Find the general solution to the differential equation y'' + 4y' + 3y = 6e^(2x) [where y' is dy/dx and y'' is d^2 y/ dx^2]

First find the general solution to the differential equation y'' + 4y' + 3y = 0 as an arbitrary number of the solution to this differential equation can be added to the solution of the differential equation in the question. Equations of this form have the solution y = Ae^(mx) (where A is an arbitrary constant) so y ' = my and y'' = (m^2)y. Cancelling y this gives m^2 + 4m + 3 = 0, solved by m = -3, -1. The solution of this is equation (the complementary function) is y = Ae^(-3x) + Be^(-x) [where A and B are arbitrary constants]Then find the particular integral, the solution to the differential equation in the question. This is found by trying a solution of the form of the right hand side of the equation, y = ce^(2x). This gives y' = 2y, y'' = 4y. Substituting y, y' and y'' in and cancelling y gives 15c = 6, so c = 0.4. The general solution is the sum of the particular integral and the complementary function, y = Ae^(-3x) + Be^(-x) + 0.4e^(2x)