Correct answer is (c) -4

For explanation I would say: (D + 1)^2y = 0

Or, (D^2 + 2D+ 1)y = 0

=> d^2y/dx^2 + 2dy/dx + y = 0……….(1)

Let y = e^mx be a trial solution of equation (1). Then,

=> dy/dx = me^mx and d^2y/dx^2 = m^2e^mx

Clearly, y = e^mx will SATISFY equation (1). Hence, we have

=> m^2.e^mx + 2M.e^mx + e^mx = 0

Or, m^2 + 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 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

So, C = 4