Differential Equations and Eigenvectors/Eigenvalues I am trying to solve $\dfrac{d\mathbf{x}}{dt} = \left[\begin{array}{cc}
-4 &1\cr
-6 &1
\end{array}\right] \mathbf{x}$ and I need to find the general solution of the system in the form $x=c_1x_1+c_2x_2$.
Finding the eigenvalues, I have $\det(P - \lambda I) = 0 \implies \lambda = \dfrac{-5 \pm \sqrt{21}}{2}$. But that can't be right since I'm going outside of the integers and into the irrationals and my vectors $x_1$ and $x_2$ are supposed to be integer-valued vectors. Any help would be appreciated. Thanks in advance.
 A: OK, I'll give the "dumb" way to do this--the way I used to do it in my Diff. EQ. class because I didn't understand eigenvalues or linear algebra at the time.  Assume you have the following:
$$
x = Ae^{rt} \\
y = Be^{rt}
$$
Plug in:
$$
\dot{x} = rAe^{rt} = -4Ae^{rt} + Be^{rt} \\
\dot{y} = rBe^{rt} = -6Ae^{rt} + Be^{rt}
$$
Now, we can factor out $e^{rt}$ because this is never zero (but even if it was, we could just set this equal to zero and factor out $e^{rt}$...so long as $e^{rt}$ is not identically zero), this gives:
$$
rA = -4A + B \\
rB = -6A + B
$$
Which leads to:
$$
0 = (-4 - r)A + B \\
0 = -6A + (1 - r)B
$$
Now, the "correct" way is to recognize that $r$ represents the eigenvalues of the original matrix $\begin{pmatrix}-4 & 1 \\ -6& 1\end{pmatrix}$ (i.e. by setting the above matrix's determinate to $0$).  But, hey, I didn't understand that at the time, so instead I simply solved for $B$ and then plugged into the second equation to find $r$ (or we could solve for $A$ and plug into the second equation or we could solve for $B$ in the second and plug into the first or we could solve for $A$ in the second and plug into the first...there isn't a "single" correct way to approach this):
$$
B = (4 + r)A \\
0 = -6A + (1 - r)(4 + r)A
$$
We can factor out $A$ to give:
$$
0 = -6 + (1 - r)(4 + r) = 0 \\
-6 + 4 - 3r - r^2 = 0 \\
r^2 + 3r + 2 = 0 \\
(r + 1)(r + 2) = 0
$$
This gives that $r = -1, -2$--two solutions.  The nice thing is that I already figured out what $B$ was in terms of $A$ and $r$ which means I get the general solution now:
$$
B_{-1} = (4 + -1)A = 3A \\
B_{-2} = (4 + -2)A = 2A
$$
Which finally leads to the general solution:
$$
x = A_1e^{-t} + A_2e^{-2t} \\
y = 3A_1e^{-t} + 2A_2e^{-2t}
$$
or
$$
\vec{x}(t) = \begin{pmatrix} 1 & 1 \\
3 & 2 \end{pmatrix}\begin{pmatrix}A_1e^{-t} \\
A_2e^{-2t}\end{pmatrix}
$$
A: I think you messed up the eigenvalues.. If $A \in \Bbb R^{2 \times 2}$, then $p_A(t) = t^2 - {\rm tr}(A)\,t + \det(A)$, so we have to solve: $$t^2 + 3t +2 = 0,$$which clearly has roots $-1$ and $-2$. Now you find eigenvectors $\bf v$ and $\bf w$, respectively. Your solution will be: $${\bf x} = e^{-t}{\bf v} + e^{-2t}{\bf w}.$$However, even if the eigenvalues weren't integers, you would do the same computations.
