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

(A) The given differential equation is $(\frac{d^3y}{dx^3})^2 + (\frac{d^2y}{dx^2})^3 + (\frac{dy}{dx})^4 + y^5 = 0$. The order of a differential equa

Vedclass · Accepted Answer

Order $3$,Degree $2$

Vedclass · Answer

(A) The given differential equation is $(\frac{d^3y}{dx^3})^2 + (\frac{d^2y}{dx^2})^3 + (\frac{dy}{dx})^4 + y^5 = 0$. The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^3y}{dx^3}$,so the order is $3$. The degree of a differential equation is the power of the highest order derivative,provided the equation is a polynomial in its derivatives. Here,the highest order derivative $\frac{d^3y}{dx^3}$ is raised to the power of $2$. Therefore,the degree is $2$.

Determine the order and degree (if defined) of the differential equation: $(\frac{d^3y}{dx^3})^2 + (\frac{d^2y}{dx^2})^3 + (\frac{dy}{dx})^4 + y^5 = 0$.

Similar Questions

For the differential equation $\frac{d^3y}{dx^3} + \cos(\frac{d^2y}{dx^2}) = 0$, which of the following is true?

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

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

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

The order of the differential equation $y \left( \frac{dy}{dx} \right) = \frac{x}{\frac{dy}{dx} + \left( \frac{dy}{dx} \right)^3}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module