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 $\sqrt{\frac{dy}{dx}} - 4\frac{dy}{dx} - 7x = 0$. Rearranging the terms,we get $\sqrt{\frac{dy}{dx}} = 4\frac{dy}{d

Vedclass · Accepted Answer

$1$ and $2$

Vedclass · Answer

(D) 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 remove the radical sign,we get: $\frac{dy}{dx} = (4\frac{dy}{dx} + 7x)^2$ $\frac{dy}{dx} = 16(\frac{dy}{dx})^2 + 49x^2 + 56x\frac{dy}{dx}$. The highest order derivative present is $\frac{dy}{dx}$,so the order is $1$. The exponent of the highest order derivative after making the equation a polynomial in derivatives 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 number of arbitrary constants in the general solution of a differential equation of fourth order is:

Which of the following equations is non-linear?

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:

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

The differential equation whose solution is $Ax^2 + By^2 = 1$,where $A$ and $B$ are arbitrary constants,is of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module