solutions fort-order homog. system Question: Can one of the following functions be a solution of a first-order autonomous homogeneuous system?
(1) $x(t)=(3e^{t}+e^{-t},e^{2t})$
(2) $x(t)=(3e^{t}+e^{-t},e^{t})$
(3) $x(t)=(3e^{t}+e^{-t},t e^{t})$
(4) $x(t)=(3e^{t},t^2 e^{t})$
I know that every solution can be written as a linear combination of $t^j exp(\alpha t)$, therefore I would say that only (4) can be a solution, true?
 A: Its a first order system with a 2nd order characteristic equation for each of the variables. These two solutions show up as linear combinations of exp in the answer. 
It is also possible that the two roots are in fact, repeated roots, in which case you do not have two distinct combinations of exp, but what you will have is exp of the repeated root with an extra factor of t. Possible answers would therefore look like:
$Ae^a + Be^b, Ce^a + De^b$
OR
$Ae^a , t Be^{a}$
None, but 2) of the options matches the form above. 
4) cannot be an answer because, only a triple repeated root would have a t^2, which is not possible for a 1st order system.
A: I assume by "homogeneous system" you mean "homogeneous linear system". 
A $2 \times 2$ matrix can either have two eigenvalues or one eigenvalue of multiplicity $2$.  In the first case the solutions are linear combinations of vectors times two exponentials $e^{\lambda_1 t}$ and $e^{\lambda_2 t}$.  That's what you have in one of your four examples.
In the second case you have one exponential and $t$ times that exponential. 
But you can't have three different $\lambda$'s, or $e^{\lambda_1 t}$ and $t e^{\lambda_2 t}$ with $\lambda_2 \ne \lambda_1$, or $t^2 e^{\lambda t}$.  
