If $\alpha \cosh (x/\alpha)$ is a solution to $f'f''^2=f'''(1+f'^2)$, can I conclude that this is the only answer? While solving a physics problem, this equation was the result. 
$$f'f''^2=f'''(1+f'^2)$$
The problem wanted the function describing the shape of a hanging chain. By seeing $1+f'^2$ I immediately thought about the identity $\cosh^2x=1+\sinh^2x$ and that $(d/dx)\cosh x=\sinh x$. So I tested the function $A\cosh(Bx)$ to see if this can be the answer:
After plugging in, you will see this:
$$A^2B^2\cosh^2x=1+A^2B^2\sinh^2x$$
So if I choose $AB:=1$, then $A\cosh(Bx)$ is an answer. So for all $\alpha$, $\alpha\cosh(x/\alpha)$ is an answer to the above differential equation.
I know that I must use a theorem to conclude uniqueness. But, this equation is describing a natural phenomenon which is unique, If you shake the chain many times, after a long time, the final shape will not change!
In such cases, provided that the answer is reasonable (here it is like a parabola which is reasonable for a hanging chain.), can we conclude the uniqueness of the answer without using any theorem?
 A: $$f'f''^2=(1+f'^2)f'''$$
Let $f'(x)=y(x) \quad\to\quad yy'^2=(1+y^2)y''$
$$\frac{y''}{y'}=\frac{yy'}{1+y^2}$$
$$2\ln|y'|=\int \frac{2y}{1+y^2}y'=\ln(1+y^2)+C$$
$$y'^2=c_1^2(1+y^2) \quad \text{with}\quad c_1=\pm e^{C/2}$$
$$y'=c_1\sqrt{1+y^2}$$
$$\frac{dy}{\sqrt{1+y^2}}=c_1dx$$
$$\sinh^{-1}(y)=c_1x+c_2$$
$$y=\sinh(c_1x+c_2)$$
$f(x)=\int ydx$
$$f(x)=\frac{1}{c_1}\cosh(c_1x+c_2)+c_3$$
Let $c_1=\alpha \quad\to\quad f(x)=\frac{1}{\alpha}\cosh(\alpha x+c_2)+c_3$
Thus $\quad f(x)=\frac{1}{\alpha}\cosh(\alpha x)\quad$ isn't the only solution since they are an infinity of other solutions with $c_2\neq 0$ and/or $c_3\neq 0$
But if some well posed boundary conditions are specified which allow to determine a unique value of $c_2$ and a unique value of $c_3$, in that case the solution $\quad f(x)=\frac{1}{\alpha }\cosh(\alpha x+c_2)+c_3\quad $ is unique.
Moreover, if those boundary conditions are particular so that they lead to $c_2=0$ and $c_3=0$ , then the solution $\quad f(x)=\frac{1}{\alpha }\cosh(\alpha x)\quad $ is unique.
One cannot answer about the uniqueness of the solution if the boundary conditions are not explicitly given.
Of course, all particular solutions belonging to the general solution of the ODE are related one to another by translation. This means that the pattern is unique for one $\alpha$ given.
