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

(D) The given differential equation is $\frac{d^2y}{dx^2} = \cos \left( \frac{dy}{dx} \right) + xy$. The order of a differential equation is the order

Vedclass · Accepted Answer

$2, 	ext{not defined}$

Vedclass · Answer

(D) The given differential equation is $\frac{d^2y}{dx^2} = \cos \left( \frac{dy}{dx} \right) + xy$. The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^2y}{dx^2}$,so the order is $2$. The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in its derivatives. In this equation,the term $\cos \left( \frac{dy}{dx} \right)$ involves a transcendental function of the derivative $\frac{dy}{dx}$,which means the equation cannot be expressed as a polynomial in its derivatives. Therefore,the degree of the differential equation is not defined.

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

Similar Questions

The order and degree of the differential equation $\left(\frac{d^3 y}{d x^3}\right)^{\frac{1}{2}}-2\left(\frac{d y}{d x}\right)^{\frac{1}{4}}+x y=0$ are respectively

The degree of the differential equation $\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^3 + 6y = 0$ is

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

The degree of the differential equation $\frac{d^2y}{dx^2} + \sqrt{1 + \left( \frac{dy}{dx} \right)^3} = 0$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module