Solving a symmetric pair of differential equations 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);

 A: The general solution for this system should be: $$x(t)=C_1 (c+e^{-(1+c)t})+C_2 (c-ce^{-(1+c)t}) \\ y(t)=C_1 (1-e^{-(1+c)t})+C_2 (1+ce^{-(1+c)t})$$ Note that $c$ is inside the brackets in some places, but not others. 
$C_1,C_2$ should be found from initial conditions. 
Also note that $x+y$ really doesn't depend on $t$ here.
$$x(0)=(1+c)C_1 \\ y(0)=(1+c)C_2$$
A: 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.
A: The general method uses eigenvalues and eigenvectors, but they are not really needed here. 
You have, from the second equation, that $x=y'+cy$. So $x'=y''+cy'$. Going back to the first equation, 
$$
y''+cy'+y'+cy=cy.
$$
That is, 
$$
y''+(c+1)y'=0. 
$$
This is first order linear in $y'$, with $y'=y_1\,e^{-(1+c)t}$. So, unles $c=-1$, 
$$
y(t)=-\frac{y_1}{1+c}\,e^{-(1+c)t}+y_0, 
$$
and
\begin{align}
x(t)&=y'+cy=y_1\,e^{-(1+c)t}-\frac {c\,y_1}{1+c}\,e^{-(1+c)t}+c\,y_0\\ \ \\
&=y_1\,\frac{1}{1+c}e^{-(1+c)t}+c\,y_0.
\end{align}
As $y_1$ is an arbitrary constant, we may replace it by $y_1(1+c)$, and we get 

$$
x(t)=y_1\, e^{-(1+c)t}+c\,y_0,
$$
  $$
y(t)=-y_1\,e^{-(1+c)t}+y_0.
$$

If you rewrite your candidate solutions as 
$$
x=x_\infty+(x_0-x_\infty)e^{-t/\tau},\ \ \ \ y=y_\infty+(y_0-y_\infty)e^{-t/\tau},
$$
indeed you need $x_0+y_0=x_\infty+y_\infty$, but you also need $x_\infty=cy_\infty$, and $\tau=1/(1+c)$. 

For the particular case $c=-1$, we have $y''=0$, so $y(t)=y_1\,t+y_0$, and $x(t)=-y_1\, t+y_1-y_0$. 
A: We have the following pair of coupled linear ODEs
\begin{align}
\dot{x} &= - x + cy\\
\dot{y} &= \,\,\,\, x - cy\\
\end{align}
Such ODEs are very common in chemical kinetics, for example. Note that $\dot{x} + \dot{y} = 0$. Integrating, we obtain $x(t) + y(t) = x_0 + y_0$, where $x_0$ and $y_0$ are initial conditions. Therefore, the 1st ODE can be decoupled from the 2nd ODE as follows
$$\dot{x} = - x + c \left( (x_0 + y_0) - x \right) = - (1+c) \, x + c \,  (x_0 + y_0)$$
or, alternatively,
$$\dot{x} + (1+c) \, x = c \, (x_0 + y_0)$$
which is a standard non-homogeneous 1st order linear ODE. Integrating, I obtained the following
$$x (t) = \color{blue}{\left( \left( \frac{c}{1+c} \right) x_0 + \left( \frac{c}{1+c} \right) y_0 \right) + \left( \left( \frac{1}{1+c} \right) x_0 - \left( \frac{c}{1+c} \right) y_0 \right) e^{- (1+c) \, t}}$$
and, since $y(t) = (x_0 + y_0) - x(t)$,
$$y (t) = \color{blue}{\left( \left( \frac{1}{1+c} \right) x_0 + \left( \frac{1}{1+c} \right) y_0 \right) - \left( \left( \frac{1}{1+c} \right) x_0 - \left( \frac{c}{1+c} \right) y_0 \right) e^{- (1+c) \, t}}$$
