Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Differential Equation | Integrating Factor | ||
| (A) | \(y\,dx + (x - y^3)dy = 0\) | (I) | \(e^{-x}\) |
| (B) | \(x\displaystyle\frac{dy}{dx} + y = x^2\) | (II) | \(\displaystyle\frac{1}{x}\) |
| (C) | \(\displaystyle\frac{dy}{dx} - y = e^x\) | (III) | \(y\) |
| (D) | \(x\,dy - y\,dx = x^3dx\) | (IV) | \(x\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
For a linear differential equation \(\frac{dy}{dx} + Py = Q\), the integrating factor is \(e^{\int P\,dx}\). When the equation is linear in x, it is written as \(\frac{dx}{dy} + P_1x = Q_1\) and the integrating factor is \(e^{\int P_1\,dy}\).
(A) \(y\,dx + (x - y^3)dy = 0\)
This matches (III) \(y\). The equation is linear in x:
\(\displaystyle y\frac{dx}{dy} + x - y^3 = 0\)
\(\displaystyle \frac{dx}{dy} + \frac{1}{y}x = y^2\)
\(\displaystyle \text{I.F.} = e^{\int \frac{1}{y}\,dy} = e^{\log y} = y\)
(B) \(x\frac{dy}{dx} + y = x^2\)
This matches (IV) \(x\). Dividing by x:
\(\displaystyle \frac{dy}{dx} + \frac{1}{x}y = x\)
\(\displaystyle \text{I.F.} = e^{\int \frac{1}{x}\,dx} = e^{\log x} = x\)
(C) \(\frac{dy}{dx} - y = e^x\)
This matches (I) \(e^{-x}\). Here P = −1:
\(\displaystyle \text{I.F.} = e^{\int (-1)\,dx} = e^{-x}\)
(D) \(x\,dy - y\,dx = x^3dx\)
This matches (II) \(\frac{1}{x}\). Dividing by x dx:
\(\displaystyle \frac{dy}{dx} - \frac{1}{x}y = x^2\)
\(\displaystyle \text{I.F.} = e^{-\int \frac{1}{x}\,dx} = e^{-\log x} = \frac{1}{x}\)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).