Match each family of curves in List-I, where C is an arbitrary constant, with the differential equation in List-II whose general solution it is.
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(y = x - 1 + Ce^{-x}\) | (I) | \(\displaystyle\frac{dy}{dx} + y = x\) |
| (B) | \(y = x + 1 + Ce^{x}\) | (II) | \(\displaystyle\frac{dy}{dx} - y = x\) |
| (C) | \(y = 1 - x + Ce^{-x}\) | (III) | \(\displaystyle\frac{dy}{dx} + y = -x\) |
| (D) | \(y = Ce^{x} - x - 1\) | (IV) | \(\displaystyle\frac{dy}{dx} - y = -x\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(I), (B)-(IV), (C)-(III), (D)-(II)
The term \(Ce^{-x}\) belongs to an equation with \(+y\) (integrating factor \(e^{x}\)), and \(Ce^{x}\) to one with \(-y\). Then substituting the family into \(\frac{dy}{dx} \pm y\) shows the right side; the C terms cancel.
(A) \(y = x - 1 + Ce^{-x}\)
This matches (I): \(y' = 1 - Ce^{-x}\), so \(y' + y = 1 + x - 1 = x\).
(B) \(y = x + 1 + Ce^{x}\)
This matches (IV): \(y' = 1 + Ce^{x}\), so \(y' - y = 1 - x - 1 = -x\).
(C) \(y = 1 - x + Ce^{-x}\)
This matches (III): \(y' = -1 - Ce^{-x}\), so \(y' + y = -1 + 1 - x = -x\).
(D) \(y = Ce^{x} - x - 1\)
This matches (II): \(y' = Ce^{x} - 1\), so \(y' - y = -1 + x + 1 = x\).