Method of Frobenius to solve x2Y″(x)+xY′(x)=c2Y(x)x^2 Y”(x) + x Y'(x) = c^2 Y(x)

I’m trying to
x2Y″(x)+xY′(x)=c2Y(x)x^2 Y”(x) + x Y'(x) = c^2 Y(x)
using the method of Frobenius. I assume the solution is of the form Y=xk∑∞n=0ansnY = x^k \sum_{n=0}^\infty a_n s^n and then put that into my ODE and found 0=∑∞n=0an[(n+k)(n+k−1)+(n+k)−c2]xk+n0 = \sum_{n=0}^\infty a_n[(n+k)(n+k-1)+(n+k)-c^2]x^{k+n} . Then I found the indicial polynomial to be a0[k2−c2]xk=0a_0[k^2 – c^2]x^k = 0 which means k=ck = c. This is where I’m stuck. I’m not sure how to solve the equation after knowing this part.
I was trying to follow




I suppose that one of the Y″Y” is Y′Y’.
– Claude Leibovici
2 days ago



Maple gives y(x)=c1x−c+c2xcy(x)=c_1 x^{-c} + c_2 x^c. You may see how to get there.
– Federiko
2 days ago


1 Answer


Actually, k=±ck= \pm c. This equation is really an Euler equation, so using Frobenius is overkill. You should get that a0a_0 and a1a_1 are arbitrary constants and that all the other ana_n’s are zero.



I still can’t figure this out, so I’m not using Frobenius anymore.
– TheStrangeQuark
2 days ago