Question

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

Answer 1

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

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