1+（dy/dx）^2=y^2怎么解

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1+（dy/dx）^2=y^2怎么解
求y

(dy/dx)=sqrt(y^2-1) which implies:dy/sqrt(y^2-1)=dx integrate both sides:integrate (1/sqrt(y^2-1))dy = x which leads to:ln|y+sqrt(y^2-1)|=x Please pay attention to the +/- after each steps.I didnt consider this issue here.

1+（dy/dx）^2=y^2怎么解常微分：dy/dx=(x+y)^2怎么解 (x^2)dy+（y^2)dx=dx-dy 高数中若dy/dt=(dy/dx)*cost 那么d^2 y/dx^2 怎么求微分方程dy/dx=(y^2-1)/2如何解? dy／dx=（x-y）^2+1通解微分方程(d^2y/dx^2)^2=x^2*(dy/dx)怎么解解方程y^2+x^2*(dy/dx)=x*y*(dy/dx) 2*x*y^2（dy／dx）- x^3（dy／dx）=2y^3 解微分方程y^2+(x^2)(dy/dx)=xy(dy/dx) 试从dx/dy=1/y'导出：d^2x/dy^2=-y''/（y')^3 题目中关于d[1/y']/dx}*[dx/dy dy/dx=x(1+y^2)/y通解