= Solution
{c}
Let $W_r,W_\theta,W_\phi$ denote derivatives of the <strain energy density> with respect to its three <principal stretches>. The <principal stress formula for incompressible hyperelasticity> is
$$
t_r=\lambda_rW_r-p,\qquad
t_\theta=\lambda_\theta W_\theta-p,\qquad
t_\phi=\lambda_\phi W_\phi-p.
$$
It follows by applying $\mathbf T=\mathbf F(\partial_{\mathbf F}W)^T-p\mathbf1$ in principal axes. Equal tangential <principal stretches> and <material isotropy> give $W_\theta=W_\phi$ at $(\lambda^{-2},\lambda,\lambda)$. Differentiating the reduced <strain energy density> by the <chain rule> gives
$$
\begin{aligned}
h'(\lambda)&=-2\lambda^{-3}W_r+W_\theta+W_\phi\\
&=2\left(W_\theta-\lambda^{-3}W_r\right)
=\frac2\lambda(t_\theta-t_r).
\end{aligned}
$$
The unknown <pressure> cancels in this difference. Substituting in part (i), and then using the supplied derivative of $\lambda$, gives
$$
\frac{dt_r}{dr}=\frac{\lambda h'(\lambda)}r=\frac{h'(\lambda)}R,
\qquad
\frac{d\lambda}{dr}=\frac{1-\lambda^3}{R}.
$$
Consequently the desired <ordinary differential equation> is
$$
\boxed{\frac{dt_r}{d\lambda}=\frac{h'(\lambda)}{1-\lambda^3}.}
$$
For completeness, the supplied derivative follows from $R^3=r^3-(a^3-A^3)$: $dR/dr=r^2/R^2=\lambda^2$, and differentiating $\lambda=r/R$ gives $(1-\lambda^3)/R$. At the undeformed value $\lambda=1$ the division must be interpreted by a limit; the <spherically symmetric deformation> with $a=A$ has $\lambda\equiv1$ and should be handled directly by the radial equilibrium equation.
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