求解以下齐次方程的解y∧2+(x∧2-xy)dy/dx=0

y^2+(x^2-xy)dy/dx=0
dy/dx=-y^2/(x^2-xy)
dx/dy=-(x^2-xy)/y^2=-(x/y)^2+x/y
x/y=u x=uy x'=u+yu'
u+yu'=-u^2+u
u'/u^2=-dy/y
1/u+lnC=lny
y/x+lnC=lny
y=Ce^(y/x)