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Differential EquationsChallengeFrobenius & Special Functions

Frobenius & Special Functions · Challenge problem

The equation 2x2y′′+x(1+2x)y′−y=02x^2y'' + x(1+2x)y' - y = 0 has indicial roots 11 and −12-\tfrac12. Let y=x(1+a1x+a2x2+⋯ )y = x(1+a_1x+a_2x^2+\cdots) be the Frobenius solution for the larger root r=1r=1, normalised by a0=1a_0=1. Find a1+a2a_1 + a_2. Express your answer as a common fraction.