Assertion $(A)$: The degree of the differential equation $y'' + 2xy' + \log_e\left(\frac{dy}{dx}\right) = 0$ is $2$.
Reason $(R)$: The degree of a differential equation is the highest power of the highest order derivative occurring in the equation, after the equation is expressed in the form of a polynomial in differential coefficients.
The correct option among the following is:

  • A
    $(A)$ is true, $(R)$ is true and $(R)$ is the correct explanation for $(A)$
  • B
    $(A)$ is true, $(R)$ is true but $(R)$ is not the correct explanation for $(A)$
  • C
    $(A)$ is true but $(R)$ is false
  • D
    $(A)$ is false but $(R)$ is true

Explore More

Similar Questions

Determine the order and degree (if defined) of the differential equation $\left(\frac{ds}{dt}\right)^{4} + 3s \frac{d^{2}s}{dt^{2}} = 0$.

The number of arbitrary constants in the particular solution of a differential equation of fourth order is . . . . . .

For the differential equation given below,determine its order and degree (if defined):
$\frac{d^{2} y}{d x^{2}}+5 x\left(\frac{d y}{d x}\right)^{2}-6 y=\log x$

For the differential equation $\frac{d^3 y}{d x^3}=0$,$y=a x^2+b x+c$ is

If the order and degree of the differential equation $x \frac{d^2 y}{d x^2} = \left(1 + \left(\frac{d^2 y}{d x^2}\right)^2\right)^{-1/2}$ are $k$ and $l$ respectively,then $k, l$ are the roots of

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