Atılım Üniversitesi

MATH 276

(Q1.) (20 pts.) Using the method of undetermined coefficients solve the differential equation y ′′ + y ′ − 6y = 7e 2x − 50 sin x. Solution: The characteristic equation is r 2 + r − 6 = 0 which gives (r + 3)(r − 2) = 0. The roots are r1 = −3 and r2 = 2. So, a complementary function is yc = C1e −3x + C2e 2x . The right function is the sum of f1(x) = 7e 2x and f2(x) = −50 sin x. f1 is of the form Ae2x . Since r = 2 is simple root of the characteristic equation, we modify this form and take yp1 = Axe2x . f2 is of the form B sin x + C cos x. Since r = πi is not root of the characteristic equation, we do not modify this form and take yp2 = B sin x + C cos x. Thus, yp = Axe2x + B sin x + C cos x. Then, y ′ p = Ae2x + 2Axe2x + B cos x − C sin x, y′′ p = 4Ae2x + 4Axe2x − B sin x − C cos x. Put this into the given equation to get 5Ae2x + (−7B − C) sin x + (−7C + B) cos x = 7e 2x − 50 sin x. This is possible only if 5A = 7, −7B − C = −50 and −7C + B = 0. Hence, A = 7/5, B = 7 and C = 1. Therefore, a particular solution is yp = 7 5 xe2x + 7 sin x + cos x