Diff Eq. : Find an explicit solution of $y^2 - 1 = y'$ $$y' = y^2 - 1$$
I think this cannot be solved using the integration factor method because of the square power of the y.
Attempted: 
$$ \frac{y'}{y^2} = -1 $$
$$ y^{-1} = t + C $$
$$ \frac{1}{y} =  t + C$$
It's not the right answer.
How do I go about solving this?
Thanks.
 A: There was an algebra error at the beginning. From $y'=y^2-1$ it is not correct to conclude that $\frac{y'}{y^2}=-1$. We could conclude that $\frac{y'}{y^2}=-\frac{1}{y^2}$, but that is not helpful.  Our DE is separable, and there is a standard method for dealing with separable equations.  
If $y^2\ne 1$, we can divide by $y^2-1$, and obtain
$$\frac{y'}{y^2-1}=1.$$
The following notation is more common in beginning DE courses:
$$\frac{dy}{y^2-1}=dt.$$
Integrate on both sides. On the left, use the partial fraction decomposition
$$\frac{1}{y^2-1}=\left(\frac{1}{y-1}-\frac{1}{y+1}\right).$$
when we integrate, we get
$$\frac{1}{2}\left(\ln(|y-1|)-\ln(|y+1|)\right)=t+C.$$
Multiply through by $2$, and take the exponential of both sides. After a while you should be able to find an explicit expression for $y$.  Don't forget the two special solutions $y\equiv 1$ and $y\equiv -1$. (You may have an initial condition that lets you not bother with them.)
A: $$ y = 1$$
$$ y = -1$$
There are also non-constant  solutions, for which
$$  \frac{y'}{y^2 - 1} = 1.  $$
A: $$\frac{y^\prime}{y^2-1} = 1$$
Then you integrate. This is a Ricatti equation.
The integration will involve argth.
A: I quite like @GHL's idea of going straight to the hyperbolic trigonometric functions. 
I cannot say I ever really used those in a beginning ODE course, but it is a good idea.
So I thought of putting that information here as community wiki. 
We have $$ \frac{d}{dx} \cosh x = \sinh x,$$
$$ \frac{d}{dx} \sinh x = \cosh x,$$ and
$$  \cosh^2 x - \sinh^2 x = 1.  $$
They do have Latex for the hyperbolic tangent and cotangent, but not secant and cosecant, ALRIGHT I'M USING OPERATORNAME.
$$ \frac{d}{dx} \tanh x =  \operatorname{sech}^2  x,$$
$$ \frac{d}{dx}  \operatorname{sech}  x = -  \operatorname{sech}  x \; \tanh x,$$
$$  \tanh^2  x +   \operatorname{sech}^2  x = 1.   $$ 
Next
$$ \frac{d}{dx} \coth x = -  \operatorname{csch}^2  x,$$
$$ \frac{d}{dx}  \operatorname{csch}  x = -  \operatorname{csch}  x \; \coth x,$$
$$  \coth^2  x -  \, \operatorname{csch}^2 \, x = 1.   $$
For the problem at hand, $y' = y^2 - 1,$ for $|y| < 1$ we use
$$  \tanh^2  x - 1 = -   \operatorname{sech}^2  x  $$
and
$$ \frac{d}{dx} (- \tanh x) = -  \operatorname{sech}^2  x,$$
where we may also shift $x$ to any $x - C,$ and $y = - \tanh(x-C).$
For $y > 1,$
we use
$$  \coth^2  x - 1 =  \operatorname{csch}^2  x ,   $$
and
$$ \frac{d}{dx} ( - \coth x) =   \operatorname{csch}^2  x.$$
We may also shift $x$ to any $x - C,$ and $y = - \coth(x-C),$ but only for $x < C$ which is where the particular solution goes to infinity.
For $y < -1,$
we use
$$  \coth^2  x - 1 =  \operatorname{csch}^2  x ,   $$
and
$$ \frac{d}{dx} ( - \coth x) =   \operatorname{csch}^2  x.$$
We may also shift $x$ to any $x - C,$ and $y = - \coth(x-C),$ but only for $x > C.$ In this case the curve begins with $y$ at negative infinity and eventually it gets close to the $y = -1$ constant solution in forward time. 
