双参数函数族(x?c1)^2+(y?c2)^2=1(c1,c2为参数)所满足的微分方程

解：∵(x-c1)^2+(y-c2)^2=1
==>2(x-c1)+2(y-c2)y'=0 (等式两端对x求导)
==>x-c1-(1+(y')^2)y'/y"=0 (等式两端对x求导)
∴又上两式，得 x-c1=(1+(y')^2)y'&#