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 degree of the differential equation $x\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{3}}+2 x^2\left(\frac{d^2 y}{d x^2}\right)^{\frac{5}{3}}+7 \frac{d y}{d x}+y=0$ is:

The order and degree of the differential equation $\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[3]{\left(\frac{d y}{d x}\right)^4+2}$ are . . . . . . and . . . . . . .

The order of the differential equation whose solution is ${x^2} + {y^2} + 2gx + 2fy + c = 0$ is:

The order and degree of the differential equation $y = x \frac{dp}{dx} + \sqrt{a^{2} p^{2} + b^{2}}$,where $p = \frac{dy}{dx}$ (here $a$ and $b$ are arbitrary constants) respectively are:

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