∵y'-xy'=2(y²+y') ==>y'-xy'=2y²+2y'
==>(x+1)y'=-2y²
==>dy/y²=-2dx/(x+1)
==>1/y=2ln│x+1│+ln│C│ (C是非零积分常数)
==>e^(1/y)=C(x+1)²
∴原方程的通解是e^(1/y)=C(x+1)² (C是非零积分常数).
y'-xy'=2(y^2+y') 求通解或特解
y'-xy'=2y²+2y'==>(x+1)y'=-2y²==>dy/y²=-2dx/(x+1)==>1/y="}}}'>