Given system is x vector^11 = [- 5/3 1 1 - 1]x vector + [3 cos (4t) 0] -1 we need to find a particular solution of the form x vector [x_1 (t) x_2 (t)] v vector cos (4t) Then x vector = v vector middot 4 (- sin 4t) = - 4 v vector sin (4t) x vector^11 = - 16 v vector cos (4t) -2 By equation 1 and 2 (we put the value of x^1, x^n in eq 1) - 16 v vector cos (4t) = [- 5/3 1 1 - 1] v vector cos (4t) + [3 cos (4t) 0] ([- 5/3 1 1 - 1] + 16 I_2) nu vector cos (4t) = -[3 cos (4t) 0] ([- 5/3 1 1 - 1] + 16 [1 0 0 1]) nu vector = [- 3 0] [43/3 1 1 15]