Give the general solution to (d2y/dx2) - 2dy/dx -3y = 2sinx

Using the auxiliary equation t^2 - 2t - 3t = 0  t therefore is equal to 3 or -1. Using this value, a complementary function is derived.  Y= Ae^(3x) + Be^(-x). Finally, to fully solve, a particular integral of y = asinx + bcosx and differentiate it twice, to give equations for Dy/dx and (d2y/dx2). These can be substituted into the initial differential equations to find the values of a and b, Which are -2/5 and 1/5 respectively. The answer is then the complementary function plus the solution to the particular integral y = Ae^(3x) + Be^(-x) + (1/5)cosx - (2/5)sinx