The solution of the differential equation $(x^2 + 1) \frac{dy}{dx} + (y^2 + 1) = 0$ is . . . . . .

  • A
    $(A) \ x + y = c$
  • B
    $(B) \ (x^2 + 1)(y^2 + 1) = c$
  • C
    $(C) \ x^2 = y^2 + c$
  • D
    $(D) \ \tan^{-1} x + \tan^{-1} y = c$

Explore More

Similar Questions

If $\frac{dy}{dx} + \frac{2^{x-y}(2^y - 1)}{2^x - 1} = 0$,$x, y > 0$,and $y(1) = 1$,then $y(2)$ is equal to

The general solution of the differential equation $\frac{dy}{dx} + \frac{1}{\sqrt{1-x^2}} = 0$ is

Find the general solution of the differential equation: $\left(x+y \frac{dy}{dx}\right)=1$.

Difficult
View Solution

The general solution of the differential equation $(x-2y+1)dy-(3x-6y+2)dx=0$ is

The general solution of the differential equation $\cos (x+y) dy = dx$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo