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?

Category: QuestionsWhat 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?
Editor">Editor Staff asked 11 months ago

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?
 
(a) y = 4xe^-x
 
(b) y = 4xe^x
 
(c) y = -4xe^-x
 
(d) y = -4xe^x
 
This question was addressed to me in a national level competition.
 
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1 Answers
Editor">Editor Staff answered 11 months ago

Right answer is (a) y = 4xe^-x
 
To explain: (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