Question

求通解为(x-c1)^2+(y-c2)^2=1微分方程,答案是（y’’）^2=［（y’）^2+1］^3,

Answer 1

∵(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.