Question

y" – 3y' + 2y = 2 sin x cos x​

Answer 1

Step-by-step explanation:

The general solution to the homogeneous equation is y = c1 ex + c2 2x Since the r(x) = ex sin x, we assume the particular solution of the form yp = m ex sin x + n ex cos x Substituting the above equation into the differential equation and equating the coefficients of ex sin x and ex cos x, we have yp = ex/2 (cos x - sin x) and y(x) = c1 ex + c2 2x + ex/2 (cos x - sin x)