Prove The Theorem 7.7.2 down below Let Ly = x^2(alpha_0 + alpha_2x^2)y\" + x(bet

Question

Prove The Theorem 7.7.2 down below

Let Ly = x^2(alpha_0 + alpha_2x^2)y" + x(beta_0 + beta_2x^2)y' + (gamma_0 + gamma_2x^2)y, where alpha_0 notequalto 0, and define p_0(r) = alpha_0r(r - 1) + beta_0r + gamma_0, p_2(r) = alpha_2r(r - 1) + beta_2r + gamma_2. Suppose r is a real number such that p_0 (2m + r) is nonzero for all positive integers m, and define a_0(r) = 1 a_2m(r) = p_2(2m + r - 2)/p_0(2m + r)a_2m-2(r), m greaterthanorequalto 1. Let r_1 and r_2 be the roots of the indicial equation P_0(r) = 0, and suppose r_1 = r_2 + 2k, where k is a positive integer. Then y_1 = x^r_1 sigma^infinity_m=0 a_2m(r_1)x^2m is a Frobenius solution of Ly = 0. Moreover, if we define a_0(r_2) = 1, a_2m(r_2) = p_2(2m + r_2 - 2)/p_0(2m + r_2) a2m-2(r_2), 1 lessthanorequalto m lessthanorequalto k - 1 and C = p_2(r_1 - 2)/2k alpha_0 a-2k-2(r_2), then y_2 = x^r_2 sigma^k-1_m=0 a_2m(r_2)x^2m + C(y_1 ln x + x^r_1 sigma^infinity_m=1 a'_2m(r_1)x^2m) is also a solution of Ly =0, and {y_1, y_2} is a fundamental set of solutions.

Explanation / Answer

as y2 satisfies the condition Ly=0 so it is a solution of the given condition and {y1,y2} is a fundamental set of solutions.

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