Solve $(y+1)dx+(x+1)dy=0$ Solve $(y+1)dx+(x+1)dy=0$
$$\frac{dy}{y+1}=-\frac{dx}{x+1}$$
then we get $\ln|y+1|=-\ln|x+1|+c$
$$\ln(|(y+1)(x+1)|)=c$$
$$|(y+1)(x+1)|=e^c=c_1$$
but answer is $y+1=\frac{c}{x+1}$ can you help to find where is my mistake?

Simpler:
$(y+1)dx + (x+1)dy = 0$
$ydx + xdy = -(dx+dy)$
$d(xy) = d(-(x+y))$
$xy = -(x+y) + c$
$xy + x + y = c$, which is equivalent to your given answer.
