= Solution
Choose the internal transverse force $S(x,t)$ so that the net internal transverse <force> on a slice is $[S(x+dx,t)-S(x,t)]$. Its <linear momentum> balance is then
$$
m_b(x)h_{tt}\,dx=S_x\,dx+F_y\,dx.
$$
The coordinate $x$ labels body slices, so their lateral material acceleration is $h_{tt}$; the fluid transport operator $\mathcal D$ must not be substituted for this body acceleration.
Let $G$ be the total internal <bending moment>, including both passive elastic and active muscular contributions. With the compatible moment orientation, the net couple on the slice is $(G_x-S)\,dx$. The cross-sectional rotary inertia is smaller than the translational inertia by the square of the transverse-to-longitudinal length ratio, and is neglected in this slender-beam approximation. Slice <angular momentum> balance therefore gives $S=G_x$. Eliminating $S$ from the slice <linear momentum> balance proves
$$
\boxed{G_{xx}=m_bh_{tt}-F_y.}
$$
This equation is the balance law for an <active bending beam for a swimming fish>; a constitutive relation for the passive <bending moment> or the muscle actuation is not needed to derive it. With the force-on-body convention established above it becomes
$$
G_{xx}=m_bh_{tt}+\mathcal D(m\mathcal Dh).
$$
As a check, constant $m$ and rigid lateral acceleration give the effective inertia $m_b+m$. Combining the printed positive force in part (i) with the minus sign in the beam equation would instead give $m_b-m$. That is the same sign defect, rather than a negative hydrodynamic <added mass>. If the positive quantity in part (i) is called $F_y$ as a force on the fluid, the beam equation must contain $+F_y$.
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