The order of $\sqrt{1+\left(\frac{dy}{dx}\right)^2} = \left[a\frac{d^2y}{dx^2}\right]^{\frac{1}{3}}$ is
Answer & explanation
Correct answer: option 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 = $2$