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

  • A
    $\sin^{-1} x + \sin^{-1} y = c$
  • B
    $\sin^{-1} x - \sin^{-1} y = c$
  • C
    $\cos^{-1} x + \cos^{-1} y = c$
  • D
    None of these

Explore More

Similar Questions

The equation of the curve passing through $(1, 0)$ and having a slope of $\frac{y - 1}{x^2 + x}$ is:

The particular solution of the differential equation $(1+y^2) dx - xy dy = 0$ at $x=1, y=0$,represents

Given that the slope of the tangent to a curve $y=y(x)$ at any point $(x, y)$ is $\frac{2y}{x^2}$. If the curve passes through the centre of the circle $x^2+y^2-2x-2y=0$,then its equation is

For the differential equation $x y \frac{dy}{dx} = (x+2)(y+2)$,find the solution curve passing through the point $(1, -1)$.

Difficult
View Solution

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