The degree of the differential equation frac{ | vedclass

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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

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