Final answer:
The differential equation is solved by finding the complementary solution to the associated homogeneous equation and then finding a particular solution to the non-homogeneous equation, using initial conditions to determine constants.
Step-by-step explanation:
The differential equation to solve is y'' + 2y' + y = 2e-x, with initial conditions y(0) = 3 and y'(0) = -1. This is a second-order linear non-homogeneous differential equation with constant coefficients. The solution involves finding the complementary solution (yc) from the associated homogeneous equation (y'' + 2y' + y = 0) and a particular solution (yp) to the non-homogeneous equation.
To obtain the complementary solution, we solve the characteristic equation associated with the homogeneous part, which is r2 + 2r + 1 = 0. This equation has a repeated root r = -1. Therefore, the complementary solution has the form yc = (C1 + C2x)e-x, where C1 and C2 are constants to be determined by the initial conditions.
The particular solution can be guessed to have a form similar to the non-homogeneous part. Given that the non-homogeneous term is 2e-x, and considering the form of the complementary solution, a suitable guess for the particular solution is yp = Ae-x. Plugging this into the non-homogeneous differential equation, we can solve for A.
Finally, we add yc and yp to get the general solution of the differential equation. Using the initial conditions, we can find the specific values for C1, C2, and A, giving us the particular solution to the initial value problem.