MathCastle
Differential EquationsHardFrobenius & Special Functions

Frobenius & Special Functions · Hard problem

The equation xy′′+y′+y=0xy'' + y' + y = 0 has a double indicial root r=0r = 0 at x=0x = 0, with Frobenius solution y1=∑n≥0(−1)n(n!)2xny_1 = \displaystyle\sum_{n\ge0}\frac{(-1)^n}{(n!)^2}x^n. A second solution has the form y2=y1ln⁡x+∑n≥1bnxny_2 = y_1\ln x + \displaystyle\sum_{n\ge1}b_nx^n. Find b2b_2. Express your answer as a common fraction.