Question

Решить дифференцированные уравнения а) x+xy+y'(y+xy)=0 б) x^+y^=2xyy'

Answer 1

$x+xy+y'(y+xy)=0\\x(1+y)+\frac{ydy(1+x)}{dx}=0|*\frac{dx}{(1+y)(1+x)}\\\frac{ydy}{1+y}=-\frac{xdx}{1+x}\\\int(1-\frac{1}{1+y})dy=-\int(1-\frac{1}{1+x})dx\\y-ln|1+y|=-x+ln|1+x|+C\\y+x-ln|(1+y)(1+x)|=C\\y'+1-\frac{y'}{1+y}-\frac{1}{1+x}=0;\\y=-1$

$x^2+y^2=2xyy'\\y=tx;y'=t'x+t\\x^2+t^2x^2=2x^2t(t'x+t)|:x^2\\1+t^2=2t(t'x+t)\\1+t^2=2tt'x+2t^2\\\frac{2txdt}{dx}=1-t^2|*\frac{dx}{x(1-t^2)}\\\frac{2tdt}{1-t^2}=\frac{dx}{x}\\1-t^2=0\\t=^+_-1\\y=^+_-x\\x^2+x^2=2x*(^+_-x)*(^+_-1)\\\\-\int\frac{d(1-t^2)}{1-t^2}=\int\frac{dx}{x}\\-ln|1-t^2|=ln|x|+ln|C|\\\frac{1}{1-t^2}=Cx\\\frac{x}{x^2-y^2}=C;y=^+_-x$