We have found the complementary function for the given nonhomogeneous differential equation. Now we must find…

We have found the complementary function for the given nonhomogeneous differential equation. Now we must find the particular solution, which will be based on the form of the function of x that makes the equation nonhomogeneous, g(x) = 200×2 – 78xe. The idea is that when the partial solution y, (and its derivatives) is substituted into the equation, it must be equal exactly to g(x). Therefore, we can assume the solution contains a quadratic expression and an exponential term. Assume that the solution contains Ax2 + Bx + C corresponding to 200×2, and that it contains (DX + E)ex, corresponding to – 78xe. Therefore, y, and its derivatives are as follows. Yo = Ax2 + Bx + C + (DX + E)e Yo’ = 24X + B + (Dx + ex + De ex + Det ex + 2Det Yo” = 24 + (Dx +