Question

X+xy+y'(y+xy)=0 ХЕЛП

Answer 1

$\displaystyle x+xy+y'(y+xy)=0\\\\x(y+1)+y'(x+1)y=0\\\\y'(x+1)y=-x(y+1)\\\\\frac{dy}{dx}*\frac{y}{y+1}=-\frac{x}{x+1}\\\\\\\frac{ydy}{y+1}=-\frac{xdx}{x+1}\\\\\\ \int\limits \frac{ydy}{y+1}= \int\limits -\frac{xdx}{x+1}\\\\\\ \int\limits(1-\frac{1}{y+1})dy=- \int\limits(1-\frac{1}{x+1})dx\\\\\\ \int\limits dy-\int\limits\frac{dy}{1+y}=-\int\limits dx +\int\limits \frac{dx}{1+x}\\\\\\\int\limits dy-\int\limits\frac{d(y+1)}{y+1}=-\int\limits dx +\int\limits \frac{d(x+1)}{x+1}$

$\boxed{y-ln|y+1|=-x+ln|x+1|+C}$