The order of $\sqrt{1+\left(\frac{dy}{dx}\right)^2} = \left[a\frac{d^2y}{dx^2}\right]^{\frac{1}{3}}$ is

2

The correct answer is Option (2) → 2

$\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \left[a \frac{d^2y}{dx^2} \right]^{\frac{1}{3}}$

To determine the order, all fractional powers must be removed.

Raise both sides to the power $6$:

$\left(\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\right)^6 = \left(\left[a \frac{d^2y}{dx^2} \right]^{\frac{1}{3}}\right)^6$

$\left(1 + \left(\frac{dy}{dx}\right)^2\right)^3 = a^2 \left(\frac{d^2y}{dx^2}\right)^2$

Now all derivatives are in polynomial form with integer powers.

Definition: The order of a differential equation is the order (i.e., the highest number of differentiations) of the highest order derivative present in the equation.

The highest order derivative present is $\frac{d^2y}{dx^2}$.

Order = $2$