Match List-I with List-II
| List-I | List-II | ||
| Differential equation | Integrating factor | ||
| (A) | \( \displaystyle\frac{dy}{dx}+\displaystyle\frac{y}{x}=1 \) | (I) | \( e^x \) |
| (B) | \( x\displaystyle\frac{dy}{dx}-y=x^2 \) | (II) | \( x^2 \) |
| (C) | \( \displaystyle\frac{dy}{dx}+y=x^3+1 \) | (III) | \( x \) |
| (D) | \( x^2\displaystyle\frac{dy}{dx}+2xy=x^5 \) | (IV) | \( \displaystyle\frac{1}{x} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Each equation is written in the form \( \frac{dy}{dx}+Py=Q \); the integrating factor is \( e^{\int P\,dx} \).
(A) \( \frac{dy}{dx}+\frac{y}{x}=1 \)
This matches (III) x.
\( \displaystyle P=\frac{1}{x} \)
\( \displaystyle \text{I.F.}=e^{\log x}=x \)
(B) \( x\frac{dy}{dx}-y=x^2 \)
This matches (IV) \( \frac{1}{x} \). Dividing by x first:
\( \displaystyle \frac{dy}{dx}-\frac{1}{x}y=x \)
\( \displaystyle P=-\frac{1}{x} \)
\( \displaystyle \text{I.F.}=e^{-\log x}=\frac{1}{x} \)
(C) \( \frac{dy}{dx}+y=x^3+1 \)
This matches (I) \( e^x \).
\( \displaystyle P=1 \)
\( \displaystyle \text{I.F.}=e^{x} \)
(D) \( x^2\frac{dy}{dx}+2xy=x^5 \)
This matches (II) \( x^2 \). Dividing by \( x^2 \) first:
\( \displaystyle \frac{dy}{dx}+\frac{2}{x}y=x^3 \)
\( \displaystyle P=\frac{2}{x} \)
\( \displaystyle \text{I.F.}=e^{2\log x}=x^2 \)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).