Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Differential Equation | Integrating Factor | ||
| (A) | \(\displaystyle\frac{dy}{dx} + 2xy = 1\) | (I) | \(x\) |
| (B) | \(x\displaystyle\frac{dy}{dx} + 2xy = 1\) | (II) | \(e^{2x}\) |
| (C) | \(x\displaystyle\frac{dy}{dx} + y = 1\) | (III) | \(x^2\) |
| (D) | \(x\displaystyle\frac{dy}{dx} + 2y = 2\) | (IV) | \(e^{x^2}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
Each equation is put in the linear form \(\frac{dy}{dx} + Py = Q\). The integrating factor is then \(e^{\int P\,dx}\) (taking x > 0).
(A) \(\displaystyle\frac{dy}{dx} + 2xy = 1\)
This matches (IV) \(e^{x^2}\).
\(P = 2x\)
\(\text{IF} = e^{\int 2x\,dx} = e^{x^2}\)
(B) \(x\displaystyle\frac{dy}{dx} + 2xy = 1\)
This matches (II) \(e^{2x}\).
\(\displaystyle \frac{dy}{dx} + 2y = \frac{1}{x}\)
\(P = 2\)
\(\text{IF} = e^{\int 2\,dx} = e^{2x}\)
(C) \(x\displaystyle\frac{dy}{dx} + y = 1\)
This matches (I) \(x\).
\(\displaystyle \frac{dy}{dx} + \frac{1}{x}\,y = \frac{1}{x}\)
\(\displaystyle \text{IF} = e^{\int \frac{1}{x}dx}\)
\(= e^{\log x} = x\)
(D) \(x\displaystyle\frac{dy}{dx} + 2y = 2\)
This matches (III) \(x^2\).
\(\displaystyle \frac{dy}{dx} + \frac{2}{x}\,y = \frac{2}{x}\)
\(\displaystyle \text{IF} = e^{\int \frac{2}{x}dx}\)
\(= e^{2\log x} = x^2\)
So the matching is (A)-(IV), (B)-(II), (C)-(I), (D)-(III).