Consider the differential equation x2 y + x(1 + bx)y + [b(1 N)x N2]y = 0, x > 0

Chapter 11, Problem 19

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Consider the differential equation x2 y + x(1 + bx)y + [b(1 N)x N2]y = 0, x > 0, (11.7.13) where N is a positive integer and b is a constant. (a) Show that the roots of the indicial equation are r = N. (b) Show that the Frobenius series solution corresponding to r = N is y1(x) = a0x N n=0 (2N)!(b)n (2N + n)! xn and that by an appropriate choice of a0, one solution to (11.7.13) is y1(x) = xN ebx 2 N1 n=0 (bx)n n! . (c) Show that Equation (11.7.13) has a second linearly independent Frobenius series solution that can be taken as y2(x) = xN 2 N1 n=0 (bx)n n! . Hence, conclude that Equation (11.7.13) has linearly independent solutions y1(x) = xN ebx , y2(x) = xN 2 N1 n=0 (bx)n n! . 20.