Question

separate the differential equation into variables

xydx + (x+1)dy=0

Answer 1

$xydx + (x+1)dy=0\\ xydx =- (x+1)dy\\ \frac{xdx}{x+1}=-\frac{dy}{y}$

Let, $x+1=t\Rightarrow dx=dt$

$\frac{t-1}{t}dt=-\frac{dy}{y}$

$(1-\frac{1}{t})dt=$ $-\frac{dy}{y}$

Integrating both sides, we get

$t-\ln t=-\ln y+c$

$x+1-\ln (x+1)=-\ln y+c$

$x-\ln (x+1)+\ln y+1=c$