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

(D) Given differential equation is $\sqrt{\frac{dy}{dx}} - 4 \frac{dy}{dx} - 7x = 0$. To find the degree,we must eliminate the radical sign. Rewrite t

Vedclass · Accepted Answer

$1$ and $2$

Vedclass · Answer

(D) Given differential equation is $\sqrt{\frac{dy}{dx}} - 4 \frac{dy}{dx} - 7x = 0$. To find the degree,we must eliminate the radical sign. Rewrite the equation as $\sqrt{\frac{dy}{dx}} = 4 \frac{dy}{dx} + 7x$. Squaring both sides,we get: $(\frac{dy}{dx}) = (4 \frac{dy}{dx} + 7x)^2$. $(\frac{dy}{dx}) = 16(\frac{dy}{dx})^2 + 49x^2 + 56x \frac{dy}{dx}$. Rearranging the terms,we get $16(\frac{dy}{dx})^2 + (56x - 1)\frac{dy}{dx} + 49x^2 = 0$. 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

Similar Questions

Determine the order and degree (if defined) of the differential equation $\left(\frac{ds}{dt}\right)^{4} + 3s \frac{d^{2}s}{dt^{2}} = 0$.

The order of the differential equation,whose solution is $y=(C_1+C_2) e^x+C_3 e^{x+C_4}$,is

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

The degree of the differential equation $x\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{3}}+2 x^2\left(\frac{d^2 y}{d x^2}\right)^{\frac{5}{3}}+7 \frac{d y}{d x}+y=0$ is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module