Solution to system of linear ODE's Let $\Delta_n$ be  the closed unit simplex in $\mathbb R^n$. For any $a,b \in \Delta_n$, define the differential equation:
$$
a'(u) = b-a(u) \quad\quad\quad a(0) = a
$$
How does one go about solving this system of ODEs? Solving the case where $n=1$ is simple, does this carry over to the $n$ dimensional case?
I've looked at using eigen values but is there a simpler way when dealing with a large choice of $n$ ? Any hints appreciated
Attempt for the Matrix Differential Equation:
$$
\frac{dx}{dt} = Ax + f(t), \quad\quad\quad x(0) = x_0
$$
where $A$ is a $n\times n$ matrix, $x\in \mathbb R^n$
the solution is given by:
$$
x(t) = e^{tA} \int_0^te^{-sA}f(s)ds + e^{tA}x_0
$$
Using the fact that for a diagonal matrix : $D=\text{diag}(a_1,...,a_n)$, $e^D = \text{diag}(e^{a_1},....e^{a_n})$
then for my problem:
$$
\frac{da}{dt} = \text{diag}(-1,...-1)a(u) + b
$$
then
\begin{align*}
a(t) &= \text{diag}(e^{-t},...,e^{-t}) \int_0^t \text{diag}(e^{-s},...,e^{-s})~ b~ ds + \text{diag}(e^{-t},...,e^{-t}) a\\
&=\text{diag}(e^{-t},...,e^{-t})\text{diag}(1-e^{-t},...1-e^{-t}) b +\text{diag}(e^{-t},...e^{-t}) a\\
&=\text{diag}(e^{-t}-e^{-2t},...e^{-t}-e^{-2t}) b + \text{diag}(e^{-t},...e^{-t}) a\\
\end{align*}
$$
   =\begin{pmatrix}
        e^{-t}(a_1(0)+b_1) -b_1e^{-2t}   \\
          \vdots\\
           \vdots \\
         e^{-t}(a_n(0)+b_n) -b_ne^{-2t} \\
        \end{pmatrix}
$$
Is this correct?
 A: (You have $e^{-sA}$ not converted correctly when you plugged your special $A$ into the general recipe.)
Your system
$$x'(t)=b-x(t),\qquad x(0)=a,$$
separates without any further ado into the $n$ IVPs
$$x_k'(t)=b_k-x_k(t),\qquad x_k(0)=a_k\qquad(1\leq k\leq n)\ .\tag{1}$$
The solutions of $(1)$ are given by
$$x_k(t)=(a_k-b_k)e^{-t}+b_k\qquad(1\leq k\leq n)\ ,$$
and we can combine them to the vector solution
$$x(t)=e^{-t} (a-b) +b\ .$$
A: Well, the answer is already there, I would just like to add a typo correction. The answer via your method also yields the correct solution. 
\begin{align*}
a(t) &= \text{diag}(e^{-t},...,e^{-t}) \int_0^t \text{diag}(e^{\color{blue}{+}t},...,e^{\color{blue}{+}t})~ b~ ds + \text{diag}(e^{-t},...,e^{-t}) a\\
&=\text{diag}(e^{-t},...,e^{-t})\text{diag}(\color{blue}{e^{+t}-1},...,\color{blue}{e^{+t}-1}) b +\text{diag}(e^{-t},...e^{-t}) a\\
&=b+ + \text{diag}(e^{-t},...e^{-t}) (a-b)\\
\end{align*}
Which is  the result you can find above, as Christian said, the problem was when you plugged the $A$ in your recipe.
Hope this helps
