Euler-Lagrange equation with higher derivatives in the functional Find a function $\phi$ of class $C^2$ (first and second derivatives exist and are continuous) that minimice the functional: 
$I(\phi)= \int_0^1 \frac {\phi''(t)} {\phi(t)} dt$
and $\phi(0)=1$, $\phi(1)=4$.
I think i have to use Euler lagrange equations to solve this. But how? The functional has second derivative of $\phi$! Any help would be greatly appreciated!
 A: Hint: Euler-Lagrange equations for a Lagrangian $L(\phi,\dot{\phi},\ddot{\phi},\dots)$ are given by 
$$\frac{\partial L}{\partial\phi}-\frac{d}{dt}\Big(\frac{\partial L}{\partial\dot{\phi}}\Big)+\cdots+(-1)^n\frac{d^n}{dt^n}\Big(\frac{\partial L}{\partial\phi^{(n)}}\Big)=0.$$
A: Warning: the functional seems to be unbounded below, even if we include the natural condition $\phi(x)>0$ for $0\leq x\leq 1$ (which is, perhaps, understood?). In fact, if we set $u_s = 1+3x+sx(1-x)$, where $s>0$, then each $u_s$ is nonnegative and satisfy the end point conditions. Moreover
\begin{equation*}
\int_0^1\frac{u_s''(x)}{u_s(x)}\,dx \to -\infty\quad\text{as $x\to\infty$}.
\end{equation*}
Maximizing the integral therefore seems to be a more interesting problem.
A: HINT:
Taking hint from yngabI further. I am not quite sure of this but if it helps to find correct way further for higher order derivatives, am indicating for a start. I did it once or twice a long time ago. 
No $x$ here explicitly. For convenience adopting $y=f(x).$ 
So,
$$ F= y''/y$$
$$ F - y' F_{y'} + y'' F_{y''} = const1 $$
$$ y''/y - 0 + y'' (1/y) = const1 $$
$$ y''/y = const2 =  \pm \omega^2 $$
$$ \phi^{\prime\, \prime} (t) \pm  \omega^2 \phi(t) =0 . $$
Apply boundary values of DE the usual way and evaluate constants.
