The differential equation frac{d^2y}{dx^2} + | vedclass

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

Question

(D) The given differential equation is $\frac{d^2y}{dx^2} + x\frac{dy}{dx} + \sin y + x^2 = 0$.$1$. The order of a differential equation is the order

Vedclass · Accepted Answer

Both $(b)$ and $(c)$

Vedclass · Answer

(D) The given differential equation is $\frac{d^2y}{dx^2} + x\frac{dy}{dx} + \sin y + x^2 = 0$.$1$. The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $\frac{d^2y}{dx^2}$,so the order is $2$.$2$. The degree of a differential equation is the power of the highest derivative when the equation is expressed as a polynomial in derivatives. Since $\sin y$ is present,the equation is non-linear,but the highest derivative $\frac{d^2y}{dx^2}$ has an exponent of $1$. Thus,the degree is $1$.Therefore,the equation is of order $2$ and degree $1$. Hence,option $(d)$ is correct.

The differential equation $\frac{d^2y}{dx^2} + x\frac{dy}{dx} + \sin y + x^2 = 0$ is of the following type:

Similar Questions

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

For the differential equation $\frac{d^3 y}{d x^3}=0$,$y=a x^2+b x+c$ is

Order of the differential equation of the family of all concentric circles centered at $(h, k)$ 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 . . . . . . .

Degree of the differential equation $e^{\frac{dy}{dx}} + (\frac{dy}{dx})^3 = x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module