What is solution of given equation (D + 1)²y = 0 given y = 2 log_e 2 when x = log_e 2 and y = (4/3) log_e3 when x = log_e3?

(D + 1)²y = 0
Or, (D² + 2D+ 1)y = 0
=> d²y/dx² + 2dy/dx + y = 0 ???.(1)
Let y = e^mx be a trial solution of equation (1). Then,
=> dy/dx = me^mx and d²y/dx² = m²e^mx
Clearly, y = e^mx will satisfy equation (1). Hence, we have
=> m².e^mx + 2m.e^mx + e^mx = 0
Or, m² + 2m +1 = 0 (as, e^mx ≠ 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 log_e 2 when x = log_e 2
Therefore, from (3) we get,
2 log_e 2 = (A + B log_e2)e^-x
Or, 1/2(A + B log_e2) = 2 log _e2
Or, A + B log_e2 = 4 log_e2 ???.(4)
Again y = (4/3) log_e3 when x = log_e3
So, from (3) we get,
4/3 log_e3 = (A + Blog_e3)
Or, A + Blog_e3 = 4log_e3 ???.(5)
Now, (5) ? (4) gives,
B(log_e3 ? log_e2) = 4(log_e3 ? log_e2)
=> B = 4
Putting B = 4 in (4) we get, A = 0
Thus the required solution of (1) is y = 4xe^-x