Question

The solution of $\frac{xdy}{x+1}=(\frac{y}{xy+y}-1)dx$ is:-

Answer 1

Consider the given equation.

$\frac{xdy}{x+1}=(\frac{y}{xy+y}-1)dx$

$\frac{xdy}{x+1}=(\frac{y}{y(x+1)}-1)dx$

$\frac{xdy}{x+1}=(\frac{1}{(x+1)}-1)dx$

$\frac{xdy}{x+1}=\left(\frac{1-x-1}{x+1}\right)dx$

$xdy=-xdx$

$dy=-dx$

On taking integration both sides, we get

$\int dy=-\int dx$

$y=-x+c$

$x+y=c$

Hence, this is the answer.