Rct33 said:
Homework Statement
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
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.
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
<br />
y_{4(n+1)+k} = - \frac{y_{4n + k} (4n + k +1)}{(4n + k +2)(4n + k + 3)(4n + k + 4)}<br />
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
<br />
a_{n+1}(k) = - \frac{a_n(k) (4n + k +1)}{(4n + k +2)(4n + k + 3)(4n + k + 4)}<br />
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
<br />
b_{n+1} = (An + B)b_n<br />
doesn't have a solution other than the not closed-form
<br />
b_n = b_0\prod_{r = 0}^{n-1} (Ar + B)<br />
and similarly
<br />
b_{n+1} = \frac{b_n}{An + B}<br />
doesn't have a solution other than
<br />
b_n = b_0\prod_{r = 0}^{n-1} \frac{1}{Ar + B}<br />
(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
<br />
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)<br />
since the initial condition y_3 = 0 requires that a_n(3) = 0 for all n.
