(x²+y²)dx-2xydy=0
(1+y²/x²)dx-2(y/x)•dy=0
2(y/x)•y'=(1+y²/x²).
Однородное уравнение.
y=tx; y’=t’x+t;
2t•(t’x+t)=1+t² => 2t•t’x+2t²=1+t²
2t•t’x=1-t²
2t•dt/(1-t²)=dx/x => -INT[1/(1-t²)]d(1-t²)=INTdx/x
ln|1-t²|=-ln|x|+C => 1-t²=C/x => t²=1-C/x.
y²/x²=1-C/x => y²=x²-Cx.