Question

SOLVE. y''+3y'+2y=4e^x

Answer 1

This a second order linear differential equation with the solution given by the general solution + the particular solution.
The general solution is given by Ae^mx + Be^nx; where m and n are solutions to the quadratic equation t^2 + 3t + 2 = 0
(t + 1)(t + 2) = 0
t = -1 or t = -2
general solution = Ae^(-x) + Be^(-2x)
For the particular solution, Ce^x + 3Ce^x + 2Ce^x = 4e^x
6Ce^x = 4e^x
C = 4/6 = 2/3

Therefore, solution is y = Ae^(-x) + Be^(-2x) + 2/3e^x