= Solution
Let $y_\epsilon=y+\epsilon\eta$, with $\eta=\eta'=0$ at both endpoints. At an extremal the <first variation> is zero:
$$
0=\int_{x_1}^{x_2}(F_y\eta+F_{y'}\eta'+F_{y''}\eta'')\,dx.
$$
Integrate the second term by parts once and the third term twice. The boundary term is
$$
\left[F_{y'}\eta+F_{y''}\eta'-\frac d{dx}F_{y''}\eta\right]_{x_1}^{x_2}=0.
$$
The given fundamental lemma therefore gives
$$
\boxed{F_y-\frac d{dx}F_{y'}+\frac{d^2}{dx^2}F_{y''}=0}.
$$
This is the classical form under the usual smoothness hypotheses ensuring the displayed derivatives exist; without that extra regularity the variation first yields its distributional form.
For the concrete functional the equation is $y^{(4)}-y=0$. Write $L=\pi/2$, $C=\cosh L$ and $S=\sinh L$. A general solution is $A\cos x+B\sin x+D\cosh x+E\sinh x$. The two initial conditions give $D=-A$ and $B+E=1$. The final derivative condition gives $A=EC/(1+S)$, and the final value condition then gives $E=(1+S)/2$. Thus an extremal is
$$
\boxed{y(x)=\frac C2(\cos x-\cosh x)+\frac{1-S}{2}\sin x+\frac{1+S}{2}\sinh x}.
$$
It solves the differential equation, has initial value zero and initial derivative one, and at $L$ both its value and derivative vanish, using $C^2-S^2=1$. These checks verify all four prescribed endpoint conditions.
\Image[/past-exam-of-the-options-i-course-of-the-university-of-oxford/2007-clamped-extremal.png]
{title=The clamped fourth-order extremal}
{description=The computed curve satisfies both endpoint values, with initial slope one and final slope zero.}
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