Math_334__Assignment_5_Solutions

1. Find the solution to the following initial value problems. (a) y ′′′ − y ′′ − 4y ′ + 4y = 0; y(0) = −4, y′ (0) = −1, y′′(0) = −19. (b) y ′′′ − 4y ′′ + 7y ′ − 6y = 0; y(0) = 1, y′ (0) = 0, y′′(0) = 0. Solution (a) The characteristic polynomial for this d.e. is: C(r) = r 3 − r 2 − 4r + 4 = (r + 2)(r − 1)(r − 2), whose roots are r = −2, 1, 2. The general solution to the equation is: y(x) = c1e −2x + c2e x + c3e 2x . Apply the initial conditions:    y(0) = −4 y ′ (0) = −1 y ′′(0) = −19 =⇒    c1 + c2 + c3 = −4 −2c1 + c2 + 2c3 = −1 4c1 + c2 + 4c3 = −19 =⇒    c1 = −2 c2 = 1 c3 = −3 Therefore the solution to the initial value problem is y(x) = −2e −2x + e x − 3e 2x . (b) The characteristic polynomial for this d.e. is: C(r) = r 3 − 4r 2 + 7r − 6 = (r − 2)(r 2 − 2r + 3), whose roots are r = 2, 1 ± √ 2i. The general solution to the equation is: y(x) = c1e 2x + e x (c2 cos √ 2x + c3 sin √ 2x). Apply the initial conditions:    y(0) = 1 y ′ (0) = 0 y ′′(0) = 0 =⇒    c1 + c2 = 1 2c1 + c2 + 21/2 c3 = 0 4c1 − c2 + 23/2 c3 = 0 =⇒    c1 = 1 c2 = 0 c3 = − √ 2 Therefore the solution to the initial value problem is y(x) = e 2x − √ 2e x sin √ 2x) .