JEE Exam > JEE Questions > The solution of the equation (1+x2)(1+y)dy+(1...
Start Learning for Free
The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 is
- a)tan⁻1x+log(1+x2)+tan⁻1y+log(1+y2)=c
- b)tan⁻1x-(1/2)log(1+x2)+tan⁻1y-(1/2)log(1+y2)=c
- c)tan⁻1x+(1/2)log(1+x2)+tan⁻1y+(1/2)log(1+y2)=c
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan1x+l...
The given differential equation is nonlinear, so it cannot be solved using standard techniques like separation of variables. To solve this equation, we can try to rearrange it into a standard form that can be recognized as a known differential equation.
We can rewrite the given equation as:
(1+y)(1+y^2)dy + (1+x)(1+x^2)dx = 0
This equation can be rearranged as follows:
(1+y^2)dy + (1+x^2)dx = -(1+y)(1+x)
This equation can be rewritten as:
dy/dx = (1+x)/(1+y) - (1+y^2)/(1+x^2)
This equation is in the form dy/dx = f(x,y), which is the standard form of a first-order differential equation. This equation can be recognized as a separable differential equation, which means that it can be solved using separation of variables.
To solve this equation using separation of variables, we can isolate the dy term on one side of the equation and the dx term on the other side, and then integrate both sides. This gives:
∫(1+y^2)dy = -∫(1+x)(1+x^2)dx
Solving for y, we get:
tan⁻1y + (1/2)log(1+y^2) = -∫(1+x)(1+x^2)dx + C
Where C is an integration constant.
Similarly, we can solve for x to get:
tan⁻1x + (1/2)log(1+x^2) = -∫(1+y)(1+y^2)dy + D
Where D is an integration constant.
Therefore, the solution of the differential equation is:
tan⁻1x + (1/2)log(1+x^2) + tan⁻1y + (1/2)log(1+y^2) = C + D
This is equivalent to:
c) tan⁻1x+(1/2)log(1+x^2)+tan⁻1y+(1/2)log(1+y^2)=c
Free Test
FREE
| Start Free Test |
Community Answer
The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan1x+l...
Given equation is (1-x²)(1-y)dy - (1-x)(1-y²)dx = 0.
Separating the variables, we get (1-y)dy/(1-x²) - (1-y²)dx/(1-x) = 0.
Integrating both sides, we get ∫(1-y)dy/(1-x²) = ∫(1-y²)dx/(1-x) + c.
Solving the integrals, we get tan⁻¹x - (1/2)ln(1-x²) = tan⁻¹y - (1/2)ln(1-y²) + c.
Rearranging the terms, we get tan⁻¹x + (1/2)ln(1-x²) - tan⁻¹y - (1/2)ln(1-y²) = c.
Therefore, the correct option is (C) tan⁻¹x + (1/2)ln(1-x²) tan⁻¹y + (1/2)ln(1-y²) = c.
Separating the variables, we get (1-y)dy/(1-x²) - (1-y²)dx/(1-x) = 0.
Integrating both sides, we get ∫(1-y)dy/(1-x²) = ∫(1-y²)dx/(1-x) + c.
Solving the integrals, we get tan⁻¹x - (1/2)ln(1-x²) = tan⁻¹y - (1/2)ln(1-y²) + c.
Rearranging the terms, we get tan⁻¹x + (1/2)ln(1-x²) - tan⁻¹y - (1/2)ln(1-y²) = c.
Therefore, the correct option is (C) tan⁻¹x + (1/2)ln(1-x²) tan⁻¹y + (1/2)ln(1-y²) = c.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
Question Description
The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan1x+log(1+x2)+tan1y+log(1+y2)=cb)tan1x-(1/2)log(1+x2)+tan1y-(1/2)log(1+y2)=cc)tan1x+(1/2)log(1+x2)+tan1y+(1/2)log(1+y2)=cd)none of theseCorrect answer is option 'C'. Can you explain this answer? for JEE 2025 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 (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan1x+log(1+x2)+tan1y+log(1+y2)=cb)tan1x-(1/2)log(1+x2)+tan1y-(1/2)log(1+y2)=cc)tan1x+(1/2)log(1+x2)+tan1y+(1/2)log(1+y2)=cd)none of theseCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan1x+log(1+x2)+tan1y+log(1+y2)=cb)tan1x-(1/2)log(1+x2)+tan1y-(1/2)log(1+y2)=cc)tan1x+(1/2)log(1+x2)+tan1y+(1/2)log(1+y2)=cd)none of theseCorrect answer is option 'C'. Can you explain this answer?.
The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan1x+log(1+x2)+tan1y+log(1+y2)=cb)tan1x-(1/2)log(1+x2)+tan1y-(1/2)log(1+y2)=cc)tan1x+(1/2)log(1+x2)+tan1y+(1/2)log(1+y2)=cd)none of theseCorrect answer is option 'C'. Can you explain this answer? for JEE 2025 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 (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan1x+log(1+x2)+tan1y+log(1+y2)=cb)tan1x-(1/2)log(1+x2)+tan1y-(1/2)log(1+y2)=cc)tan1x+(1/2)log(1+x2)+tan1y+(1/2)log(1+y2)=cd)none of theseCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan1x+log(1+x2)+tan1y+log(1+y2)=cb)tan1x-(1/2)log(1+x2)+tan1y-(1/2)log(1+y2)=cc)tan1x+(1/2)log(1+x2)+tan1y+(1/2)log(1+y2)=cd)none of theseCorrect answer is option 'C'. Can you explain this answer?.
Solutions for The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan1x+log(1+x2)+tan1y+log(1+y2)=cb)tan1x-(1/2)log(1+x2)+tan1y-(1/2)log(1+y2)=cc)tan1x+(1/2)log(1+x2)+tan1y+(1/2)log(1+y2)=cd)none of theseCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan1x+log(1+x2)+tan1y+log(1+y2)=cb)tan1x-(1/2)log(1+x2)+tan1y-(1/2)log(1+y2)=cc)tan1x+(1/2)log(1+x2)+tan1y+(1/2)log(1+y2)=cd)none of theseCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan1x+log(1+x2)+tan1y+log(1+y2)=cb)tan1x-(1/2)log(1+x2)+tan1y-(1/2)log(1+y2)=cc)tan1x+(1/2)log(1+x2)+tan1y+(1/2)log(1+y2)=cd)none of theseCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan1x+log(1+x2)+tan1y+log(1+y2)=cb)tan1x-(1/2)log(1+x2)+tan1y-(1/2)log(1+y2)=cc)tan1x+(1/2)log(1+x2)+tan1y+(1/2)log(1+y2)=cd)none of theseCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan1x+log(1+x2)+tan1y+log(1+y2)=cb)tan1x-(1/2)log(1+x2)+tan1y-(1/2)log(1+y2)=cc)tan1x+(1/2)log(1+x2)+tan1y+(1/2)log(1+y2)=cd)none of theseCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The solution of the equation (1+x2)(1+y)dy+(1+x)(1+y2)dx=0 isa)tan1x+log(1+x2)+tan1y+log(1+y2)=cb)tan1x-(1/2)log(1+x2)+tan1y-(1/2)log(1+y2)=cc)tan1x+(1/2)log(1+x2)+tan1y+(1/2)log(1+y2)=cd)none of theseCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily