Which of the following differential equations are of order 2 and degree 1?
(A). \( \displaystyle\frac{d}{dx}\left(\displaystyle\frac{dy}{dx}\right)+x\ \displaystyle\frac{dy}{dx}=y \)
(B). \( \displaystyle\frac{d^2y}{dx^2}+x\ =\sqrt{\displaystyle\frac{dy}{dx}-y} \)
(C). \( \left(\displaystyle\frac{dy}{dx}\right)^2+\displaystyle\frac{d^2y}{d\,x^2}=\ \log x \)
(D). \( \left(\displaystyle\frac{dy}{dx}\right)^2+\displaystyle\frac{dy}{dx}=3 \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
The order is the highest derivative present. The degree is the power of that highest derivative, once the equation is free of roots and fractional powers of derivatives.
In (A), the first term is a second derivative:
\( \displaystyle \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d^2y}{dx^2} \)
In (B), squaring removes the root:
\( \displaystyle \left(\frac{d^2y}{dx^2}+x\right)^2=\frac{dy}{dx}-y \)
So the order and degree are: (A) 2 and 1, (B) 2 and 2, (C) 2 and 1, (D) 1 and 2.
Checking the statements
(A) \( \displaystyle\frac{d}{dx}\left(\displaystyle\frac{dy}{dx}\right)+x\ \displaystyle\frac{dy}{dx}=y \) — Correct. Order 2, degree 1.
(B) \( \displaystyle\frac{d^2y}{dx^2}+x\ =\sqrt{\displaystyle\frac{dy}{dx}-y} \) — Incorrect. After squaring, \( \frac{d^2y}{dx^2} \) appears squared, so the degree is 2.
(C) \( \left(\displaystyle\frac{dy}{dx}\right)^2+\displaystyle\frac{d^2y}{d\,x^2}=\ \log x \) — Correct. Order 2, degree 1; the square is on the lower derivative.
(D) \( \left(\displaystyle\frac{dy}{dx}\right)^2+\displaystyle\frac{dy}{dx}=3 \) — Incorrect. Only \( \frac{dy}{dx} \) appears, so the order is 1.