Question

The solution of $\frac{dy}{dx}=\frac{xy+y}{xy+x}$ is

• A. $y-x+\log \left(xy\right)=c$
• B. $y+x+\log \left(xy\right)=c$
• C. $y-x+\log \left(\frac{x}{y}\right)=c$
• D. $y-x-\log \left(\frac{x}{y}\right)=c$

Answer 1

The correct option is D $y-x-\log \left(\frac{x}{y}\right)=c$

$\frac{dy}{dx}=\frac{xy+y}{xy+x}=\frac{y\left(x+1\right)}{x\left(y+1\right)}$

$\int \left(\frac{y+1}{y}\right)dy=\int \left(\frac{x+1}{x}\right)dx$

$\int \left(1+\frac{1}{y}\right)dy=\int \left(1+\frac{1}{x}\right)dx$

$y+\log \left|y\right|=x+\log \left|x\right|+c$

$y-x+\log \left|y\right|-\log \left|x\right|=c$

$y-x+\log \left(\frac{y}{x}\right)=c$

$y-x-\log \left(\frac{x}{y}\right)=c$