微分方程求初值

Question

如题所述

Answer 1

令p=dy/dx
(x+p^2)*dp/dx=p
pdx-(x+p^2)dp=0
(1/p)dx-(x/p^2+1)dp=0
d(x/p-p)=0

x/p-p=C
因为p|(x=1)=dy/dx|(x=1)=1，所以C=0
x/p-p=0
p=±√x
y=±(2/3)*x^(3/2)+B
因为y|(x=1)=1，所以B=1/3或5/3
y=(2/3)*x^(3/2)+1/3
或y=(-2/3)*x^(3/2)+5/3