Question

Find the solution of $\frac{dy}{dx}=\frac{xy+y}{xy+x}$

• A. $y-x=\log \frac{kx}{y}$
• B. $y+x=\log \frac{ky}{x}$
• C. $y+x=\log \frac{y}{kx}$
• D. $y-x=\log \frac{ky}{x}$

Answer 1

The correct option is A $y-x=\log \frac{kx}{y}$

$\frac{dy}{dx}=\frac{xy+y}{xy+x}$
or $\frac{y+1}{y}dy=\frac{x+1}{x}dx,$
or $(1+\frac{1}{y})dy=(1+\frac{1}{x})dx$
or $\int (1+\frac{1}{y})dy=\int (1+\frac{1}{x})dx$
$y+\log y=x+\log x+\log k$
$\therefore y-x=\log \frac{kx}{y}$