I've found that the only regular singular point of this differential equation:
$$x^2 (x-1)^2 y'' + 4 (x-1)y' - 4x^2 y = 0$$
is $x = 1$.
How do I determine the characteristic exponents for it?
2026-03-26 23:11:11.1774566671
Finding Characteristic Exponents for $x^2 (x-1)^2 y'' + 4 (x-1)y' - 4x^2 y = 0$
46 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in ORDINARY-DIFFERENTIAL-EQUATIONS
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