u = y' = dy/dx
y'' = u' = du/dx = (dy/dx) * (du/dy) = u*(du/dy)
2u^2 = (y-1)*u*(du/dy)
∫ dy/(y-1) = ∫ du/(2u)
ln|y-1| = (1/2)*ln|2u| + C
y-1 = A*√(2u) ........ A=e^C
(1/B) * (y-1)^2 = u = y' = dy/dx .......... B=2A^2
∫ dx = ∫ B*(y-1)^(-2) dy
x = D - B/(y-1)
y = 1 + B/(D-x)
x=1, y=2, y' = (1/B) * (y-1)^2 = 1/B = -1
B = -1
x=1, y=2,
2=1+B/(D-1)
D = B+1 = 0
y = 1+1/x
( y' = -x^(-2), y'' = 2x^(-3) )
内容全部来源于网络收集，如有侵权，请联系网站删除！