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 $3 - (\frac{d^3y}{dx^3})^{7/3} = (\frac{dy}{dx})^5$ are respectively

The number of arbitrary constants that appear in the general solution of the differential equation $\left(\frac{d^4 y}{d x^4}+\frac{d^2 y}{d x^2}\right)^{3 / 2}=5 \frac{d^3 y}{d x^3}$ is

The order and degree of the differential equation $\frac{d^3 y}{d x^3} = \left[1 + \left(\frac{d y}{d x}\right)^2\right]^{5/2}$ are respectively:

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module