求通解为(x-c1)^2+(y-c2)^2=1微分方程,答案是（y’’）^2=［（y’）^2+1］^3,
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∵(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'/y".(1)
x-c2=-(1+(y')^2)/y".(2)
把(1)和(2)式代入通解,得
((1+(y')^2)y'/y")^2+(-(1+(y')^2)/y")^2=1
==>(1+(y')^2)^2(y')^2/(y")^2+(1+(y')^2)^2/(y")^2=1
==>(1+(y')^2)^2(y')^2+(1+(y')^2)^2=(y")^2
==>(1+(y')^2)(1+(y')^2)^2=(y")^2
==>(1+(y')^2)^3=(y")^2
故通解为(x-c1)^2+(y-c2)^2=1微分方程是(y")^2=(1+(y')^2)^3.
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