Question

The general solution of the differential equation $ydx-(x+y)dy=0$ is

• A. $\frac{x}{y^2}=\ln |y|+C$
• B. $\frac{x^2}{y^2}=2\ln |y|+C$
• C. $\frac{x}{y}=\ln |x|+C$
• D. $\frac{x}{y}=\ln |y|+C$

Answer 1

The correct option is D $\frac{x}{y}=\ln |y|+C$

$ydx-(x+y)dy=0\Rightarrow \frac{dx}{dy}-\frac{x}{y}-1=0\Rightarrow \frac{dx}{dy}-\frac{x}{y}=1$
Integrating factor
$=e^{\int -\frac{1}{y}dy}=e^{-\ln y}=\frac{1}{y}$
Solution of D.E,
$x\cdot \frac{1}{y}=\int \frac{1}{y}dy+C\Rightarrow \frac{x}{y}=\ln |y|+C$