The general solution|Differential Equations|JEE-MAIN

JEE MAIN 2020

06-09-2020_S1

Question

The general solution of the differential equation $\sqrt{1+x^2+y^2+x^2 y^2}+x y \frac{d y}{d x}=0$ is : (where C is a constant of integration)

Select the correct option:

$\sqrt{1+\mathrm{y}^2}+\sqrt{1+\mathrm{x}^2}=\frac{1}{2} \log _{\mathrm{e}}\left(\frac{\sqrt{1+\mathrm{x}^2}+1}{\sqrt{1+\mathrm{x}^2}-1}\right)+\mathrm{C}$

$\sqrt{1+\mathrm{y}^2}+\sqrt{1+\mathrm{x}^2}=\frac{1}{2} \log _{\mathrm{e}}\left(\frac{\sqrt{1+\mathrm{x}^2}-1}{\sqrt{1+\mathrm{x}^2}-1}\right)+\mathrm{C}$

$\sqrt{1+\mathrm{y}^2}-\sqrt{1+\mathrm{x}^2}=\frac{1}{2} \log _{\mathrm{e}}\left(\frac{\sqrt{1+\mathrm{x}^2}-1}{\sqrt{1+\mathrm{x}^2}-1}\right)+\mathrm{C}$

$\sqrt{1+\mathrm{y}^2}-\sqrt{1+\mathrm{x}^2}=\frac{1}{2} \log _{\mathrm{e}}\left(\frac{\sqrt{1+\mathrm{x}^2}+1}{\sqrt{1+\mathrm{x}^2}-1}\right)+\mathrm{C}$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Mathematics

Hard

Start Preparing for JEE with Competishun

Report Issue