∵y'-xy'=a(y'+y²) ==>(1-a-x)y'=ay²
==>dy/y²=adx/(1-a-x)
==>-1/y=-aln│1-a-x│-ln│C│(C是积分常数)
==>e^(1/y)=C(1-a-x)^a
∴原方程的通解是e^(1/y)=C(1-a-x)^a(C是积分常数).