Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Differential Equations | Order and degree | ||
| (A) | \(y\,dx + x\log(y/x)\,dy - 2x\,dy = 0\) | (I) | Order: 2, degree: 1 |
| (B) | \(\displaystyle \left(\frac{d^3y}{dx^3}\right)^2 + 3\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^4 = y^2\) | (II) | Order: 1, degree: 1 |
| (C) | \(\displaystyle \frac{dy}{dx} + \log\left(\frac{dy}{dx}\right) + x = y\) | (III) | Order: 3, degree: 2 |
| (D) | \(\displaystyle \left(\frac{ds}{dt}\right)^4 + 2s\frac{d^2s}{dt^2} = 0\) | (IV) | Order: 1, degree: Not defined |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
The order is the order of the highest derivative present. The degree is the power of that highest derivative, and it is defined only when the equation is a polynomial in its derivatives.
(A) \(y\,dx + x\log(y/x)\,dy - 2x\,dy = 0\)
This matches (II) Order: 1, degree: 1. Dividing by dx:
\( \displaystyle y + \left(x\log\frac{y}{x} - 2x\right)\frac{dy}{dx} = 0 \)
The only derivative is \(\frac{dy}{dx}\), to the power 1. The log contains only x and y, not a derivative.
(B) \(\displaystyle \left(\frac{d^3y}{dx^3}\right)^2 + 3\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right)^4 = y^2\)
This matches (III) Order: 3, degree: 2. The highest derivative is \(\frac{d^3y}{dx^3}\), and its power is 2. The power 4 belongs to a lower derivative and does not decide the degree.
(C) \(\displaystyle \frac{dy}{dx} + \log\left(\frac{dy}{dx}\right) + x = y\)
This matches (IV) Order: 1, degree: Not defined. The highest derivative is \(\frac{dy}{dx}\), but it appears inside a logarithm, so the equation is not a polynomial in \(\frac{dy}{dx}\).
(D) \(\displaystyle \left(\frac{ds}{dt}\right)^4 + 2s\frac{d^2s}{dt^2} = 0\)
This matches (I) Order: 2, degree: 1. The highest derivative is \(\frac{d^2s}{dt^2}\), to the power 1.
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).