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

(B) Given differential equation is $\sqrt{\frac{dy}{dx}}-4 \frac{dy}{dx}-7x=0$. Rearranging the terms,we get $\sqrt{\frac{dy}{dx}}=4 \frac{dy}{dx}+7x$

Vedclass · Accepted Answer

$1$ and $2$

Vedclass · Answer

(B) Given differential equation is $\sqrt{\frac{dy}{dx}}-4 \frac{dy}{dx}-7x=0$. Rearranging the terms,we get $\sqrt{\frac{dy}{dx}}=4 \frac{dy}{dx}+7x$. Squaring both sides to eliminate the radical,we obtain $\frac{dy}{dx} = (4 \frac{dy}{dx} + 7x)^2$. Expanding the right side,we get $\frac{dy}{dx} = 16(\frac{dy}{dx})^2 + 56x(\frac{dy}{dx}) + 49x^2$. The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$. The highest power of the highest order derivative is $2$,so the degree is $2$. Thus,the order is $1$ and the degree is $2$.

The order and degree of the differential equation $\sqrt{\frac{dy}{dx}}-4 \frac{dy}{dx}-7x=0$ are respectively.

Similar Questions

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

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

The degree of the given differential equation $\left( \frac{d^2y}{dx^2} \right)^3 = \left( 1 + \frac{dy}{dx} \right)^{1/2}$ is:

The order of the differential equation,whose general solution is given by $y = (c_1 + c_2) \cos (x + c_3) - c_4 e^{x + c_5}$,where $c_1, c_2, c_3, c_4$ and $c_5$ are arbitrary constants,is

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 . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module