xy'-y-√(y²-x²)=0

Answer 1

求微分方程 xy'-y-√(y²-x²)=0的通解
解：xy'-y-y√[1-(x/y)²]=0
两边同除以y得：
(x/y)y'-1-√[1-(x/y)²]=0..............(1)
令x/y=u，得y=x/u.............(2)
对(2)取导数得：y'=(u-xu')/u².............(3)
将(2)和(3)代入方程(1)得：
(u-xu')/u-1-√(1-u²)=0
化简得：-(x/u)u'-√(1-u²)=0
分离变量得：du/[u√(1-u²)]=-dx/x
积分之得：-ln{[1+√(1-u²)]/u}=-lnx+lnc=ln(c/x)
即ln{[1+√(1-u²)]/u}=-ln(c/x)=ln(x/c)
于是得 [1+√(1-u²)]/u=x/c
1+√(1-u²)=ux/c
√(1-u²)=(ux/c)-1
1-u²=(ux/c)²-2ux/c+1
[(x/c)²+1]u²-2ux/c=0
[(x/c)²+1]u-2x/c=0
故u=2cx/(x²+c²)
代入(2)式即得原方程的通解为：
y=(x²+c²)/(2c)