y'' + 5y' + 6y = 2e^t is a second order differential equation with solution y = Ae^mt + Be^nt + p(t); where m and n are the roots of the quadratic equation x^2 + 5x + 6 = 0, and p(t) is the particular solution.
x = -2 and x = -3
y = Ae^-2t + Be^-3t + p(t)
For the particular solution, Ke^t + 5Ke^t + 6Ke^t = 2e^t
12K = 2
K = 1/6
Therefore, solution is
y = Ae^-2t + Be^-3t + 1/6 e^t
