How do you solve $\dot{X} = UX$? Let $X(t) = \left[\begin{matrix}a(t) & b(t) \\c(t) & d(t)\end{matrix}\right]$ and let $U$ be a nonsingular matrix. How do you solve
$$\frac{d}{dt} X(t)=UX(t)$$
I presume there is some general method to solve these kinds of ODEs but I cannot find anything about it, online. 
BTW: -----------------
I know that you can get 4 ODE equations for the four unknown functions of $t$. The problem is that each ODE equation includes other functions as so
$$\frac{d a}{d t}=u_{11}a+u_{12}c$$

You are correct, it can be solved column-wise, and in analogy to the scalar form of this differential equation.
The solution is 
$$X(t)=\exp\left(U t\right) X(0)$$
where $X(0)$ is the initial condition (matrix) and 
$$\exp(Ut)=\sum_{n=0}^\infty \frac{(Ut)^n}{n!}$$ is the matrix exponential, which can be computed explicitly if eigenvectors and eigenvalues are known.
Let $U=W \Lambda V^T$ with diagonal matrix $\Lambda$ and left and right eigenvector matrices $V$ and $W$. Then,
$$\exp(Ut)=W \exp(\Lambda t) V^T$$
where
$$\Lambda = \textrm{diag}(\lambda_1 ,\ldots,\lambda_N )$$
is the diagonal matrix of the $N$ eigenvalues of $U$
$$\exp(\Lambda t) = \textrm{diag}(\exp(\lambda_1 t),\ldots,\exp(\lambda_N t))$$
If you assume the functions $a(t)$, $b(t)$ etc. to be scalars, you have $N=2$ and can compute the eigenvalues from the trace and determinant of $U$, $\lambda_1+\lambda_2=U_{11}+U_{22}$ and $\lambda_1\lambda_2=U_{11}U_{22}-U_{21}U_{12}$.
