ln|(y-1)/(y+1)|=2*ln|x|+C → ln|(y-1)/(y+1)|=ln|x^2|+C、
ここで両辺を「eの肩の上」へ乗っけると、(y-1)/(y+1)=±(e^C)*x^2、±(e^C)=aと置くと、
(y-1)/(y+1)=ax^2、y-1=ax^2(y+1)、y(1-ax^2)=1+ax^2、y=(1+ax^2)/(1-ax^2)