Match List - I with List - II. Match the integrating factors:
| List - I | List - II | ||
| (Differential Equation) | (Integrating factor) | ||
| (A) | \(\displaystyle\frac{dy}{dx} + 3y = e^{-2x}\) | (I) | \(\displaystyle\frac{1}{x}\) |
| (B) | \(x\displaystyle\frac{dy}{dx} + y = 3x^2\) | (II) | \(e^{-x}\) |
| (C) | \(x\displaystyle\frac{dy}{dx} - y = 3x^2\) | (III) | \(x\) |
| (D) | \(\displaystyle\frac{dy}{dx} - y = x\) | (IV) | \(e^{3x}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
For a linear equation \(\frac{dy}{dx} + Py = Q\), the integrating factor is \(e^{\int P\,dx}\). Each equation must first be put in this form (coefficient of \(\frac{dy}{dx}\) equal to 1).
(A) \(\frac{dy}{dx} + 3y = e^{-2x}\)
This matches (IV) \(e^{3x}\).
\(P = 3,\quad e^{\int 3\,dx} = e^{3x}\)
(B) \(x\frac{dy}{dx} + y = 3x^2\)
This matches (III) \(x\).
\( \displaystyle \frac{dy}{dx} + \frac{1}{x}\,y = 3x,\quad P = \frac{1}{x} \)
\(e^{\int \frac{1}{x}dx} = e^{\log x} = x\)
(C) \(x\frac{dy}{dx} - y = 3x^2\)
This matches (I) \(\frac{1}{x}\).
\( \displaystyle \frac{dy}{dx} - \frac{1}{x}\,y = 3x,\quad P = -\frac{1}{x} \)
\( \displaystyle e^{-\int \frac{1}{x}dx} = e^{-\log x} = \frac{1}{x} \)
(D) \(\frac{dy}{dx} - y = x\)
This matches (II) \(e^{-x}\).
\(P = -1,\quad e^{\int (-1)\,dx} = e^{-x}\)