= Solution
Substitute $y=y_0+y_F$ into the energy <functional> and integrate the resulting <polynomial>:
$$
\begin{aligned}
E
&=\int_0^L
\left[\frac A2(y'')^2+\rho gy\right]\,dx\\
&=-\frac{\rho^2g^2L^5}{40A}
+\frac{F^2L^3}{6A}.
\end{aligned}
$$
The term linear in $F$ cancels. Hence
$$
\boxed{E_0=-\frac{\rho^2g^2L^5}{40A}},
\qquad
\boxed{\mathcal E=\frac{F^2L^3}{6A}},
\qquad
\boxed{E=E_0+\mathcal E}.
$$
As required, $E_0$ is independent of the <force> and $\mathcal E$ is a <quadratic function> of it.
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