求解微分方程xy'-y-(y^2-x^2)^1/2=0_数学解答

求解微分方程xy'-y-(y^2-x^2)^1/2=0

人气：283 ℃　时间：2020-05-20 04:22:52

解答

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