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The solution of the equation (2x + y + 1) dx + (4x + 2y – 1) dy = 0 is
  • a)
    log (2x + y – 1) = C + x + y
  • b)
    log (4x + 2y – 1) = C + 2x + y
  • c)
    log (2x + y + 1) + x + y = C
  • d)
    log (2x + y – 1) + x + 2y = C
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
The solution of the equation (2x + y + 1) dx + (4x + 2y – 1) dy ...
Put 2x + y = X ⇒Therefore, the given equation is reduced to

 = x + constant
2(2x + y) + log (2x + y – 1) = 3x + constant
x + 2y + log (2x + y – 1) = C
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Most Upvoted Answer
The solution of the equation (2x + y + 1) dx + (4x + 2y – 1) dy ...

Explanation:

Given equation:
(2x + y + 1) dx + (4x + 2y - 1) dy = 0

Using the exact differential equation:
For the given equation to be exact, we need to check if ∂(2x + y + 1)/∂y = ∂(4x + 2y - 1)/∂x

Calculating the partial derivatives:
∂(2x + y + 1)/∂y = 1
∂(4x + 2y - 1)/∂x = 4

Since the partial derivatives are not equal, the given equation is not exact.

Integrating factor:
To solve this non-exact equation, we need to find the integrating factor.
Let the integrating factor be μ(x, y)

μ(x, y) = e^∫(∂Q/∂x - ∂P/∂y)dx
μ(x, y) = e^∫(0 - 0)dx
μ(x, y) = 1

Multiplying the given equation by the integrating factor:
(2x + y + 1) dx + (4x + 2y - 1) dy = 0
2x dx + y dx + dx + 4x dy + 2y dy - dy = 0
(2x dx + y dx + dx) + (4x dy + 2y dy - dy) = 0
d(2x + y + 1) + d(4x + 2y - 1) = 0

Final solution:
Integrating both sides, we get:
log (2x + y + 1) + x = C
log (2x + y + 1) + x + 2y = C

Therefore, the correct answer is option D.
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The solution of the equation (2x + y + 1) dx + (4x + 2y – 1) dy = 0 isa)log (2x + y – 1) = C + x + yb)log (4x + 2y – 1) = C + 2x + yc)log (2x + y + 1) + x + y = Cd)log (2x + y – 1) + x + 2y = CCorrect answer is option 'D'. Can you explain this answer?
Question Description
The solution of the equation (2x + y + 1) dx + (4x + 2y – 1) dy = 0 isa)log (2x + y – 1) = C + x + yb)log (4x + 2y – 1) = C + 2x + yc)log (2x + y + 1) + x + y = Cd)log (2x + y – 1) + x + 2y = CCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The solution of the equation (2x + y + 1) dx + (4x + 2y – 1) dy = 0 isa)log (2x + y – 1) = C + x + yb)log (4x + 2y – 1) = C + 2x + yc)log (2x + y + 1) + x + y = Cd)log (2x + y – 1) + x + 2y = CCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The solution of the equation (2x + y + 1) dx + (4x + 2y – 1) dy = 0 isa)log (2x + y – 1) = C + x + yb)log (4x + 2y – 1) = C + 2x + yc)log (2x + y + 1) + x + y = Cd)log (2x + y – 1) + x + 2y = CCorrect answer is option 'D'. Can you explain this answer?.
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