What is the solution of the given equation (D + 1)2y = 0 given y = 2 loge 2 when x = loge 2 and y = (4/3) loge3 when x = loge3?

What is the solution of the given equation (D + 1)2y = 0 given y = 2 loge 2 when x = loge 2 and y = (4/3) loge3 when x = loge3? Correct Answer y = 4xe-x

n:(D + 1)2y = 0 Or, (D2 + 2D+ 1)y = 0 => d2y/dx2 + 2dy/dx + y = 0 ……….(1) Let y = emx be a trial solution of equation (1). Then, => dy/dx = memx and d2y/dx2 = m2emx Clearly, y = emx will satisfy equation (1). Hence, we have => m2.emx + 2m.emx + emx = 0 Or, m2 + 2m +1 = 0 (as, emx ≠ 0) ………..(2) Or, (m + 1)2 = 0 => m = -1, -1 So, the roots of the auxiliary equation (2) are real and equal. Therefore, the general solution of equation (1) is y = (A + Bx)e-x where A and B are two independent arbitrary constants ……….(3) Given, y = 2 loge 2 when x = loge 2 Therefore, from (3) we get, 2 loge 2 = (A + B loge2)e-x Or, 1/2(A + B loge2) = 2 log e2 Or, A + B loge2 = 4 loge2 ……….(4) Again y = (4/3) loge3 when x = loge3 So, from (3) we get, 4/3 loge3 = (A + Bloge3) Or, A + Bloge3 = 4loge3 ……….(5) Now, (5) – (4) gives, B(loge3 – loge2) = 4(loge3 – loge2) => B = 4 Putting B = 4 in (4) we get, A = 0 Thus the required solution of (1) is y = 4xe-x