Match each differential equation in List-I with its general solution in List-II (C is an arbitrary constant, \(x \ne 0\), \(y \ne 0\)).
| List-I | List-II | ||
| (A) | \(\displaystyle\frac{dy}{dx} = 3y\) | (I) | \(xy = C\) |
| (B) | \(\displaystyle\frac{dy}{dx} = -\frac{y}{x}\) | (II) | \(y = Cx^2\) |
| (C) | \(\displaystyle\frac{dy}{dx} = \frac{2y}{x}\) | (III) | \(y = Ce^{3x}\) |
| (D) | \(\displaystyle\frac{dy}{dx} = \frac{x}{4y}\) | (IV) | \(4y^2 - x^2 = C\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
Each equation is variables separable.
(A) \(\frac{dy}{dx} = 3y\)
This matches (III) \(y = Ce^{3x}\).
\( \displaystyle \frac{dy}{y} = 3\,dx \)
\(\log|y| = 3x + C_1\), so \(y = Ce^{3x}\)
(B) \(\frac{dy}{dx} = -\frac{y}{x}\)
This matches (I) \(xy = C\).
\( \displaystyle \frac{dy}{y} = -\frac{dx}{x} \)
\(\log|y| + \log|x| = C_1\), so \(xy = C\)
(C) \(\frac{dy}{dx} = \frac{2y}{x}\)
This matches (II) \(y = Cx^2\).
\( \displaystyle \frac{dy}{y} = \frac{2\,dx}{x} \)
\(\log|y| = 2\log|x| + C_1\), so \(y = Cx^2\)
(D) \(\frac{dy}{dx} = \frac{x}{4y}\)
This matches (IV) \(4y^2 - x^2 = C\).
\(4y\,dy = x\,dx\)
\( \displaystyle 2y^2 = \frac{x^2}{2} + C_1\), so \(4y^2 - x^2 = C\)