The degree of the differential equation left( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given differential equation is $\left(\frac{d^2 y}{d x^2}\right)^{\frac{4}{3}}+x\left(\frac{d y}{d x}\right)^2-y \cos \left(\frac{d y}{d x}\ri

Vedclass · Accepted Answer

not defined

Vedclass · Answer

(D) The given differential equation is $\left(\frac{d^2 y}{d x^2}\right)^{\frac{4}{3}}+x\left(\frac{d y}{d x}\right)^2-y \cos \left(\frac{d y}{d x}\right)=0$. To define the degree of a differential equation, it must be a polynomial equation in terms of its derivatives. In this equation, the term $\cos \left(\frac{d y}{d x}\right)$ involves a transcendental function of the derivative, which means the equation cannot be expressed as a polynomial in terms of its derivatives. Therefore, the degree of this differential equation is not defined.

The degree of the differential equation $\left(\frac{d^2 y}{d x^2}\right)^{\frac{4}{3}}+x\left(\frac{d y}{d x}\right)^2-y \cos \left(\frac{d y}{d x}\right)=0$ is

Similar Questions

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

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

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

Determine the order and degree (if defined) of the differential equation $y''' + 2y'' + y' = 0$.

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module