The order and degree of the differential equa | vedclass

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

Question

(A) The given differential equation is $\left(y^{\prime \prime \prime}\right)^3+\left(y^{\prime \prime}\right)^4+\left(y^{\prime}\right)^4+y=7$. The o

Vedclass · Accepted Answer

$3$ and $3$

Vedclass · Answer

(A) The given differential equation is $\left(y^{\prime \prime \prime}\right)^3+\left(y^{\prime \prime}\right)^4+\left(y^{\prime}\right)^4+y=7$. The order of a differential equation is the order of the highest derivative present in the equation. Here,the highest derivative is $y^{\prime \prime \prime}$,which is of order $3$. Thus,the order of the differential equation is $3$. The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in derivatives. The highest order derivative is $y^{\prime \prime \prime}$ and its exponent is $3$. Therefore,the degree of the differential equation is $3$. Hence,the order and degree are $3$ and $3$ respectively.

The order and degree of the differential equation $\left(y^{\prime \prime \prime}\right)^3+\left(y^{\prime \prime}\right)^4+\left(y^{\prime}\right)^4+y=7$ are . . . . . . respectively.

Similar Questions

The number of arbitrary constants in the particular solution of a differential equation of third order is:

Find the order and degree,if defined,of the following differential equation:
$\frac{dy}{dx} - \cos x = 0$

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 and degree of the differential equation $(\frac{d^3y}{dx^3}+y\frac{dy}{dx})^{\frac{7}{5}} =x^3\frac{d^2y}{dx^2}$ are $m$ and $n$ respectively,then $(m + n)$ is equal to-

Determine the order and degree (if defined) of the differential equation $y^{\prime} + y = e^{x}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module