微分方程通解y"+sqr(1-(y'))=0

y"=-sqr(1-(y'))
设u=sqr(1-(y')) 两边平方
u^2=1-(y')
求导
2udu/dx=-y''
y''=-2udu/dx 代入原方程
-2udu/dx+u=0
du=1/2dx
两边积分
u=1/2x+c1
(1/2x+c1)^2=1-dy/dx
dy=(-1/4x^2-c1x-c1^2+1)dx
两边积分 得微分方程的通解
y=-1/12x^3-c1/2x^2+(1-X1^2)x+C2

设u=sqrt(1-y'),则y'=1-u^2,则y''=(1-u^2)'=(-u^2)'=-2uu'
故原式化为:
-2uu'+u=0
==>u(1-2u')=0
==>u=0或u=(x+C1)/2.
==>y'=1或y'=1-x^2/4-C1*x/2-C1^2/4.
==>y= x+C1或y = -1/12*x^3-1/4*C1*x^2+x-1/4*x*C1^2+C2.

令p=y'，则y''=p'
p' + √(1-p) = 0
dp/dx = -√(1-p)
p=1时上式成立，y = x+C

p不等于1时，-dp/√(1-p) = dx
积分得2√(1-p) = x + C
1-p = x²/4 + Cx/2 + C²/4
y' = p = 1 - C²/4 - Cx/2 - x²/4
积分得
y = x(1 - C²/4) - Cx²/4 - x³/12 + C'

y'＝1是方程的解，所以y＝x＋C是原微分方程的解

y'≠1时，方程化为－y''/√(1－y'