题目

xdy-ylnydx=0
还有
y'=1+y^2-2x-2xy^2,y(0)=0

提问时间：2020-11-23

答案

1.xdy-ylnydx=0
∵xdy-ylnydx=0 ==>xdy=ylnydx
==>dy/(ylny)=dx/x
==>d(lny)/lny=dx/x
==>ln│lny│=ln│x│+ln│C│ (C是积分常数)
==>lny=Cx
==>y=e^(Cx)
∴原微分方程的通解是y=e^(Cx) (C是积分常数)
2.y'=1+y^2-2x-2xy^2,y(0)=0
∵y'=1+y²-2x-2xy² ==>y'=1-2x+y²(1-2x)
==>y'=(1-2x)(1+y²)
==>dy/(1+y²)=(1-2x)dx
==>arctany=x-x²+C (C是积分常数)
==>y=tan(x-x²+C)
又y(0)=0,则把它带入上式,得0=tanC,即C=0
∴原微分方程的解是y=tan(x-x²)