TITLE: Solving a symmetric pair of differential equations
QUESTION [5 upvotes]: Looking at graphical solutions of
\begin{align}
\dot{x} &= - x + cy \\
\dot{y} &= - cy + x,
\end{align}
which for example describe how a two-level toy system approaches thermal equilibrium:

plot created with Octave (1)
one might be tempted to think that 
\begin{align}
x(t) &= x_\infty (1-e^{-t/\tau}) + x_0e^{-t/\tau} \\
y(t) &= y_\infty (1-e^{-t/\tau}) + y_0e^{-t/\tau},
\end{align}
would be canonical solutions – provided that $x_\infty + y_\infty = x_0 + y_0$. Then one might want to find out how $\tau$ depends on $c$.
But unfortunately, it seems not to be true:
$$\dot{x} = -\frac{x_\infty-x_0}{\tau}e^{-t/\tau} \neq - x + cy$$
My questions are:

Did I make a mistake, and it is a solution? 
If not so: What are the canonical solutions to the pair of equations above? Is there a closed form? How do I derive it?


(1) Octave code:
function dx = f(x,t,c)
   dx(1) = - x(1) + c * x(2);
   dx(2) = -c * x(2) + x(1);
endfunction
g = @(x,t) f(x, t, 5);
xs = lsode(g,[1,2],0:0.01:1);
plot(xs);

REPLY [3 votes]: It is quite easy to see that
$$
\dot x+\dot y=0\implies x(t)+y(t)=C=x_0+y_0=x_\infty+y_\infty
$$
and
$$
\dot x-c\dot y = -(c+1)(x-cy)
$$
so that for $c\ne -1$ 
$$
x(t)-cy(t)=e^{-(c+1)t}(x_0-cy_0)
$$
The first confirms that the sum is a conserved quantity, and both partial solutions together allow to compute the exact solution.
One might say that this is the eigenvalue-eigenvector approach, but with such a simple structure of the system an explicit computation of them is not necessary.