Encuentra el extremo de ∫d0x¨2−αxdt∫0dx¨2−αxdt\int_0^d \ddot{x}^2-\alpha x \: dt

Nos piden encontrar el extremo de$\displaystyle \int_0^d \ddot{x}^2-\alpha x \: dt$dónde$x=x(t),\: \alpha$constante,$d>0$,$x(0)=0, \: x(d)=0$,$\dot x(0) = 0, \: \dot x(d) = 0$,

considerando:$\displaystyle 0 = \frac{\partial f}{\partial x}- \frac{d}{dt}\frac{\partial f}{\partial\dot x} + \frac{d^2}{dt^2}\frac{\partial f}{\partial \ddot x}$

lo que tengo es:$f(t,x,\dot x, \ddot x) = \ddot{x}^2-\alpha x$

y resolviendo:

$$\begin{align*} 0 &= \frac{\partial f}{\partial x}- \frac{d}{dt}\frac{\partial f}{\partial \dot x} + \frac{d^2}{dt^2}\frac{\partial f}{\partial\ddot x}\\ &= -\alpha + 0 + \frac{d^2}{dt^2}2\ddot x\\ x^{(4)}&=\frac{\alpha}{2}\\ \Longrightarrow x &= \frac{\alpha}{48}t^4 + \beta t^3 + \gamma t^2 + \delta t + \varepsilon, \quad \beta, \gamma, \delta, \varepsilon \in \mathbb R \end{align*}$$

De$x(0)=0 \Longrightarrow \varepsilon=0$,$\dot x(0)=0 \Longrightarrow \delta = 0$.

$x(d)=0 \Longrightarrow 0 = \frac{\alpha}{48} d^4 + \beta d^3 + \gamma d^2 \Longrightarrow 0 = \frac{\alpha}{48} d^2 + \beta d + \gamma \Longrightarrow \gamma = - \frac{\alpha}{48} d^2- \beta d$

$\dot x(d)=0 \Longrightarrow 0 = \frac{\alpha}{12} d^3 + 3\beta d^2 + 2\gamma d$

$\Longrightarrow 0 = \frac{\alpha}{12} d^2 + 3\beta d + 2\gamma = \frac{\alpha}{12} d^2 + 3\beta d + 2(- \frac{\alpha}{48} d^2- \beta d)= \frac{\alpha}{24} d^2 + \beta d \Longrightarrow 0 =\frac{\alpha}{24} d^2 + \beta d$

$\Longrightarrow \beta = - \frac{\alpha}{24} d$

$\Longrightarrow \gamma = - \frac{\alpha}{48} d^2- \beta d = - \frac{\alpha}{48} d^2 - (-\frac{\alpha}{24} d)d =\frac{\alpha}{48} d^2$

Entonces,$\displaystyle x(t) = \frac{\alpha}{48}t^4 -\frac{\alpha}{24} d t^3 + \frac{\alpha}{48} d^2t^2$es el extremo.

Solo estoy comprobando si esto es correcto.