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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?

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    Any constant $f$ will solve this ODE. – 2017-01-29
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    I mean the only reasonable answer @user1337. A constant $f$ cannot be the answer obviously. That's a chain, not a rod! – 2017-01-29
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    You are leaving out the initial conditions -- I assume these are what will make the solution unique. And this is exactly what the [Picard–Lindelöf theorem](https://en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem) makes mathematically rigorous. – 2017-01-29
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    @ClementC. we can, at the bottom, put both the function value and its first derivative zero. And we can specify a chain length and also the difference in the heights of the two fixed points of the chain. These provide enough constraints to the problem. But translating these to some usable math will be a tough job :( and I don't understand a word of that theorem, unfortunately. – 2017-01-29
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    i think you have right the equation has hyberbolic functions as solutions – 2017-01-29

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$$f'f''^2=(1+f'^2)f'''$$ 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.

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    Great solution! Nearly identical to what I had in mind. – 2017-01-29