I want to solve the paired ordinary differential equations: $$ \frac{du}{dt} = 2\Omega v \sin\phi \\ \frac{dv}{dt} = -2\Omega u \sin \phi $$ The book* I am working from says the solutions are:$$ u = V_H\sin(2\Omega \sin \phi t) \\ v= V_H\cos(2\Omega \sin \phi t) $$ where: $u$ and $v$ represent horizontal velocities in the $x-$ and $y-$ planes respectively $$\Omega ~\text{is the angular rotation of Earth}$$ $$\phi ~\text{is the latitude}$$ $$ t ~\text{is time}$$ $$V_H^2 = u^2 + v^2$$
I'd be very grateful if someone could demonstrate how to solve these differential equations.$$ \text{*Pond and Pickard, "Introductory Dynamical Oceanography", 1983, Second Edition, p.64}$$
We are given the following system of linear differential equation:
$$ \frac{d}{dt} \begin{Bmatrix} u\\ v \end{Bmatrix} = \begin{bmatrix} 0 & 2\Omega\sin(\theta)\\ -2\Omega\sin(\theta) & 0 \end{bmatrix} \begin{Bmatrix} u\\ v \end{Bmatrix} $$
The eigenvalues and their respective eigenvectors of the square matrice on the right hand side are $\lambda_{1,2}=\pm i\phantom{.}2\Omega\sin(\theta)$ and $v_{1,2}=\{\mp i\phantom{x}1\}^{T}$. Now we define:
$$ \begin{Bmatrix} m\\ n \end{Bmatrix} = \begin{bmatrix} -i & i\\ 1 & 1 \end{bmatrix}^{-1} \begin{Bmatrix} u\\ v \end{Bmatrix} $$
We then have the following differential equations:
$$ \begin{align} \frac{d}{dt} \begin{Bmatrix} m\\ n \end{Bmatrix} &= \begin{bmatrix} -i & i\\ 1 & 1 \end{bmatrix}^{-1} \frac{d}{dt} \begin{Bmatrix} u\\ v \end{Bmatrix}\\ \\ &= \begin{bmatrix} -i & i\\ 1 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 2\Omega\sin(\theta)\\ -2\Omega\sin(\theta) & 0 \end{bmatrix} \begin{Bmatrix} u\\ v \end{Bmatrix}\\ \\ &= \begin{bmatrix} -i & i\\ 1 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 0 & 2\Omega\sin(\theta)\\ -2\Omega\sin(\theta) & 0 \end{bmatrix} \begin{bmatrix} -i & i\\ 1 & 1 \end{bmatrix} \begin{Bmatrix} m\\ n \end{Bmatrix} \\ \\ &= \begin{bmatrix} i\phantom{.}2\Omega\sin(\theta) & 0\\ 0 & -i\phantom{.}2\Omega\sin(\theta) \end{bmatrix} \begin{Bmatrix} m\\ n \end{Bmatrix} \end{align} $$
Notice that this is simply two separate first order differential equations:
$$ \begin{align} \frac{dm}{dt}&=\phantom{-}i\phantom{.}2\Omega\sin(\theta)\phantom{x}m\\ \frac{dn}{dt}&=-i\phantom{.}2\Omega\sin(\theta)\phantom{x}n\\ \\ m&=k_{m}\exp\left(\phantom{-}i\phantom{.}2\Omega\sin(\theta)\phantom{.}t\right)\\ n&=k_{n}\exp\left(-i\phantom{.}2\Omega\sin(\theta)\phantom{.}t\right) \end{align} $$
Now we solve for $u$ and $v$ to obtain the result you want. I leave the rest to you.
$$ \begin{align} \begin{Bmatrix} u\\ v \end{Bmatrix} &= \begin{bmatrix} -i & i\\ 1 & 1 \end{bmatrix} \begin{Bmatrix} m\\ n \end{Bmatrix} \end{align} $$