dy╱dx=1-y╱y-x 求通解
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解：∵dy╱dx=(1-y)╱(y-x)
==>(y-x)dy+(y-1)dx=0
==>ydy+[(y-1)dx-xdy]=0
==>ydy/(y-1)²+[(y-1)dx-xdy]/(y-1)²=0 (等式两端同除(y-1)²)
==>∫ydy/(y-1)²+∫[(y-1)dx-xdy]/(y-1)²=0 (积分)
==>ln│y-1│-1/(y-1)+x/(y-1)=ln│C│ (C是积分常数)
==>(y-1)*e^((x-1)/(y-1))=C
∴此方程的通解是(y-1)*e^((x-1)/(y-1))=C。
==>(y-x)dy+(y-1)dx=0
==>ydy+[(y-1)dx-xdy]=0
==>ydy/(y-1)²+[(y-1)dx-xdy]/(y-1)²=0 (等式两端同除(y-1)²)
==>∫ydy/(y-1)²+∫[(y-1)dx-xdy]/(y-1)²=0 (积分)
==>ln│y-1│-1/(y-1)+x/(y-1)=ln│C│ (C是积分常数)
==>(y-1)*e^((x-1)/(y-1))=C
∴此方程的通解是(y-1)*e^((x-1)/(y-1))=C。
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