如何解下列微分方程dh/(hdx)=-dp/(pdx),而p=10-x^2/2

由p=10-x^2/2 得,dp/dx=-x,-dp/(pdx)=x/(10-x^2/2)
即dh/(hdx)=x/(10-x^2/2),分离变量,
dh/h=（x/(10-x^2/2))dx,左右两边积分得lnIhI=-lnI10-x^2/2I+C===>lnIhI=-lnIpI+C1,hp=C2 (lnC2=C1)