y'/y=1/x
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y'/y=1/x
y'/y=1/x
y'/y=1/x
dy/dx=y/x
dy/y=dx/x
同时积分
ln|y|=ln|x|+c
e为底同时幂次方
|y|=|x|+c
这就是通解
y'/y=1/x
y'-(1/x)y=xe^-x
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2y(x+二分之一y)-[(x+y)(x-y)+2y(y+x)],其中|x-1|=2
x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2
常微分方程y'=(x+y)ln(x+y)-1
y'=(x-y+1)/(x+y-3)通解
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