常微分方程求解：（1）1+y'=e^y (2)xy'+y=y^2
问题描述：
常微分方程求解：（1）1+y'=e^y (2)xy'+y=y^2
答
1.∵1+y'=e^y ==>y'=e^y-1
==>dy/(e^y-1)=dx
==>e^(-y)dy/(1-e^(-y))=dx
==>d(1-e^(-y))/(1-e^(-y))=dx
==>ln│1-e^(-y)│=x+ln│C│ (C是积分常数)
==>1-e^(-y)=Ce^x
∴原方程的通解是1-e^(-y)=Ce^x (C是积分常数)；
2.∵xy'+y=y² ==>xy'=y(y-1)
==>dy/[y(y-1)]=dx/x
==>[1/(y-1)-1/y]dy=dx/x
==>ln│y-1│-ln│y│=ln│x│+ln│C│ (C是积分常数)
==>(y-1)/y=Cx
==>y=Cxy+1
∴原方程的通解是y=Cxy+1 (C是积分常数)