The order of \(\sqrt{1 + \left(\displaystyle\frac{dy}{dx}\right)^2} = \left[a\displaystyle\frac{d^2y}{dx^2}\right]^{\frac{1}{3}}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
The order of a differential equation is the order of the highest derivative in it.
The derivatives present are \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\). The highest is \(\frac{d^2y}{dx^2}\), so the order is 2.
The square root and the power \(\frac{1}{3}\) affect only the degree, never the order.