$y''(t)=-g+y'(t)^2/y(t)$ unique solution I am looking for a theorem of proof that tells me, that
$$
y(t)=r \left( 1-\cos \left( \sqrt{\frac{g}{r}}t \right) \right)
$$ 
is a unique solution to the differential equation 
$$y''(t)=-g+\frac{y'(t)^2}{y(t)}, \quad y(0) = 0, \quad y'(0) = 0,$$ 
where $r$ and $g$ are constants. Unfortunately, I am not a mathematician, so I do not know how to show this. 
 A: There is an analytical solution, obtained as follows:
You can rewrite the equation as 
$$y \frac{d}{dt} \left(\frac{y'}{y}\right) = -g$$
A few more manipulations give us the following equation:
$$\frac{y'}{y} \frac{d}{dt} \left(\frac{y'}{y}\right)  = -\frac{g \, y'}{y^2}$$
This equation is integrable; doing so produces
$$\frac{1}{2} \left ( \frac{y'}{y}\right)^2 = \frac{g}{y} + c_1$$
where $c_1$ is an integration constant.  Let's just use the positive solution for now.  A little more manipulation produces the following integration:
$$\sqrt{c_1} t + c_2 = \int \frac{dy}{\sqrt{y^2 + a y}} = \int \frac{du}{\sqrt{u^2+a}}$$
where $a=2 g/c_1$ and $u^2=y$.  This integral is well-known and may be attacked with the substitution $u=\sqrt{a} \tan{t}$, etc.  We then arrive at the result of integrating and back-substituting:
$$\sqrt{c_1} t + c_2 = 2 \log{\left ( \sqrt{\frac{c_1 y}{2 g}} + \sqrt{1+\frac{c_1 y}{2 g}} \right )}$$
From here, some clever algebra is needed, which of course I will leave to the reader.  (Rest assured, I have done it.) The result is
$$y(t) = \frac{g}{2 c_1} \left ( c_2' e^{\sqrt{c_1} t/2} - \frac{1}{c_2'} e^{-\sqrt{c_1} t/2} \right )^2$$
where $c_2'$ is some function of $c_2$ and is just a constant.  Of course, now you must apply the initial conditions which are, as stated, $y(0) = y'(0) = 0$.  The condition $y(0)=0$ implies that $c_2' = \pm 1$.  The condition $y'(0)=0$, though, is problematic in that it generates no new information. 
Is this solution unique?  Let's see. When I took a square root of $\left ( \frac{y'}{y}\right)^2$, I only used the positive sign.  A negative sign just reverses the sign of the square root, which does nothing to alter the solution.
That said, I made no progress in evaluating $c_1$, and this may indicate the discrepancy, if any, between my solution and yours.  I suspect that $c_1 < 0$ would bring about something similar to what was shown above.  A different initial condition is needed to show this.
ADDENDUM
In fact, I can reproduce the OP's result by assuming $c_1 = -|c_1| < 0$ and $c_2'=1$.  Then
$$\frac{g}{2 c_1} \left ( c_2' e^{\sqrt{c_1} t/2} - \frac{1}{c_2'} e^{-\sqrt{c_1} t/2} \right )^2 =   2 \frac{g}{|c_1|} \sin^2{\left(\frac{\sqrt{|c_1|} t}{2} \right)} = \frac{g}{|c_1|} (1-\cos{\sqrt{|c_1|} t})$$
The OP's formula results from setting $|c_1| = g/r$.  Unfortunately, I am still not sure of what $r$ is, nor the initial condition that produces it.
