Match List-I with List-II (C is an arbitrary constant)
| List-I | List-II | ||
|---|---|---|---|
| Family of curves | Differential equation | ||
| (A) | \(y = Ce^{-x^2}\) | (I) | \(x\displaystyle\frac{dy}{dx} + y = 0\) |
| (B) | \(y = C(x + 1)^2\) | (II) | \(y\displaystyle\frac{dy}{dx} = x\) |
| (C) | \(xy = C\) | (III) | \((x + 1)\displaystyle\frac{dy}{dx} = 2y\) |
| (D) | \(y^2 - x^2 = C\) | (IV) | \(\displaystyle\frac{dy}{dx} = -2xy\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Each family has one arbitrary constant. Differentiating once and eliminating C gives its differential equation.
(A) \(y = Ce^{-x^2}\)
This matches (IV) \(\frac{dy}{dx} = -2xy\):
\( \displaystyle \frac{dy}{dx} = -2xCe^{-x^2} = -2xy \)
(B) \(y = C(x + 1)^2\)
This matches (III) \((x + 1)\frac{dy}{dx} = 2y\):
\( \displaystyle \frac{dy}{dx} = 2C(x + 1) \)
\( \displaystyle (x + 1)\frac{dy}{dx} = 2C(x + 1)^2 = 2y \)
(C) \(xy = C\)
This matches (I) \(x\frac{dy}{dx} + y = 0\). By the product rule:
\( \displaystyle x\frac{dy}{dx} + y = 0 \)
(D) \(y^2 - x^2 = C\)
This matches (II) \(y\frac{dy}{dx} = x\):
\( \displaystyle 2y\frac{dy}{dx} - 2x = 0 \)
\( \displaystyle y\frac{dy}{dx} = x \)