Match each differential equation with the given condition in List-I with its particular solution in List-II.
| List-I | List-II | ||
| (A) | \(\displaystyle\frac{dy}{dx} = 2x + 1\), \(y(0) = 1\) | (I) | \(y = e^{x} + 1\) |
| (B) | \(\displaystyle\frac{dy}{dx} = y\), \(y(0) = 2\) | (II) | \(y = 1 + \sin x\) |
| (C) | \(\displaystyle\frac{dy}{dx} = e^{x}\), \(y(0) = 2\) | (III) | \(y = x^2 + x + 1\) |
| (D) | \(\displaystyle\frac{dy}{dx} = \cos x\), \(y(0) = 1\) | (IV) | \(y = 2e^{x}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
In each case the general solution is found first, and the condition then fixes the constant.
(A) \(\frac{dy}{dx} = 2x + 1\), y(0) = 1
This matches (III) \(y = x^2 + x + 1\).
\(y = x^2 + x + C\), and y(0) = 1 gives C = 1
(B) \(\frac{dy}{dx} = y\), y(0) = 2
This matches (IV) \(y = 2e^{x}\).
\( \displaystyle \frac{dy}{y} = dx \), so \(y = Ce^{x}\), and y(0) = 2 gives C = 2
(C) \(\frac{dy}{dx} = e^{x}\), y(0) = 2
This matches (I) \(y = e^{x} + 1\).
\(y = e^{x} + C\), and y(0) = 2 gives \(1 + C = 2\), so C = 1
(D) \(\frac{dy}{dx} = \cos x\), y(0) = 1
This matches (II) \(y = 1 + \sin x\).
\(y = \sin x + C\), and y(0) = 1 gives C = 1
Both \(2e^{x}\) and \(e^{x} + 1\) equal 2 at x = 0, so the condition alone cannot separate (B) and (C): \(2e^{x}\) has derivative equal to itself, while \(e^{x} + 1\) has derivative \(e^{x}\).