Parallel transport for a scalar linear ode

Put $C(t)={\int }_a^{t}c(u)du$ and $v(t)=e^{-C(t)}{v}_0$. The ordinary fundamental theorem of calculus on each continuity piece and chain rule give $v^{\mathrm{\prime }}=-cv$. The function $C$ is continuous across the finite subdivision, so $v$ matches at every corner. Also $C(a)=0$, whence $v(a)=v_0$. By [F1] and [F2] this is the parallel solution.

Evaluating at $b$ proves the formula. If $c=2$ and $[a,b]=[0,1]$, the multiplier is $e^{-2}$. Zero initial value remains zero, $c=0$ gives the identity, and $a=b$ gives the empty integral and identity. The exponential multiplier is always positive and nonzero. For the reversed curve $h(u)=a+b-u$, its coefficient is $-c(h(u))$; substitution changes the integral's sign, giving the reciprocal multiplier. These finite scalar integrations require no selection of solution branches.