Q. The solution of differential equation $(1 + x)y\ dx + (1 - y)x\ dy = 0$ is
A) $\log_{e}{(xy)} + x - y = c$
B) $\log_{e}\left( \frac{x}{y} \right) + x + y = c$
C) $\log_{e}\left( \frac{x}{y} \right) - x + y = c$
D) $\log_{e}{(xy)} - x + y = c$
Explanation
(a)
Given, $(1 + x)y\ dx + (1 - y)x\ dy = 0$
$\Rightarrow$ $\frac{(1 - y)}{y}\ dy + \frac{(1 + x)}{x}\ dx = 0$
$\Rightarrow$ $\int_{}^{}\left( \frac{1}{y} - 1 \right)dy + \int_{}^{}\left( \frac{1}{x} + 1 \right)dx = 0$
$\Rightarrow$ $\log_{e}y - y + \log_{e}x + x = c$
$\Rightarrow$ $\log_{e}{(xy)} + x - y = c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.