Frobenius Solution to 2xy'' +5y' -4xy = 0 at x = 0

[Gwozdzilla]

After determining that x = 0 is a regular singular point of this equation, the frobenius method allows you to assume that y = Σa_nx^{n + r}. Then I can take the first and second derivative of this assumption and plug it into the DE and begin solving with the general method:

1. Multiply the coefficients inside
2. Raise all of the x's to the same power
3. Remove all of the "early terms" such that each summation is indexed starting at the same point
4. Combine all of the like summations, set it equal to zero, and solve for the recurrence relation, or the highest indexed a.

In step 2, when I raise all of the x's to the same power, does this power have to be (n+r) or can it be something like (n+r-1) if that is more convenient? If I choose (n+r-1), do I have to change it back when I'm solving for the recurrence relation?

 

[LCKurtz]

After determining that x = 0 is a regular singular point of this equation, the frobenius method allows you to assume that y = Σa_nx^{n + r}. Then I can take the first and second derivative of this assumption and plug it into the DE and begin solving with the general method:

1. Multiply the coefficients inside
2. Raise all of the x's to the same power
3. Remove all of the "early terms" such that each summation is indexed starting at the same point
4. Combine all of the like summations, set it equal to zero, and solve for the recurrence relation, or the highest indexed a.

In step 2, when I raise all of the x's to the same power, does this power have to be (n+r) or can it be something like (n+r-1) if that is more convenient? If I choose (n+r-1), do I have to change it back when I'm solving for the recurrence relation?

Yes, you can use a different index. At the end of the day, it doesn't matter whether your recurrence gives $a_{n+1}$ in terms of $a_n$ or $a_n$ in terms of $a_{n-1}$. You just need to be consistent as you work the problem.