The differential equation $\left[\frac{1+\left(\frac{dy}{dx}\right)^2}{\left(\frac{d^2y}{dx^2}\right)^{\frac{3}{2}}}\right]^2 = kx$ is of

  • A
    order $= 2$,degree $= 3$
  • B
    order $= 3$,degree $= 2$
  • C
    order $= 2$,degree $= 2$
  • D
    order $= 3$,degree $= 3$

Explore More

Similar Questions

The order and degree of the differential equation $(\frac{d^3y}{dx^3})^{2/3} - 3 \frac{d^2y}{dx^2} + 5 \frac{dy}{dx} + 4 = 0$ are respectively:

The order of the differential equation $\frac{d^2 y}{d x^2}=\sqrt{1-\left(\frac{d y}{d x}\right)^2}$ is

The degree and order of the differential equation $\left[1+\left(\frac{dy}{dx}\right)^3\right]^{\frac{7}{3}}=7\left(\frac{d^2y}{dx^2}\right)$ respectively are

The differential equation $x(\frac{d^2y}{dx^2})^3 + (\frac{dy}{dx})^4 + y = x^2$ is of

Consider the following differential equations.
$D_1: y=4 \frac{dy}{dx}+3x \frac{dx}{dy}$
$D_2: \frac{d^2y}{dx^2}=\left(3+\left(\frac{dy}{dx}\right)^2\right)^{\frac{4}{3}}$
$D_3: \left[1+\left(\frac{dy}{dx}\right)\right]^2=\left(\frac{dy}{dx}\right)^2$
The ratio of the sum of the orders of $D_1, D_2$ and $D_3$ to the sum of their degrees 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