Question

Find a particular solution of the recurrence equation.

Answer 1

Given recurrence equation is t_n = 4t_n - 1 + t_n - 2 + 3.2^n rightwardsdoublearrow t_n - 4t_n - 1 - t_n - 2 = 3.2^n ___(1) Now associated homogeneous equation of eqn (1) is t_n - 4t_n - 1 - t_n - 2 = 0 ___(2) So the characteristic equation of eqn (2) is alpha^2 - 4 alpha - alpha = 0. on solving this equation, we get alpha = -(-4) plusminus squareroot (-4)^2 - 4 (1) (-1)/2 times 1 = 4 plusminus squareroot 16 + 4/2 = 4 plusminus squareroot 20/2 = 2 plusminus squareroot 5. So the roots are 2 + squareroot 5 and 2 - squareroot 5. Again we know that if f (n) is of the form (F_1 n^t + F_2 n^t - 1 + Ellipsis + F_t + 1) beta^n, then corresponding particular solution is of the form (p_1 n^t + p_2 n^t - 1 + Ellipsis + p_t + 1) beta^n ___(3)