xy'+y-2y^3=0微分方程的解?
问题描述：
xy'+y-2y^3=0微分方程的解?
答
xy'=(2y^3-y)
dy/(2y^3-y)=dx/x
dy/[y(2y^2-1)]=dx/x
∫dy/[y(2y^2-1)]=∫[2y^2-(2y^2-1)]dy/[y(2y^2-1)]=∫2ydy/(2y^2-1)-∫dy/y
=(1/2)ln(2y^2-1)-lny+C0
d[(1/2)ln(2y^2-1)-lny]=dlnx
(1/2)ln(2y^2-1)-lny+C0=lnx
C1√|2y^2-1| /y=x
通解C1√|2y^2-1| /y=x
答
伯努利方程
xy'+y=2y^3
->
x/y^3*y'+1/y^2=2
令1/y^2=t
-x/2*dt/dx+t=2
解这个一阶方程得
(2x^(-2)+c)*x^2