ODE problem reading do Carmo's book of Riemannian geometry I'm reading do Carmo's book, Riemannian geometry. I have a problem at the Jacobi fields. He talks about the case of constant curvature. He gets to the ODE above and my problem is how dose he solve it? Can some one fill in the details? Thanks a lot!

As a result, the Jacobi equation can be written as
$$\frac{D^2J}{dt^2}+KJ~=~0$$
Let $\omega(t)$ be a parallel field along $\gamma$ with $\langle\gamma'(t),\omega(t)\rangle=0$ and $|\omega(t)|=1$. It is easy to verify that
$$J(t)~=~\begin{cases}\frac{\sin(t\sqrt{K})}{\sqrt{K}}\omega(t),~~~&\text{if}~K>0\\t\omega(t),~~~&\text{if}~K=0\\\frac{\sinh(t\sqrt{-K})}{\sqrt{-K}}\omega(t),~~~&\text{if}~K<0\end{cases}$$
is a solution of $(2)$ with initial conditions $J(0)=0,J'(0)=\omega(0)$

 A: Let me clear the relation between the covariant derivative formalism and the standard ODE formalism. Given a parallel vector field $\omega(t)$ along $\gamma(t)$ which satisfies the conditions written, let us try and find a solution for the Jacobi equation of the form $J(t) = f(t) \omega(t)$ where $f \colon I \rightarrow \mathbb{R}$ is a scalar function. By the product rule and the fact that $\omega$ is parallel, we have
$$ \frac{DJ}{dt}(t) = f'(t) \omega(t) + f(t) \frac{D\omega}{dt}(t) = f'(t) \omega(t),\\
\frac{D^2J}{dt}(t) = f''(t) \omega(t)
$$
so the equation becomes
$$ f''(t) \omega(t) + K f(t) \omega(t) = (f''(t) + Kf(t)) \omega(t) = 0. $$
Since $\omega(t) \neq 0$ for all $t \in I$, we must have $f''(t) + Kf(t) = 0$ for all $t \in I$ and in addition, by the initial conditions, we must also have
$$ J(0) = 0 \iff f(0) = 0, J'(0) = f'(0) \omega(0) = \omega(0) \iff f'(0) = 1. $$
Hence, to find a solution for the Jacobi equation of the form above we must solve the second order constant coefficients scalar ODE
$$ f''(t) + K f(t) = 0 $$
with initial conditions
$$ f(0) = 0, f'(0) = 1. $$
