Match List-I with List-II, where \(x \gt 0\) and C is an arbitrary constant.
| List-I | List-II | ||
|---|---|---|---|
| Differential equation | General solution | ||
| (A) | \(\displaystyle\frac{dy}{dx} + \frac{y}{x} = 1\) | (I) | \(y = \displaystyle\frac{x}{3} + \frac{C}{x^2}\) |
| (B) | \(\displaystyle\frac{dy}{dx} + \frac{2y}{x} = 1\) | (II) | \(y = \displaystyle\frac{x^2}{3} + \frac{C}{x}\) |
| (C) | \(\displaystyle\frac{dy}{dx} + \frac{3y}{x} = 1\) | (III) | \(y = \displaystyle\frac{x}{2} + \frac{C}{x}\) |
| (D) | \(\displaystyle\frac{dy}{dx} + \frac{y}{x} = x\) | (IV) | \(y = \displaystyle\frac{x}{4} + \frac{C}{x^3}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
For \(\frac{dy}{dx} + \frac{n}{x}\,y = Q\), the integrating factor is
\( e^{n\log x} = x^n \)
and the solution is \(x^n y = \int Q\,x^n\,dx + C\).
(A) \(\frac{dy}{dx} + \frac{y}{x} = 1\)
This matches (III). With IF x:
\( \displaystyle xy = \frac{x^2}{2} + C \)
\( \displaystyle y = \frac{x}{2} + \frac{C}{x} \)
(B) \(\frac{dy}{dx} + \frac{2y}{x} = 1\)
This matches (I). With IF \(x^2\):
\( \displaystyle x^2y = \frac{x^3}{3} + C \)
\( \displaystyle y = \frac{x}{3} + \frac{C}{x^2} \)
(C) \(\frac{dy}{dx} + \frac{3y}{x} = 1\)
This matches (IV). With IF \(x^3\):
\( \displaystyle x^3y = \frac{x^4}{4} + C \)
\( \displaystyle y = \frac{x}{4} + \frac{C}{x^3} \)
(D) \(\frac{dy}{dx} + \frac{y}{x} = x\)
This matches (II). With IF x:
\( \displaystyle xy = \int x^2\,dx = \frac{x^3}{3} + C \)
\( \displaystyle y = \frac{x^2}{3} + \frac{C}{x} \)