Solve non-linear differential equation with boundary conditions Consider this problem with a differential equation and border conditions. I have to find a differentiable function $f:[0,1]\rightarrow \mathbb{R}$ such that it will satisfy the following conditions: $$f(0)=a,f(1)=b,\ddot f=\dfrac{\dot f^2-c^2f^4}{f}$$
with $a,b,c\in \mathbb{R}$ given.
Differential equations are really not my field and I really don't know how to find this function $f$ (in particular how to solve this differential equation), so I hope someone will point me out a way to proceed.
As I said in the comments this problem arises from riemannian geometry: these are the conditions that a component of a geodesic must satisfy.
I've tried to solve the differential equation imposing $v:=\dot f$. In this way I get
$$\dfrac{\partial v}{\partial f}v=\dfrac{v^2-c^2f^4}{f}$$
but this isn't an ordinary separable differential equation and unfortunately I don't know how to proceed.
Thank you.
EDIT: user zwim found a solution $f=\frac{k}{c\cdot ch(kx+\phi)}$ which checks out. My question now is: how can I be sure that there aren't other solutions? If there are other solutions, which are they?
 A: Let try another change 
$u=1/cf$.
$u'=-f'/cf^2$
$u''=2f'^2/cf^3-f''/cf^2=(2f'^2-ff'')/cf^3$
Thus we get : $uu''-u'^2=(2f'^2-ff'')/c^2f^4-f'^2/c^2f^4=(f'^2-ff'')/c^2f^4=1$.
The new equation is $$uu''-u'^2=1$$
According to this link : How to solve $1+y'^2=yy''$?
If we derivate this and integrate we get $u=Ae^{kx}+Be^{-kx}$.
Let's report in the initial equation [...] and I get $4ABk^2=1$, let's call $\lambda=2Ak$ which is arbitrary.
We get $u=\frac{1}{2k}(\lambda e^{kx}+\frac{1}{\lambda}e^{-kx})$
And finally $f=\frac{2k}{c\space(\lambda e^{kx}+\frac{1}{\lambda}e^{-kx})}$
Note: there is a lot of questionnable stuff done in the way $u$ is solved, so we may have miss some solutions, but at least since we reported in the initial equation these we found, we are sure that what we found is at least correct.
Addendum Jan 02:
The substitution $u=1/cf$ does not pose problem, since there is already $f$ at denominator in original expression. The only delicate thing there is, comes from the resolution of $uu′′−u′^2=1$ because we derivate (assuming $u$ is 3 times derivable). We also divide by $uu′′$ but this is not an issue because $uu′′=1+u′^2>0$. 
As stated by Han.d.Bruijn we then get a linear ODE so unicity theorem applies. 
Yet $u''$ has an expression that depends only of $f,f',f''$ so $u'''$ exists if $f'''$ exists, but $f''$ itself has an expression that depends only from $f,f'$ so we can carry on the derivation. By the same process if one supposes $f$ is $C^1$ then it becomes automatically $C^\infty$ and so does $u$. Consequently we can be quite confident that we found all solutions at least $C^1$ and which do not annulate. 
