Is there a way to solve systems of linear differential equations without using eigenvector/eigenvalues? I'm teaching basic differential equations this semester and I was wondering if there were methods of solving systems of linear differential equations that don't use (directly) eigenvectors/eigenvalues.  Is there such a way?
 A: I know my question was a bit malformed.  What I'm thinking is how would one approach solving a system of linear homogeneous differential equations without knowledge of linear algebra. 
I was experimenting with something along the lines of looking at lines through the origin as solutions and from these constructing the general solution.  Lines through the origin can be characterized as
$x_{i}(t) = m_{i} x_{1}(t)$
for $i$ from 2 to $n$. Now use that 
$\frac{dx_{i}}{dx_{1}} = m_{i}$
and that
$\frac{dx_{i}}{dt} = \left(\sum a_{ij} m_{i}\right)x_{1}(t)$
with $m_{1}=1$ to get a system of equations in the $m_{i}$.  Note that $x_{1}(t)$ falls out.  Once one finds the solutions to the $m_{i}$ we can find a solution to the system of equations.  This will lead to finding the eigenvalues but without use of them explicitly.
An example.  Consider
A = {{1, -1}, {1, 1}} (in Mathematica notation).
$y(t) = m x(t)$
$\frac{dx}{dt} = x(t) - m x(t)$
$\frac{dy}{dt} = x(t) + m x(t)$
Thus
$\frac{dy}{dx} = m = \frac{x(t) +m x(t)}{x(t) - m x(t)}=\frac{1+m}{1-m}$
Solutions to $m = \frac{1+m}{1-m}$ are $-i$ and $i$.  This gives us the equation
$\frac{dx}{dt} = x(t)+i x(t)$
Which has a solution of $e^{(1+i)t}$ and $1+i$ is an eigenvalue of $A$. 
I'm finding eigenvalues of $A$ by looking at solutions to the system of differential equation that go through the origin.  I run into trouble with eigenvalues of multiplicity greater than 1.  I was wondering if there was a general technique along these lines that is more formal that what I'm doing.
