y'=p.dp/(1+p^2)+dx/(1+x^2)=0
arctanp+arctanx=arctanc1
y'=p=tan(arctanc1-arctanx)
=(c1-x)/(1-c1x)=(1/c1)(c1^2-1+1-c1x)/(1-c1x)
y=(-1/c1^2)(c1^2-1)ln|1-c1x|+x/c1+c2
求微分方程的通解(1+x^2)y"+(y')^2+1=0
y'=p.dp/(1+p^2)+dx/(1+x^2)=0
arctanp+arctanx=arctanc1
y'=p=tan(arctanc1-arctanx)
=(c1-x)/(1-c1x)=(1/c1)(c1^2-1+1-c1x)/(1-c1x)
y=(-1/c1^2)(c1^2-1)ln|1-c1x|+x/c1+c2