On the solutions of $y''(t)=f(y(t))$. This topic has been inspired by some time spending on trying to refining my knowledge about PDE's in general. Then everybody knows how to solve $$y''(t)=\pm y(t).$$ Then I tried to slightly modify the question and I focused on $$(\therefore)\;y''(t)=f(y(t)),\; f\in C^1(\mathbb R,\mathbb R).$$ What I was trying to do was to derive some general properties about the solutions to this equation. In particular I ask to you, since I was not able to answer myself:
Must a solution of $(\therefore)$, not identically zero, have necessarily a finite number of zeroes on $[0,1]$? My idea was to derive an estimate like 
$$|y(\eta)-y(\xi)|\leq C|\eta-\xi|^p,\; p>1;$$ Moreover, if a solution $y$ were to have an infinite number of zeroes in $[0,1]$, the the set of zeroes should have an accumulation point, and in this point all the derivatives should be equal to $0$ by continuity, then maybe the function should remain to much squeezed to be different from zero.
Hope you can help me because this interests me a lot.
Many thanks for your attention. 
(I tried to post this on mathlinks as well but nobody answered me yet) 
 A: This is a familiar type of ODE. In the following I shall deal with the "purely formal" aspect of it. Questions of sign under the square root or behavior near special points have to be treated "at runtime".
Let $F$ be a primitive of the given function $y\mapsto f(y)$. Multiplying the given differential equation by $y'(t)$ we get
$$y''(t)y'(t)=f\bigl(y(t)\bigr)\,y'(t)$$
or
$${d\over dt}\left({1\over2} y'^2(t)-F\bigl(y(t)\bigr)\right)=0\ .$$
Therefore there is a constant $C$ such that
$$y'(t)=\sqrt{2F\bigl(y(t)\bigr)+C}\ .$$
Now the variables can be separated:
$${dy\over\sqrt{2F(y)+C}}= dt\ .$$
This shows that ODEs of the considered type can be solved by quadratures. (The trick of multiplying by $y'$ is absolutely standard in mechanics, where it leads to the principle of "conservation of energy".)
A: Another way to do what Christian did.  
The equation does not explicitly mention $t$ (only $y(t)$), so we can reduce to a first-order equation.  Say $u = y'$ and get an equation for $u$ as a function of $y$ like this:
$$\begin{align}
\frac{du}{dt} &= \frac{du}{dy}\cdot\frac{dy}{dt} = \frac{du}{dy}\cdot u
\\
\text{so}\qquad y'' &= f(y)\qquad\text{becomes}
\\
u \frac{du}{dy} &= f(y)
\end{align}$$
a separable first order linear equation.  Solve it.  Once you get $u$ as a function of $y$, you have an integration problem to get (implicitly) $y$ as a function of $t$.
A: As you said, if the set of zeroes has an accumulation point $x$, then the solution and its derivatives should vanish there. In particular, this implies that $f(0)=f(y(x))=y''(x)=0$. Then by ODE uniqueness theory the solution must vanish everywhere, because $y\equiv0$ is a solution.
A detailed treatment of such equations can be found in Cazenave's notes An introduction to semilinear elliptic equations. Google can find it easily for you.
