The degree of the differential equation frac{ | vedclass

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

Question

(B) Given the differential equation: $\frac{d^2y}{dx^2} + \sqrt{1 + \left( \frac{dy}{dx} \right)^3} = 0$Rearrange the equation to isolate the radical

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given the differential equation: $\frac{d^2y}{dx^2} + \sqrt{1 + \left( \frac{dy}{dx} \right)^3} = 0$Rearrange the equation to isolate the radical term:$\frac{d^2y}{dx^2} = - \sqrt{1 + \left( \frac{dy}{dx} \right)^3}$To eliminate the square root,square both sides of the equation:$\left( \frac{d^2y}{dx^2} \right)^2 = 1 + \left( \frac{dy}{dx} \right)^3$The degree of a differential equation is the highest power of the highest-order derivative when the equation is expressed as a polynomial in its derivatives.Here,the highest-order derivative is $\frac{d^2y}{dx^2}$,and its power is $2$.Therefore,the degree of the differential equation is $2$.

The degree of the differential equation $\frac{d^2y}{dx^2} + \sqrt{1 + \left( \frac{dy}{dx} \right)^3} = 0$ is

Similar Questions

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

The order and degree of the differential equation $x\frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 + y^2 = 0$ are respectively:

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

If $m$ and $n$ are the degree and order of the differential equation $\left(1+y_{1}^{2}\right)^{2 / 3}=y_{2}$,then the value of $\frac{m+n}{m-n}$ is

The differential equation $\frac{d^2y}{dx^2} + x\frac{dy}{dx} + \sin y + x^2 = 0$ is of the following type:

Vedclass Test Series

Exam Paper Generator

Online Exam Module