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

(C) To find the order and degree of the differential equation,we must first eliminate the fractional exponent by raising both sides to the power of $3

Vedclass · Accepted Answer

$3, 3$

Vedclass · Answer

(C) To find the order and degree of the differential equation,we must first eliminate the fractional exponent by raising both sides to the power of $3$.Given equation: ${\left( {1 + 3\frac{{dy}}{{dx}}} \right)^{\frac{2}{3}}} = 4\frac{{{d^3}y}}{{d{x^3}}}$Cubing both sides:${\left( {1 + 3\frac{{dy}}{{dx}}} \right)^2} = {\left( {4\frac{{{d^3}y}}{{d{x^3}}}} \right)^3}$Now,the equation is a polynomial in terms of derivatives.The order of a differential equation is the highest derivative present,which 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. Here,the power of $\frac{{{d^3}y}}{{d{x^3}}}$ is $3$.Thus,the order is $3$ and the degree is $3$.

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

Similar Questions

Order of the differential equation of the family of all concentric circles centered at $(h, k)$ is

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

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

Determine the order and degree (if defined) of the differential equation $y^{\prime} + y = e^{x}$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module