We have found that the roots of the auxiliary equation are m, = -1 and m, = -2. In the case that theauxiliary equation for a second-order linear equation has two distinct, real roots, we know the complementaryfunction is of the form y. = c,em1* + c,em2*. Therefore, the complementary function is as follows.Yc = Ce* + cze-2xLet y, = e* and y, = e-2x be the two independent solutions which are terms of the complementaryfunction. We will find functions u,(x) and u,(x) such that y, = u,y, + uzy, is a particular solution. These newfunctions are found by calculating multiple Wronskians. First, find the following Wronskian.W(y;(x), Y2(x)) = W(e-X, e-2x)%3De-2x|Y1(x) ¥2(x)|y,'(x) Y2'(x)

We have found that the roots of the auxiliary equation are m, = -1 and m, = -2. In the case that the auxiliary equation for a second-order linear equation has two distinct, real roots, we know the complementary function is of the form y, = c,em1* + c,em2*. Therefore, the complementary function is as follows. Yc = Ce* + cze-2x Let y, = eX and y, = e-2X be the two independent solutions which are terms of the complementary function. We will find functions u, (x) and u,(x) such that y, = u,y, + u,y, is a particular solution. These new functions are found by calculating multiple Wronskians. First, find the following Wronskian. W(y;(x), Y2(x)) = w(e-X, e-2x) %3D e-2x |Y1(x) Y2(x) |y,'(x) Y2'(x)