Another Frobenius method ODE

[ognik]

Frobenius method - recurrance relation question

If, using the Frobenius method, I get a 3 term recurrence relation of the form $a_{j+2} = a_j .f(k,j) + a_{j-2}. g(k,j)$ ( j even), how do I treat the $a_{j-2}$ term at first? I have found $a_1 = 0$, but how do I find a value for $a_{-2}$ so as to start the recurrance?

Addendum: I found a solution to this but it just says $a_{-2} = 0 = a_{-4}$ by definition. This is very unsatisfying, I use the indicial eqtn to find $a_1$ - but just assume $a_{-2} = 0$ 'by definition? Can anyone shed any more light on this please?

 

[gtoguy97]

The solution to this problem is to observe that the recurrence relation is homogeneous, which means that $a_n = 0$ for all $n<0$. This is because if $a_n = 0$ for all $n<0$, then the recurrence relation will evaluate to 0 for all negative values of n. Therefore, we can set $a_{-2} = 0 = a_{-4}$ to start the recurrance.