Match List-I with List-II
| List-I | List-II | ||
| Differential equation | Integrating factor | ||
| (A) | \( \displaystyle\frac{xdy}{dx}-y=x^2e^x \) | (I) | \( x \) |
| (B) | \( \displaystyle\frac{dy}{dx}-2y=e^x \) | (II) | \( x^4 \) |
| (C) | \( \displaystyle\frac{xdy}{dx}+y=1 \) | (III) | \( \displaystyle\frac{1}{x} \) |
| (D) | \( \displaystyle\frac{xdy}{dx}+4y=x^4 \) | (IV) | \( e^{-2x} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
In the form \( \frac{dy}{dx}+Py=Q \), the integrating factor is \( e^{\int P\,dx} \). Where an equation starts with \( x\frac{dy}{dx} \), it is divided by x first (x > 0).
(A) \( \frac{xdy}{dx}-y=x^2e^x \)
This matches (III) \( \frac{1}{x} \).
\( \displaystyle \frac{dy}{dx}-\frac{1}{x}y=xe^x,\quad P=-\frac{1}{x} \)
\( \displaystyle \text{IF}=e^{-\int\frac{1}{x}dx}=e^{-\log x}=\frac{1}{x} \)
(B) \( \frac{dy}{dx}-2y=e^x \)
This matches (IV) \( e^{-2x} \).
\( P=-2 \)
\( \displaystyle \text{IF}=e^{\int -2\,dx}=e^{-2x} \)
(C) \( \frac{xdy}{dx}+y=1 \)
This matches (I) \( x \).
\( \displaystyle \frac{dy}{dx}+\frac{1}{x}y=\frac{1}{x},\quad P=\frac{1}{x} \)
\( \text{IF}=e^{\log x}=x \)
(D) \( \frac{xdy}{dx}+4y=x^4 \)
This matches (II) \( x^4 \).
\( \displaystyle \frac{dy}{dx}+\frac{4}{x}y=x^3,\quad P=\frac{4}{x} \)
\( \text{IF}=e^{4\log x}=x^4 \)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).
Shortcut: (B) is the only equation with a constant P, so it must take \( e^{-2x} \); only Options (1) and (3) do this, and they differ in (C). Since \( x\frac{dy}{dx}+y=\frac{d}{dx}(xy) \), the IF of (C) is x, which gives Option (1).