Find the general solution of y'' - 3y' + 2y = 2e^x

First we need to find the complementary function - the solutions of y'' -3y' + 2y = 0. By setting y=Ae^(kx) (A is a constant) we get :A(k^2 - 3k + 2)y = 0. This means k^2 - 3k + 2 = 0 =(k-1)(k-2). If 2 things multiply to give zero then at least one of them must be zero so we can have k=1 or k=2. Therefore any solution of the form y = Ae^x + Be^(2x) where A and B are constants solves y'' -3y' + 2y = 0 .
Now we need the particular integral. Since the right hand side of y'' - 3y' + 2y = 2e^x has an e^x term we guess a solution of the form y=Ce^x ( C constant). However, from above we see that this is part of the complementary function and so will not work as the left hand side of the equation will be 0. So we guess y = Cxe^x . Substituting this in we get: (Cx+2C)e^x - 3(Cx + C)e^x +2Cxe^x = 2e^xThe coefficients of xe^x cancel out to zero so we get: -Ce^x = 2e^x Which means C = -2 and therefore out general solution is: y = Ae^x + Be^(2x) - 2xe^x