Find general solution for y'' + y'-6y=12 (e^3t+e^-2t)

Homogeneous solution: y′′ + y′ - 6y=0 yields the characteristic equation r2 + r-6=0 ⇒ r=-3,2 So homogeneous part is yc = C1 e - 3t + C2 e 2t. Non-homogeneous solution: y′′ + y′ - 6y=12 e 3t + 12 e - 2t Use the method of undetermined coefficients. Suppose yp = A e 3t + B e - 2t is a solution, so you have yp = A e 3t + B e - 2t y′p = 3A e 3t - 2B e - 2t y′p = 9A e 3t + 4B e - 2t Substitute into the original equation: (9A e 3t + 4B e - 2t) + (3A e 3t - 2B e - 2t) - 6 (A e 3t + B e - 2t) 6A e 3t - 4B e - 2t = 12 e 3t + 12 e - 2t = 12 e 3t + 12 e - 2t This tells you that A=2 and B=-3, and so your non-homogeneous part is yp = 2 e 3t - 3 e - 2t. Your final solution would be the sum of the non/homogeneous parts, or y = yc + y p

Am i correct? Your solution would be y = yc + yp