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 $\left[1+\left(\frac{dy}{dx}\right)^{5}\right]^{\frac{1}{3}}=\frac{d^{2}y}{dx^{2}}$.To find the degree,we must

Vedclass · Accepted Answer

$2, 3$

Vedclass · Answer

(D) The given differential equation is $\left[1+\left(\frac{dy}{dx}\right)^{5}\right]^{\frac{1}{3}}=\frac{d^{2}y}{dx^{2}}$.To find the degree,we must eliminate the fractional exponent by raising both sides to the power of $3$:$\left[\left[1+\left(\frac{dy}{dx}\right)^{5}\right]^{\frac{1}{3}}\right]^{3}=\left(\frac{d^{2}y}{dx^{2}}\right)^{3}$.This simplifies to $1+\left(\frac{dy}{dx}\right)^{5}=\left(\frac{d^{2}y}{dx^{2}}\right)^{3}$.The order of a differential equation is the highest derivative present,which is $\frac{d^{2}y}{dx^{2}}$,so the order is $2$.The degree is the power of the highest derivative after the equation is made free from radicals and fractions,which is $3$.Therefore,the order and degree are $2$ and $3$ respectively.

The order and degree of the differential equation $\left[1+\left(\frac{dy}{dx}\right)^{5}\right]^{\frac{1}{3}}=\frac{d^{2}y}{dx^{2}}$ are respectively

Similar Questions

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

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

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

The differential equation representing the family of curves $y^{2}=2 d(x+\sqrt{d})$, where $d$ is a parameter, is of

The order and degree of the differential equation $\sqrt{1 + \frac{1}{(dy/dx)^2}} = (d^2y/dx^2)^{3/2}$ are, respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module