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 $\left(\frac{d^{2} y}{d x^{2}}\right)^{2} = \cos 3x + \sin 3x$.The order of a differential equation is the orde

Vedclass · Accepted Answer

Order $2$,Degree $2$

Vedclass · Answer

(A) The given differential equation is $\left(\frac{d^{2} y}{d x^{2}}\right)^{2} = \cos 3x + \sin 3x$.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 the given equation,the highest order derivative $\frac{d^{2} y}{d x^{2}}$ is raised to the power of $2$.Therefore,the degree is $2$.

Determine the order and degree (if defined) of the differential equation $\left(\frac{d^{2} y}{d x^{2}}\right)^{2} = \cos 3x + \sin 3x$.

Similar Questions

The order and degree of the differential equation $(\frac{d^3y}{dx^3})^{\frac{2}{3}} - 3\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 4 = 0$ are respectively:

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

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 . . . . . . .

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module