(y')^2=-a/y^2+b/y+c=-a[1/y-b/2a]^2+c+b^2/4
=>
y'=sqrt{-a[1/y-b/2a]^2+c+b^2/4}=dy/dx
dy/sqrt{-a[1/y-b/2a]^2+c+b^2/4}=dx
然后试试两边积分.