= Solution
Expand a <Jacobi field> in the parallel orthonormal <eigenbasis>: $J(t)=\sum_i j_i(t)e_i(t)$. Part (b) reduces its equation to the scalar equations $j_i''+\lambda_i j_i=0$. Define
$$
S_\lambda(t)=\begin{cases}
\sin(\sqrt\lambda\,t)/\sqrt\lambda,&\lambda>0,\\
t,&\lambda=0,\\
\sinh(\sqrt{-\lambda}\,t)/\sqrt{-\lambda},&\lambda<0,
\end{cases}
\qquad C_\lambda(t)=S_\lambda'(t).
$$
For arbitrary initial values $J(0)=\sum_i a_ie_i$ and $D_tJ(0)=\sum_i b_ie_i$, the complete solution is
$$
J(t)=\sum_i\bigl(a_iC_{\lambda_i}(t)+b_iS_{\lambda_i}(t)\bigr)e_i(t).
$$
A point at parameter $T>0$ is a <conjugate point> of $p$ precisely when a nonzero <Jacobi field> vanishes at both $0$ and $T$. The first condition makes every $a_i=0$. The second requires $b_iS_{\lambda_i}(T)=0$ for every $i$, with at least one nonzero $b_i$. For $T>0$, the factors for zero and negative <eigenvalues> never vanish; positive <eigenvalues> give exactly the sine zeros. Therefore
$$
\boxed{T=\frac{k\pi}{\sqrt{\lambda_i}},\qquad k\in\mathbb N_{>0},\quad\lambda_i>0.}
$$
The corresponding points are $\gamma(T)$. If several positive eigenspaces have a zero at the same parameter, the conjugate multiplicity is the sum of their dimensions. This also shows that a <locally symmetric Riemannian manifold> has no <conjugate points> along a given <geodesic> when that <geodesic>'s <Jacobi curvature operator> has no positive <eigenvalues>.
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