Match List-I with List-II
| List-I | List-II | ||
| Differential equation | Integrating factor | ||
| (A) | \( \displaystyle\frac{dx}{dy}+\frac{2x}{y}=y^{2} \) | (I) | \( e^{x} \) |
| (B) | \( \displaystyle\frac{dy}{dx}+y=e^{-x} \) | (II) | \( \displaystyle\frac{1}{\sqrt{y}} \) |
| (C) | \( \displaystyle\frac{dy}{dx}-\frac{y}{2}=x \) | (III) | \( y^{2} \) |
| (D) | \( \displaystyle\frac{dx}{dy}-\frac{1}{2}\frac{x}{y}=\sin y \) | (IV) | \( e^{-x/2} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
For \( \frac{dy}{dx}+Py=Q \) the integrating factor is \( e^{\int P\,dx} \); for \( \frac{dx}{dy}+Px=Q \) it is \( e^{\int P\,dy} \).
(A) \( \frac{dx}{dy}+\frac{2x}{y}=y^{2} \)
This matches (III) \( y^2 \).
\( \displaystyle P=\frac{2}{y} \)
\( \displaystyle e^{\int\frac{2}{y}dy}=e^{2\log y} \)
\( \displaystyle =y^2 \)
(B) \( \frac{dy}{dx}+y=e^{-x} \)
This matches (I) \( e^x \).
\( \displaystyle P=1,\quad e^{\int 1\,dx}=e^x \)
(C) \( \frac{dy}{dx}-\frac{y}{2}=x \)
This matches (IV) \( e^{-x/2} \).
\( \displaystyle P=-\frac{1}{2},\quad e^{\int -\frac{1}{2}dx}=e^{-x/2} \)
(D) \( \frac{dx}{dy}-\frac{1}{2}\frac{x}{y}=\sin y \)
This matches (II) \( \frac{1}{\sqrt{y}} \).
\( \displaystyle P=-\frac{1}{2y} \)
\( \displaystyle e^{-\frac{1}{2}\int\frac{1}{y}dy}=e^{-\frac{1}{2}\log y} \)
\( \displaystyle =y^{-1/2} \)
\( \displaystyle =\frac{1}{\sqrt{y}} \)
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).
Shortcut: The integrating factors in x, (I) and (IV), can only go with (B) and (C), whose P is a constant: P = 1 gives \( e^x \) and P = −1/2 gives \( e^{-x/2} \). Only Option (1) has (B) - (I) and (C) - (IV).