Match List-I with List-II
| List-I (Differential equation) | List-II (Order, Degree) | ||
| (A) | \(\displaystyle\frac{d^2y}{dx^2} + y = \sin x\) | (I) | Order 1, Degree 2 |
| (B) | \(\left(\displaystyle\frac{dy}{dx}\right)^2 + y = x\) | (II) | Order 2, Degree 3 |
| (C) | \(\left(\displaystyle\frac{d^3y}{dx^3}\right)^2 + \frac{dy}{dx} = 0\) | (III) | Order 2, Degree 1 |
| (D) | \(\displaystyle\frac{d^2y}{dx^2} = \left(1 + \frac{dy}{dx}\right)^{1/3}\) | (IV) | Order 3, Degree 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
The order is the order of the highest derivative; the degree is its power once the equation is a polynomial in the derivatives.
(A) \(\frac{d^2y}{dx^2} + y = \sin x\)
This matches (III) Order 2, Degree 1.
The highest derivative \(\frac{d^2y}{dx^2}\) appears with power 1.
(B) \(\left(\frac{dy}{dx}\right)^2 + y = x\)
This matches (I) Order 1, Degree 2.
The only derivative is \(\frac{dy}{dx}\), and it is squared.
(C) \(\left(\frac{d^3y}{dx^3}\right)^2 + \frac{dy}{dx} = 0\)
This matches (IV) Order 3, Degree 2.
The highest derivative \(\frac{d^3y}{dx^3}\) is squared.
(D) \(\frac{d^2y}{dx^2} = \left(1 + \frac{dy}{dx}\right)^{1/3}\)
This matches (II) Order 2, Degree 3.
Cubing both sides removes the fractional power:
\( \displaystyle \left(\frac{d^2y}{dx^2}\right)^3 = 1 + \frac{dy}{dx} \)
The highest derivative \(\frac{d^2y}{dx^2}\) now has power 3.