The DE is of 'separable variables type' and can be written as...

\(\displaystyle \frac{dy}{y\cdot (1-y^{2})} = dx\) (1)

Because is...

\(\displaystyle \frac{1}{y\cdot (1-y^{2})} = \frac{1}{y} + \frac{y}{1-y^{2}}\) (2)

... is also...

\(\displaystyle \int \frac{dy}{y\cdot (1-y^{2})} = \ln y - \frac{1}{2} \cdot \ln (1-y^{2}) + \ln c\) (3)

... so that the solution of (1) is...

\(\displaystyle \frac{y}{\sqrt{1-y^{2}}} = c \cdot e^{x} \) (4)

... that can be 'explicitate' as function of x as...

\(\displaystyle y= \pm \frac{c\cdot e^{x}}{\sqrt{1+c^{2}\cdot e^{2x}}}\) (5)

If now we intend to impose the initial condition \(\displaystyle y(0)= y_{0}\) we 'discover' that we can do it only if is \(\displaystyle |y_{0}|<1\). To overcome the handycap in case \(\displaystyle |y_{0}|>1\) we write (2) as...

\(\displaystyle \frac{1}{y\cdot (1-y^{2})} = \frac{1}{y} - \frac{y}{y^{2}-1}\) (6)

... and repeat the procedure obtaining...

\(\displaystyle y= \pm \frac{c\cdot e^{x}}{\sqrt{c^{2}\cdot e^{2x}-1}}\) (7)

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)