Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Differential equation | Integrating factor | ||
| (A) | \(x\displaystyle\frac{dy}{dx}-y = 2x^2\) | (I) | \(e^{-y}\) |
| (B) | \(\displaystyle\frac{dy}{dx}+\frac{y}{x}=2x\) | (II) | \(\displaystyle\frac{1}{x}\) |
| (C) | \(x\displaystyle\frac{dy}{dx}+ 2y = x^2\log x\) | (III) | \(x\) |
| (D) | \(\displaystyle\frac{dx}{dy}-x=y\) | (IV) | \(x^2\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
For a linear equation \(\frac{dy}{dx} + Py = Q\), the integrating factor is \(e^{\int P\,dx}\); for \(\frac{dx}{dy} + P_1 x = Q_1\), it is \(e^{\int P_1\,dy}\). The equation must first be written with coefficient 1 for the derivative.
(A) \(x\displaystyle\frac{dy}{dx}-y = 2x^2\)
This matches (II) 1/x. Dividing by x:
\(\displaystyle \frac{dy}{dx} - \frac{1}{x}\,y = 2x\)
\(\displaystyle \text{I.F.} = e^{-\int \frac{1}{x}dx} = e^{-\log x} = \frac{1}{x}\)
(B) \(\displaystyle\frac{dy}{dx}+\frac{y}{x}=2x\)
This matches (III) x. Here \(P = \frac{1}{x}\):
\(\displaystyle \text{I.F.} = e^{\int \frac{1}{x}dx} = e^{\log x} = x\)
(C) \(x\displaystyle\frac{dy}{dx}+ 2y = x^2\log x\)
This matches (IV) x2. Dividing by x:
\(\displaystyle \frac{dy}{dx} + \frac{2}{x}\,y = x\log x\)
\(\displaystyle \text{I.F.} = e^{\int \frac{2}{x}dx} = e^{2\log x} = x^2\)
(D) \(\displaystyle\frac{dx}{dy}-x=y\)
This matches (I) e−y. The equation is linear in x with \(P_1 = -1\):
\(\displaystyle \text{I.F.} = e^{\int (-1)\,dy} = e^{-y}\)
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).