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**Chris L T521** Yes it is separable...but I have a feeling we can't find an explicit solution (i.e. y=...)

The DE is equivalent to $\displaystyle \frac{\,dy}{y^2-1}=\frac{\,dx}{x^2-1}\implies \frac{\,dy}{(y+1)(y-1)}=\frac{\,dx}{(x+1)(x-1)}$

Now, we see that if we integrate, we have $\displaystyle \int\frac{\,dy}{(y+1)(y-1)}=\int\frac{\,dx}{(x+1)(x-1)}$.

Note that $\displaystyle (y+1)-(y-1)=2$ and $\displaystyle (x+1)-(x-1)=2$

Therefore, $\displaystyle \int\frac{\,dy}{(y+1)(y-1)}=\int\frac{\,dx}{(x+1)(x-1)}\implies\tfrac{1}{2}\int\frac{2\,dy}{(y+1)(y-1)}=\tfrac{1}{2}\int\frac{2\,dx}{(x+1)(x-1)}$ $\displaystyle \implies\tfrac{1}{2}\int\frac{\left[(y+1)-(y-1)\right]\,dy}{(y+1)(y-1)}=\tfrac{1}{2}\int\frac{\left[(x+1)-(x-1)\right]\,dx}{(x+1)(x-1)}$.

This now becomes $\displaystyle \int\frac{\,dy}{y-1}-\int\frac{\,dy}{y+1}=\int\frac{\,dx}{x-1}-\int\frac{\,dx}{x+1}$ $\displaystyle \implies\ln\left|y-1\right|-\ln\left|y+1\right|=\ln\left|x-1\right|-\ln\left|x+1\right|+C\implies \ln\left|\frac{y-1}{y+1}\right|=\ln\left|\frac{x-1}{x+1}\right|+C$

This format is fine...

Otherwise, you can say this is the same as

$\displaystyle \left|\frac{y-1}{y+1}\right|=C\left|\frac{x-1}{x+1}\right|\implies\frac{y-1}{y+1}=A\left|\frac{x-1}{x+1}\right|$ $\displaystyle \implies \left[1-A\left|\frac{x-1}{x+1}\right|\right]y=A\left|\frac{x-1}{x+1}\right|+1\implies y=\frac{A\left|\displaystyle\frac{x-1}{x+1}\right|+1}{1-A\left|\displaystyle\frac{x-1}{x+1}\right|}$ where $\displaystyle A=\pm C$

I hope this helps...