Solve $y ^2-x(\frac{dy}{dx})^2 = 1$ using proposed change of variables I am kind of stuck with this non linear differential. I am preparing for my finals and I cannot get this one.
Full question goes like this:
Find all the solutions to the equation $y ^2-x(\frac{dy}{dx})^2 - 1= 0$ stating in each case the maximal solution interval. Hint:
Use $u=y'\sqrt{-x},\,x<0$
and $u=y'\sqrt{x},\,x>0$
Also the final solutions are also given:
$y=1$ and $y=-1 \quad\forall x $
$y(x)=cosh(2\sqrt{x}+K),\quad x>0$
$y=cos(2\sqrt{-x}+K),\quad x<0$ 
What I have done so far.
Let $u=y'\sqrt{x}\Rightarrow u'=y''\sqrt{x}+y'\frac{1}{2\sqrt{x}}$ and I plug it into the original equation and I differentiate w.r.t x:
$y^2-u^2=1\Rightarrow 2yy'-2u(y''\sqrt{x}+\frac{y'}{2\sqrt{x}})=0$.
Switching back the change $u=y'\sqrt{x},\,x>0$ we get:
$2yy'-2y'\sqrt{x}(y''\sqrt{x}+\frac{y'}{\sqrt{x}})=yy'-y'y''x-y'^2
=y'(y-y''x-y')=0$.
So we get $y'=0 \Rightarrow y=C$, which is not one of the stated solutions or $(y-y''x-y')=0$ which does not make a lot of sense to me as we ended up with a second order equation, which needs two arbitrary constants, when we actually started with a first order equation.
I actually got the first three solutions using a different approach, which is not the one hinted, but I posted nonetheless in case it might help someone in order to help me :)
$y'^2=\frac{y^2-1}{x}\Rightarrow\frac{1}{\sqrt{y^2-1}}dy=\pm\frac{1}{\sqrt{x}}dx$. Solutions $y=\pm 1$ appear at this step.
Using the substitution $y=cosht$ we get $t=\pm2\sqrt{x}+C$ which gives the third one: $y=cosh(2\sqrt{x}+C)$.
However I just cannot get the proposed substitution to work and I cannot find the last solution when x is negative.
Any help is really appreciated!!!!
Thanks!!!
 A: 
So we get $y′=0⇒y=C$, which is not one of the stated solutions

You are right. It is not one of the stated solutions. It is two of them. Namely the solutions $y=1$ and $y = -1$.
Recall that to reach this stage, you differentiated your original equation: $y^2−u^2=1$ to get the equation you have at this point. Just like squaring equations in algebra, this introduces additional solutions. So you have to check the solutions you get against the original equation to see which ones work for it. $y = C$ for an arbitrary $C$ is a solution to the differentiated equation, but when you plug that equation back into the original, you find that it reduces to $C^2 = 1$. I.e., it is only a solution to the original equation when $C = 1$ or $C = -1$. 
Similarly, not all solutions of $y−y′′x−y′=0$ will be solutions of the original equation either. As you noted, it will have two arbitrary constants. But plugging those solutions into the original equation will give you an equation relating those two arbitrary constants, reducing the degrees of freedom back down to one.
Concerning your own method of separation of variables, when you went from $y'=\pm\frac{y^2-1}{x}$ to $\frac{1}{\sqrt{y^2-1}}dy=\pm\frac{1}{\sqrt{x}}dx$, you implicitly assumed that $y^2 > 1$. If instead you examine $y^2 < 1$, then you get the cosine solution.
