If c is an arbitrary constant, then
Match the LIST-I with LIST-II
| LIST-I Differential Equation | LIST-II Solution | ||
| A. | \( \displaystyle\frac{dy}{dx}=1+x+y+xy \) | I. | \( \log_e|y|=2x+\log_e(x-1)^2+c \) |
| B. | \( (x-1)\displaystyle\frac{dy}{dx}=2xy \) | II. | \( y=cx \) |
| C. | \( \displaystyle\frac{dy}{dx}=1-x+y-xy \) | III. | \( \log_e|1+y|=x+\displaystyle\frac{x^2}{2}+c \) |
| D. | \( x\,dy=y\,dx \) | IV. | \( \log_e|1+y|=x-\displaystyle\frac{x^2}{2}+c \) |
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
Each equation is solved by separating the variables.
(A) \( \displaystyle\frac{dy}{dx}=1+x+y+xy \)
This matches III. The right side factorises:
\( \displaystyle \frac{dy}{dx}=(1+x)(1+y) \)
\( \displaystyle \frac{dy}{1+y}=(1+x)\,dx \)
\( \displaystyle \log_e|1+y|=x+\frac{x^2}{2}+c \)
(B) \( (x-1)\displaystyle\frac{dy}{dx}=2xy \)
This matches I.
\( \displaystyle \frac{dy}{y}=\frac{2x}{x-1}\,dx \)
\( \displaystyle \frac{dy}{y}=\left(2+\frac{2}{x-1}\right)dx \)
\( \log_e|y|=2x+2\log_e|x-1|+c \)
\( \log_e|y|=2x+\log_e(x-1)^2+c \)
(C) \( \displaystyle\frac{dy}{dx}=1-x+y-xy \)
This matches IV. The right side factorises:
\( \displaystyle \frac{dy}{dx}=(1-x)(1+y) \)
\( \displaystyle \frac{dy}{1+y}=(1-x)\,dx \)
\( \displaystyle \log_e|1+y|=x-\frac{x^2}{2}+c \)
(D) \( x\,dy=y\,dx \)
This matches II.
\( \displaystyle \frac{dy}{y}=\frac{dx}{x} \)
\( \log|y|=\log|x|+\log|c| \)
\( y=cx \)
So the matching is A-III, B-I, C-IV, D-II.