Solve differential equation. $2xy'$ + $y^2 - 1 =0$ Solve differential equation.
$2xy' + y^2 - 1 =0$
My work.
$2x\dfrac{dy}{dx}+y^2-1=0$
$\dfrac{dx}{2x}=\dfrac{dy}{1-y^2}$
$\int\dfrac{dx}{2x}=\int\dfrac{dy}{1-y^2}$
getting
$\frac{1}{2}\ln|x|-\frac{1}{2}(\ln|1+y|-\ln|1-y|)=0$
but solution in the book says $y=\dfrac{cx-1}{cx+1}$ or $y=1$ or $y =-1$
EDIT
$\frac{1}{2}\ln|\frac{1+y}{1-y}| = \frac{1}{2}\ln|x| +c$
from here I still can't get the answer. And $y=1$, $y=-1$ are in fact solutions.

So $\dfrac{1}{2}\ln|\dfrac{1+y}{1-y}| = \dfrac{1}{2}\ln|x| +c \implies \ln|\dfrac{1+y}{1-y}| = \ln|x| +c_1 \implies \ln|\dfrac{1+y}{1-y}| = \ln|c_2x|$ as any number can be written as logarithm of some number. Thus, $\frac{1+y}{1-y} =cx \implies \frac{2}{1-y}-1=cx,  cx+1=\frac{2}{1-y}, 1-y=\frac{2}{cx+1}, y=1-\frac{2}{cx+1}=\frac{cx-1}{cx+1}$
