What will be the value of dy/dx = (x + 2y + 3)/(2x + 3y + 4)?

Category: QuestionsWhat will be the value of dy/dx = (x + 2y + 3)/(2x + 3y + 4)?
Editor">Editor Staff asked 11 months ago

What will be the value of dy/dx = (x + 2y + 3)/(2x + 3y + 4)?
 
(a) [(2 + √3)/2√3 * (log (√3(y + 2) – (x – 1))) + (2 – √3)/2√3 * (log (√3(y + 2) – (x – 1)))]
 
(b) [(2 + √3)/2√3 * (log (√3(y + 2) + (x – 1))) – (2 – √3)/2√3 * (log (√3(y + 2) + (x – 1)))]
 
(c) [(2 + √3)/2√3 * (log (√3(y + 2) – (x – 1))) – (2 – √3)/2√3 * (log (√3(y + 2) – (x – 1)))]
 
(d) [(2 + √3)/2√3 * (log (√3(y + 2) + (x – 1))) – (2 – √3)/2√3 * (log (√3(y + 2) – (x – 1)))]
 
This question was addressed to me in an interview for internship.
 
I would like to ask this question from Linear First Order Differential Equations topic in division Differential Equations of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

Correct choice is (c) [(2 + √3)/2√3 * (log (√3(y + 2) – (x – 1))) – (2 – √3)/2√3 * (log (√3(y + 2) – (x – 1)))]
 
Best explanation: Put x = X + h, Y = Y + k,
 
We have, dY/dX = (X + 2Y +(h + 2k + 3))/ 2X + 3Y + (2h + 3k + 4)
 
So, (a – b)x = (a – b)
 
To determine h and k we set,
 
2h + 3k + 4 = 0 and h + 2k + 3 = 0
 
=> h = 1 and k = – 2
 
Therefore, dY/dX = (X + 2Y) / (2X + 3Y)
 
Putting Y = VX, we get,
 
V + X dV/dX = (1 + 2V)/(2 + 3V)
 
        = (1 + 2V)/(3V^2 – 1)*dV = -dX/X
 
=> [(2 + √3)/(2(√3V – 1)) – (2 – √3)/(2(√3V – 1))] dV = -dX/X
 
Simplifying it further, we get;
 
[(2 + √3)/2√3 * (log (√3Y – X)) – (2 – √3)/2√3 * (log (√3Y – X))]
 
Where, X = x – 1 and Y = y + 2