x(1+y)+y'(y-xy)=0

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x(1+y)+y'(y-xy)=0
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x(1+y)+y'(y-xy)=0
x(1+y)+y'(y-xy)=0

x(1+y)+y'(y-xy)=0
微分方程 x(1+y)+y'(y-xy)=0, 即 y(x-1)dy/dx=x(1+y), 为可分离变量型,
当 x≠1, 且 y≠-1 时, 得 [y/(y+1)]dy=[x/(x-1)]dx,
[1-1/(y+1)]dy=[1+1/(x-1)]dx, 两边积分,得 y-ln(y+1)=x+ln(x-1)+lnC
ln[C(x-1)(y+1)]=y-x, 通解是 C(x-1)(y+1)=e^(y-x).
当 x=1时, 1+y=0,y=-1.

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