Determine the order and degree (if defined) o | vedclass

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

Question

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

Vedclass · Accepted Answer

Order $2$,Degree not defined

Vedclass · Answer

(C) The given differential equation is $\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0$. The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^{2} y}{d x^{2}}$,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 $\cos \left(\frac{d y}{d x}\right)$ involves a transcendental function of the derivative,which means the equation cannot be expressed as a polynomial in terms of its derivatives. Therefore,the degree of this differential equation is not defined.

Determine the order and degree (if defined) of the differential equation $\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0$.

Similar Questions

The order and degree of the differential equation $\sqrt{\frac{dy}{dx}} - 4 \frac{dy}{dx} - 7x = 0$ are

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:

Find the order and degree,if defined,of the following differential equation: $y^{\prime \prime \prime} + y^2 + e^{y^{\prime}} = 0$

The order of the differential equation of all parabolas having their directrix parallel to the $x$-axis is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module