Solving a separable ODE explicitly I am stuck at solving the following separable ODE:
$$f(x)=\sqrt{1+[f'(x)]^2},$$
with the condition that $f:[0,\infty)\longrightarrow [0,\infty)$.
I first note that $f(x)\geq 1$ and that the constant function $f(x)=1$ is a solution.
By squaring both sides and separating $f'(x)$, I get that
$$[f(x)]^2=1+[f'(x)]^2 \iff \\ f'(x)=\pm\sqrt{[f(x)]^2-1}$$
Now, for $f(x)\neq 1$, using the fact that $f'(x)=\frac{df}{dx}$ and abusing the notaion $f=f(x)$, I separate the differentials as follows:
$$\frac{df}{dx}=\pm\sqrt{f^2-1} \iff\\ \frac{df}{\sqrt{f^2-1}}=\pm dx$$
Integrating both sides gives
$$\int\frac{df}{\sqrt{f^2-1}}=\int \pm dx \implies ln(f+\sqrt{f^2-1})=\pm x+C$$
We can collect the R.H.S.
\begin{cases}
x+C_1 & (1)\\
-x+C_2 & (2)
\end{cases}
For the left integral, I used the standard integral that is provided in my textbook, namely that
$$\int\frac{dx}{\sqrt{x^2+\alpha}}=\\ ln\left|x+\sqrt{x^2+\alpha}\right|+C$$
I know that I can get rid of the logarithm in the L.H.S., using $(1)$ from above,
$$ln(f+\sqrt{f^2-1})=x+C_1\iff \\ f+\sqrt{f^2-1}=e^{x+C_1}$$
Note that i have dropped the absolute value sign, due to the given condition which constraints $f=f(x)$ to the positive real numbers.
But this is where I am stuck. What am I supposed to do? I am failing to factorize $f$. I know that we can have implicit answers for separable ODEs, but I am supposed to satisfy the condition that $f:[0,\infty)\longrightarrow [0,\infty)$, i.e., I need to find all such functions $f$.
To reiterate: how do I continue to find the explicit answer to the given separable ODE?
P.S. This is my very first time posting a question, although I have read a lot of answers. Also, english is not my first language, please excuse any grammatical errors.
 A: $$
1=f(x)^2-f'(x)^2
$$
is a hyperbole equation, so you can parametrize it in hyperbolic functions, $$f(x)=\cosh(u(x)), ~~f'(x)=\sinh(u(x)).$$ In consequence of the first part, also $$f'(x)=\sinh(u(x))u'(x),$$ so that, apart from the constant solution, $u'(x)=\pm 1$. Both signs lead to the same solution family $$f(x)=\cosh(x+C).$$
A: I stumbled across my question after some time, and i noticed that it is possible to continue my work and therefore answer the question with a different approach.
From my original answer I had arrived at
$$f+\sqrt{f^2-1}=e^{x+C_1}. (1)$$
Now, squaring both sides,
$$\left(f+\sqrt{f^2-1}\right)^2=\left(e^{x+C_1}\right)^2 \iff f^2+2f\sqrt{f^2-1}+f^2-1=e^{2(x+C_1)} \\ \iff 2f^2+2f\sqrt{f^2-1}-1=e^{2(x+C_1)}$$
and now we factor out common factors from the L.H.S. and use $\mathbf{(1)}$ from above to solve for $f$,
$$2f\underbrace{\left(f+\sqrt{f^2-1}\right)}_{e^{x+C_1}}-1=e^{2(x+C_1)} \iff 2f\cdot e^{x+C_1}=e^{2(x+C_1)}+1 \\ \iff f=\frac{e^{x+C_1}+e^{-(x+C_1)}}{2}=\text{cosh}\ \left(x+C_1\right).$$
Going back to my original question, we find that there are two different cases for $\mathbf{(1)}$, the first case is the one used above, namely that $f+\sqrt{f^2-1}=e^{x+C_1}$. The second case is $f+\sqrt{f^2-1}=e^{-x+C_2}$. However, we solve this second case just like we did above for the first case. We would then arrive at a different $f$:
$$f=\text{cosh}\ \left(-x+C_2\right)=\text{cosh}\ \left(x-C_2\right).$$
Notice that the last equality is true because hyperbolic cosine is an even function.
We recall from my original post that the constant function $f=1$ is also a solution to the DE.
Now we are done since all $f$ that satisfies the original DE are found. But we could note that the solutions are shifted hyperbolic cosine and we can therefore conclude the solutions and form a general solution with a new piecewise function $f$ by introducing arbitrary constants, say $x_0\leq x_1$:
$$f=
   \begin{cases} 
      \text{cosh}\ \left(x-x_0\right) & x\leq x_0, \\
      1 & x_0\leq x\leq x_1, \\
      \text{cosh}\ \left(x-x_1\right) & x_1\leq x. 
   \end{cases}
$$
