What is the general solution to the equation d2y/dx2 + dy/dx - 2y = -3sinx + cosx (d2y/dx2 signals a second order derivative)

The first step is to find the complementary function, yc, which is done by setting up a quadratic based upon the differential equation of the form m^2 + m -2 = 0 (i.e m^2 = d2y/dx2, m = dy/dx, and y is a constant). This (luckily!) factorises to the form (x-1)(x+2) = 0, giving two distinct real roots (x=1 and x=-2). Because of this we know our yc is going to be of the form yc = Ae^x + Be^-2x - the roots of our quadratic end up in the index of the e terms. This is alas but half of our general solution! We also need the particular integral (or solution) yp. yp is found based on the function on the right hand side of the equation, so in this instance -3sinx + cosx. As there are only trig functions our yp will be of the form yp = qsinx + pcosx. By differentiating this twice we get expressions for dyp/dx (qcosx - psinx) and d2yp/dx2 (-qsinx - pcosx). By substituting these into our original equation replacing y with yp, dy/dx with dyp/dx etc. and collecting terms we get an expression of the form (-3q-p)sinx + (q-3p)cosx = -3sinx + cosx. Equating coefficients gives -3q - p = -3 and q -3p = 1. Solving by either substitution or elimination finds q =1 p =0. So yp = sinx. The general solution is found by summing the particular integral and the complementary function so: y = Ae^x+Be^-2x + sinx - Huzzah!