微分方程一道题

@vitasapple

(y^3+xy)y'=1
(y^2+x)*2yy'=2
(y^2+x)* (y^2)'=2
y^2dy^2+xdy^2=2dx
y^2+x=u
dx=du-dy^2
udy^2=2du-2dy^2
(u+2)dy^2=2du
(1/2)dy^2=du/(u+2)
(1/2)y^2 +lnC=ln|u+2|
u+2=Ce^[(1/2)y^2]
y^2+x+2=Ce^[(1/2)y^2]
x=0,y=0
C=2
特解y^2+x+2=2e^[(1/2)y^2]