Solve $\frac{dy}{dx}=\frac{y^2-1}{x^2-1}$ with initial condition $y(2)=2$ 
Solve the following differential equation: $\dfrac{dy}{dx}=\dfrac{y^2-1}{x^2-1}$, with the initial condition $y(2)=2$.

My attempt:
I notice that this is a separable differential equation, so I try to get it into the form $p(y) dy=f(x) dx$.
$\dfrac{dy}{dx}=\dfrac{y^2-1}{x^2-1}\\\implies \dfrac{1}{y^2-1} dy=\dfrac{1}{1-x^2}dx\\\implies\displaystyle\int\dfrac{1}{y^2-1} dy=\displaystyle\int\dfrac{1}{1-x^2}dx\\\implies \ln(y^2-1)=\ln(x^2-1)$
Now I tried to apply the initial condition of $y(2)=2$:
$\ln(2^2-1)=\ln(2^2-1)\implies\ln3=\ln3$
I'm fairly new to solving differential equations, so I don't know if I'm doing a step incorrectly or what the problem is. If anyone could shed any light on this, it would be appreciated. Thanks in advance.

If $\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{u(y)}{u(x)}$ for some function $u$ then $\dfrac{\mathrm dy}{u(y)}=\dfrac{\mathrm dx}{u(x)}$ hence $\displaystyle\int_{y(a)}^{y(x)}\frac{\mathrm dt}{u(t)}=\int_{a}^{x}\frac{\mathrm dt}{u(t)}$. If $y(a)=a$ this shows that $y(x)=x$ for every $x$ such that $u(t)\ne0$ for every $t$ from $a$ to $x$. 
If $a=2$ and $u(t)=t^2-1$, this shows that $y(x)=x$ for every $x\gt1$.
