The differential equation of $y = \sec(\tan^{-1}x)$ is

  • A
    $(1 + x^2)\frac{dy}{dx} = y + x$
  • B
    $(1 + x^2)\frac{dy}{dx} = y - x$
  • C
    $(1 + x^2)\frac{dy}{dx} = xy$
  • D
    $(1 + x^2)\frac{dy}{dx} = \frac{x}{y}$

Explore More

Similar Questions

The differential equation obtained by eliminating $A$ and $B$ from $y = A \cos \omega t + B \sin \omega t$ is:

The differential equation of the family of curves given by $y = a e^{2x} + b e^{5x}$,where $a$ and $b$ are parameters,is:

The differential equation having $y=(a+b) e^{cx+d}$ as its general solution,where $a, b, c, d$ are arbitrary constants,is

Statement $I$: The differential equation corresponding to the family of circles having their centres on $Y$-axis and fixed radius $k$ is $(x^2-k^2)(\frac{dy}{dx})^2+x^2=0$.
Statement $II$: The differential equation corresponding to the family of circles passing through the origin and having their centres on $X$-axis is $x^2-y^2+2xy \frac{dy}{dx}=0$.
Which of the above statements is (are) true?

The differential equation obtained by eliminating $A$ and $B$ from the equation $y = A \cos \omega t + B \sin \omega t$ 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