dy/dx=1+1/(x-y)
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dy/dx=1+1/(x-y)dy/dx=1+1/(x-y)dy/dx=1+1/(x-y)dy/dx-1=1/(x-y)(dy-dx)/dx=1/(x-y)d(y-x)/dx=-1/(y-x)令u=y
dy/dx=1+1/(x-y)
dy/dx=1+1/(x-y)
dy/dx=1+1/(x-y)
dy/dx-1=1/(x-y)
(dy-dx)/dx=1/(x-y)
d(y-x)/dx=-1/(y-x)
令 u=y-x
du/dx=-1/u 现在可以分离变量了
udu=-dx
两边积分得 (1/2)u^2=-x+C
原方程的解为 (1/.2)(y-x)^2=-x+C
或者写成 (y-x)^2=-2x+C
x-y=u
dy=1-du
1-du=dx+dx/u
sorry,我不知道啊
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