Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| Differential equations | Integrating factor | ||
| A. | \( (x+y)\displaystyle\frac{dy}{dx}=1 \) | I. | \( e^{-x} \) |
| B. | \( x\displaystyle\frac{dy}{dx}-y=x^2 \) | II. | \( x \) |
| C. | \( \displaystyle\frac{dy}{dx}-y=\sin x \) | III. | \( \displaystyle\frac{1}{x} \) |
| D. | \( x\displaystyle\frac{dy}{dx}+y=\tan x \) | IV. | \( e^{-y} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A - IV, B - III, C - I, D - II
Each equation is put in the standard linear form \( \frac{dy}{dx}+Py=Q \) (IF = \( e^{\int P\,dx} \)) or \( \frac{dx}{dy}+P_1x=Q_1 \) (IF = \( e^{\int P_1\,dy} \)).
(A) \( (x+y)\frac{dy}{dx}=1 \)
This matches (IV) \( e^{-y} \).
It is not linear in y, but it is linear in x:
\( \displaystyle \frac{dx}{dy}=x+y \)
\( \displaystyle \frac{dx}{dy}-x=y \)
\( \displaystyle \text{IF}=e^{\int -1\,dy}=e^{-y} \)
(B) \( x\frac{dy}{dx}-y=x^2 \)
This matches (III) \( \frac{1}{x} \).
Dividing by x:
\( \displaystyle \frac{dy}{dx}-\frac{1}{x}y=x \)
\( \displaystyle \text{IF}=e^{-\int\frac{1}{x}dx} \)
\( \displaystyle =e^{-\log x}=\frac{1}{x} \)
(C) \( \frac{dy}{dx}-y=\sin x \)
This matches (I) \( e^{-x} \).
\( \displaystyle P=-1\ \Rightarrow\ \text{IF}=e^{\int -1\,dx}=e^{-x} \)
(D) \( x\frac{dy}{dx}+y=\tan x \)
This matches (II) \( x \).
Dividing by x:
\( \displaystyle \frac{dy}{dx}+\frac{1}{x}y=\frac{\tan x}{x} \)
\( \displaystyle \text{IF}=e^{\int\frac{1}{x}dx} \)
\( \displaystyle =e^{\log x}=x \)
So the matching is A - IV, B - III, C - I, D - II.