Question

Solve the initial-value problem. y" (0) =-1 y(0) = 2, y'(0) = 2, у",-2y" + y,-xe* + 5,

Answer 1

Y tripleprime - 2y doubleprime + y prime - x e^x + 5 = 0 doublerightarrow y tripleprime - 2y doubleprime + y prime = x e^x - 5 Auxiliary equation becomes D^3 - 2 D^2 + D = 0 doublerightarrow D (D^2 - 2D + 1) = 0 doublerightarrow D = 0 D^2 - 2D + 1 = 0 doublerightarrow (D - 1) (D - 1) doublerightarrow D = 1, 1 Thus complementary function is y (x) = C_1 e^0x + (C_2 + C_3 x) e^x doublerightarrow y (x) = C_1 + (C_2 + C_3 x) e^x Now y prime (x) = (C_2 + C_3 x) e^x + C_3 e^x y doubleprime (x) = (C_2 + C_3x) e^x + C_3 e^x + C_3 e^x Hence y (0) = 2 doublerightarrow C_1 + C_2 = 2 y prime (0) = 2 doublerightarrow C_2 + C_3 = 2 y doubleprime (0) = -1 doublerightarrow C_2 + 2 C_3 = -1