If $x^2+y^2=1$,then which of the following is true?

  • A
    $y y^{\prime \prime}-\left(2 y^{\prime}\right)^2+1=0$
  • B
    $y y^{\prime \prime}+\left(y^{\prime}\right)^2+1=0$
  • C
    $y y^{\prime \prime}-\left(y^{\prime}\right)^2-1=0$
  • D
    $y y^{\prime \prime}+\left(2 y^{\prime}\right)^2+1=0$

Explore More

Similar Questions

If $y = \cos^2\left(\frac{5x}{2}\right) - \sin^2\left(\frac{5x}{2}\right)$,then $\frac{d^2y}{dx^2} =$

If $f$ is twice differentiable such that $f''(x) = -f(x)$,$f'(x) = g(x)$,$h'(x) = [f(x)]^2 + [g(x)]^2$,and $h(0) = 2$,$h(1) = 4$,then the equation $y = h(x)$ represents:

If $y = x + e^x$,then at $x = 1$,$\frac{d^2x}{dy^2}$ is equal to

If $U_n$ $(n=1,2)$ denotes the $n^{\text{th}}$ derivative of $U(x) = \frac{Lx+M}{x^2-2Bx+C}$ (where $L, M, B, C$ are constants),then the equation $PU_2 + QU_1 + RU = 0$ holds for:

If $x=3 \cos t$ and $y=4 \sin t$,then $\frac{d^2 y}{d x^2}$ at the point $(x_0, y_0)=(\frac{3}{2} \sqrt{2}, 2 \sqrt{2})$ 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