The order and degree of the differential equa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given differential equation is $3 x^2 \frac{d^2 y}{d x^2}-\sin \left(\frac{d^3 y}{d x^3}\right)+\cos (x y)=0$. The order of a differential equ

Vedclass · Accepted Answer

Order is $3$ and degree can't be defined

Vedclass · Answer

(B) The given differential equation is $3 x^2 \frac{d^2 y}{d x^2}-\sin \left(\frac{d^3 y}{d x^3}\right)+\cos (x y)=0$. The order of a differential equation is the highest order derivative present in the equation. Here,the highest order derivative is $\frac{d^3 y}{d x^3}$,so the order is $3$. The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in derivatives. Since the term $\sin \left(\frac{d^3 y}{d x^3}\right)$ involves a transcendental function of a derivative,it cannot be expressed as a polynomial in derivatives. Therefore,the degree of this differential equation is not defined.

The order and degree of the differential equation $3 x^2 \frac{d^2 y}{d x^2}-\sin \left(\frac{d^3 y}{d x^3}\right)+\cos (x y)=0$ are

Similar Questions

Which of the following equations is non-linear?

The number of arbitrary constants that appear in the general solution of the differential equation $\left(\frac{d^4 y}{d x^4}+\frac{d^2 y}{d x^2}\right)^{3 / 2}=5 \frac{d^3 y}{d x^3}$ is

Find the degree of the differential equation $y_3^{2/3} + 2 + 3y_2 + y_1 = 0$.

The degree of the differential equation $y = a(1 - e^{-x/a})$,where $a$ is a parameter,is:

The order and degree of the differential equation $\frac{dy}{dx} = \left(\frac{d^2y}{dx^2} + 2\right)^{1/2} + \frac{d^2y}{dx^2} + 5$ are respectively

Vedclass Test Series

Exam Paper Generator

Online Exam Module