Exactly what do you mean by "meaning"? A differential equation involving complex numbers has every bit as much meaning as an algebraic equation involving complex numbers or, indeed, as complex numbers themselves.
If dy/dt= iy then y= Ce^{it} where C can be any complex number.
The second order equation, d^2y/dt^2+ y= 0 can be written as a pair of first order differential equations by defining x= dy/dt so that d^2y/dt^2+ y= dx/dt+ y= 0 and dy/dt= x.
You ask about "two cascaded first order systems that have complex roots". First the "system" is the pair of first order equations. Second, the system does not have "complex roots", its characteristic equation has.
We can write the example above, dx/dt= -y and dy/dt= x as the matrix equation
\frac{d}{dt}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}
That matrix has "eigenvalue equation" (the characteristic equation of the system of differential equations)
\left|\begin{array}{cc}-\lambda & -1 \\ 1 & -\lambda \end{array}\right|= \lambda^2+ 1= 0
which has roots \lambda= \pm i.
So the matrix has eigenvalues i and -i and corresponding eigenvectors (1, -i) and (1, i). In particular, if we construct the matrix B= \begin{bmatrix}1 & 1 \\ -i & i\end{bmatrix}, having those eigenvectors as columns, which has inverse matrix B^{-1}= \begin{bmatrix}1/2 & i/2 \\ 1/2 & -i/2\end{bmatrix}
Then the equation dX/dt= Ax becomes d(BX)/dt= (BAB^{-1})(BX) which is
\frac{d}{dt}\begin{bmatrix}u \\ v\end{bmatrix}= \begin{bmatrix}i & 0 \\ 0 & -i\end{bmatrix}\begin{bmatrix}u \\ v \end{bmatrix}
where u and v are defined by \begin{bmatrix}u \\ v \end{bmatrix}= \begin{bmatrix}1 & 1 \\ -i & i \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix} so that u= x+ y and v= -ix+ iy.
That last system reduces to the two uncoupled equations du/dt= iu and dv/dt= -iv so that we have u(t)= Ce^{it} and v(t)= De^{-it}. Since u= x+ y and v= -ix+ iy, x= (1/2)u+ (1/2)i v= C' e^{it}+ D' e^{-it} and y= (1/2)u- (1/2)iv= C'e^{it}- D'e^{-it} where C'= (1/2)C and D'= (1/2)Di.
