum_{m=2}^\infty\left[(m+1)(m+2)(m+3)y_{m+3}+(m-1)y_{m-1}+y_{m-1}\right]x^m+6y_3+xy_0+24xy_4=0$

Therefore I have two equations which sum to 0, with one of them being this:

$6y_3+xy_0+24xy_4=0$

$\therefore y_3=0$ and $y_4=-\frac{1}{24}y_0$

Using the other equation:

$(m+1)(m+2)(m+3)y_{m+3}+(m-1)y_{m-1}+y_{m-1}=0$

$\therefore (m+1)(m+2)(m+3)y_{m+3}+y_{m-1}((m-1)+1)=0$

$\therefore y_{m+3}=-\frac{y_{m-1}m}{ (m+1)(m+2)(m+3)}$

This gives me:
$y_3=0$
$y_4=-\frac{1}{24}y_0$
$y_5=-\frac{1}{30}y_1$
$y_6=-\frac{1}{40}y_2$

The problem is I am not sure how to relate these together to solve the recursion! Any help is much appreciated.

Last edited: Nov 28, 2013
2. Nov 28, 2013

### HallsofIvy

Staff Emeritus
It is rare that one can solve recursive equations for a "closed form" solution. Do you have reason to believe you can here?

3. Nov 28, 2013

### Rct33

Hm I am thinking then that there is a limited number of free parameters? So to answer your question, I have no good reason to believe a closed form solution exists. So how would I comment on the number of free parameters? Does this mean anything?

4. Nov 28, 2013

### pasmith

You have
$$y_{m+3} = - \frac{y_{m-1} m}{(m+ 1)(m + 2)(m + 3)}$$
or
$$y_{m+4} = - \frac{y_m (m+1)}{(m+2)(m+ 3)(m+ 4)}$$

Thus you have four independent sequences $y_{4n + k}$ for $k = 0,1,2,3$. Setting $m = 4n + k$ then gives
$$y_{4(n+1)+k} = - \frac{y_{4n + k} (4n + k +1)}{(4n + k +2)(4n + k + 3)(4n + k + 4)}$$
Now, instead of having $m$ increasing by 4 at each step, we have $n$ increasing by 1, and to make things clearer we can set $y_{4n + k} = a_n(k)$, so that
$$a_{n+1}(k) = - \frac{a_n(k) (4n + k +1)}{(4n + k +2)(4n + k + 3)(4n + k + 4)}$$

One of the convenient aspects of homogenous linear recurrences is that we can deal with each factor multiplying $a_{n}$ separately. The -1 is the easiest: we get a factor of $(-1)^n$. The factors which are linear in $n$ are more difficult, and
$$b_{n+1} = (An + B)b_n$$
doesn't have a solution other than the not closed-form
$$b_n = b_0\prod_{r = 0}^{n-1} (Ar + B)$$
and similarly
$$b_{n+1} = \frac{b_n}{An + B}$$
doesn't have a solution other than
$$b_n = b_0\prod_{r = 0}^{n-1} \frac{1}{Ar + B}$$
(and we obviously require that there not exist any integer $r \geq 0$ for which $Ar + B = 0$).

If $A = 1$ (which in your problem it isn't) these products can be expressed in terms of factorials or gamma functions. If $A$ were an integer and there were $A$ factors $(Ar + B)(Ar + B + 1) \cdots (Ar + B + A)$ (or $(Ar + B)^{-1}(Ar + B + 1)^{-1} \cdots (Ar + B + A)^{-1}$) then again there would be a solution in terms of factorials or gamma functions, but your problem doesn't have sufficient factors.

However, you can say that
$$y(x) = \sum_{m = 0}^{\infty} y_m x^m = \sum_{n=0}^{\infty} x^{4n} (a_n(0) + a_n(1) x + a_n(2) x^2)$$
since the initial condition $y_3 = 0$ requires that $a_n(3) = 0$ for all $n$.

5. Nov 28, 2013

### Rct33

Thank you so much, was a good read

Last edited: Nov 28, 2013