求解微分方程xy'-y-(y^2-x^2)^1/2=0
问题描述：
求解微分方程xy'-y-(y^2-x^2)^1/2=0
答
令u=y/x y'=u'x+u
xy'-y-(y²-x²)^1/2=0 y'-(y/x)-[(y/x)²-1]^1/2=0
u'x+u-u-[u²-1]^1/2=0
du/[u²-1]^1/2=dx/x
dln[u+[u²-1]^1/2]=dlnx
u+[u²-1]^1/2=cx
y+(y²-x²)^1/2=cx²