Relaxation time in a stationary stochastic process
Considering the complexity of the question, and the rating of your reputation, i assume you are still alive. 

In the following, i am assuming $k$ is a constant, since it is not a variable -fixed or random- expressed in the question.

If you are still interested, which is a numerically strong assumption, the system can be rewritten trivially as:

$$\frac{d}{dt}
\left(\begin{array}{cc}
x_1  \\
x_2 
\end{array}\right)=
\left(\begin{array}{cc}
-(t_1+d) &amp; t_0 \\
t_1 &amp; -(t_0+d)
\end{array}\right)
\left(\begin{array}{cc}
x_1  \\
x_2 
\end{array}\right)
+
\left(\begin{array}{cc}
k  \\
0 
\end{array}\right)
=
\left(-d
\left(\begin{array}{cc}
1 &amp; 0  \\
0 &amp; 1
\end{array}\right)
+
\left(\begin{array}{cc}
-1 &amp; 1  \\
1 &amp; -1
\end{array}\right)
\left(\begin{array}{cc}
t_1  \\
t_0
\end{array}\right)\right)
\left(\begin{array}{cc}
x_1  \\
x_2 
\end{array}\right)
+
\left(\begin{array}{cc}
k  \\
0 
\end{array}\right)$$

Which has the standard form: 
$$Dx=Ax+x_0$$

The constant vector $x_0=[k;0]$ or equivalently the constant vector $A^{-1}x_0$ is a state reference.

Through the variable change:
$$x+x_0 \to x$$
you can safely discard the $x_0$ vector.

This is an autonomous system which, unless you forget to include perturbations or inputs, is a very boring system. I am sure you did.