Match List-I with List-II
| List-I | List-II | ||
| Order and Degree | Differential Equations | ||
| (A) | order=2, degree=1 | (I) | \( \left(\displaystyle\frac{d^{2}y}{dx^{2}}\right)^{3}+e^{x}=x \) |
| (B) | order=2, degree=2 | (II) | \( x\,\displaystyle\frac{d^{2}y}{dx^{2}}=\left\{1+\left(\displaystyle\frac{dy}{dx}\right)^{2}\right\}^{3} \) |
| (C) | order=2, degree=3 | (III) | \( \displaystyle\frac{d}{dx}\left(\displaystyle\frac{dy}{dx}\right)+\log\left(\displaystyle\frac{dy}{dx}\right)=y \) |
| (D) | order=2, degree=not defined | (IV) | \( \left\{\displaystyle\frac{d}{dx}\left(\displaystyle\frac{dy}{dx}\right)+y\right\}^{2}=\displaystyle\frac{dy}{dx}+x \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
The order is 2 in every equation, since \( \frac{d^{2}y}{dx^{2}} \) (also written \( \frac{d}{dx}\left(\frac{dy}{dx}\right) \)) is always the highest derivative.
The degree is the power of that highest derivative once the equation is a polynomial in the derivatives. So the degree decides each match.
(A) order=2, degree=1
This matches (II).
In (II) the highest derivative \( \frac{d^{2}y}{dx^{2}} \) appears only to the first power.
(B) order=2, degree=2
This matches (IV).
In (IV) the square, when expanded, gives \( \left(\frac{d^{2}y}{dx^{2}}\right)^{2} \).
(C) order=2, degree=3
This matches (I).
In (I) the highest derivative is cubed.
(D) order=2, degree=not defined
This matches (III).
In (III) a derivative sits inside a logarithm, so the equation is not a polynomial in its derivatives.
So the matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).