The order and degree of the differential equation \( \sqrt{1+\left(\displaystyle\frac{dy}{dx}\right)^2}=\left(\displaystyle\frac{d^2y}{dx^2}\right)^{\frac{1}{3}} \) are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → order = 2, degree = 2
The highest derivative present is \( \frac{d^2y}{dx^2} \), so the order is 2.
For the degree, the equation must first be free of fractional powers of the derivatives. Raising both sides to the power 6 (the LCM of 2 and 3),
\( \displaystyle \left[1+\left(\frac{dy}{dx}\right)^2\right]^3=\left(\frac{d^2y}{dx^2}\right)^2 \)
This is now a polynomial in the derivatives, and the highest power of the highest-order derivative \( \frac{d^2y}{dx^2} \) is 2.
So order = 2 and degree = 2.
Why the other options are incorrect
Option (2) order = 2, degree = \( \displaystyle\frac{1}{3} \) — Incorrect. The power \( \frac{1}{3} \) is read before removing the radicals; a degree is always a positive integer.
Option (3) order = 3, degree is not defined — Incorrect. The highest derivative is the second derivative, so the order is 2, not 3; the degree is defined because the equation becomes a polynomial in the derivatives.
Option (4) order = 2, degree = 3 — Incorrect. Cubing both sides still leaves the square root on the left side; both radicals must be removed, which needs the power 6 and gives degree 2.