Solution to System of Linear Differential Equations with Variable Coefficients I'm stuck trying to solve the system $\frac{dx(t)}{dt} = A(t)x(t)$ with $A(t) =\frac{1}{2}\begin{pmatrix}
ln(t+1) & ln(t-1)\\
ln(t-1) & ln(t+1)
\end{pmatrix}$.
I think I could try computing $e^{\int A(z) dz}$ and that should give me a fundamental matrix for the solutions? (I failed miserably at that though.)
 A: We shall consider our initial time at $t_0 = 1$ for this post.
Since
\begin{align}
A(t) = \frac{1}{2}
\begin{pmatrix}
\ln (t+1) & \ln (t-1)\\
\ln (t-1) & \ln (t+1)
\end{pmatrix}
\end{align}
then it follows
\begin{align}
\int^t_{1} A(s)\ ds = \frac{1}{2}
\begin{pmatrix}
(t+1)\ln (t+1) +(t+1) & (t-1)\ln(t-1) + (t-1)\\
(t-1)\ln(t-1) + (t-1) & (t-1)\ln(t+1) + (t+1)
\end{pmatrix}+I
\end{align}
where $I$ is the identity matrix. Now, let us rewrite the above solution as
\begin{align}
\int^t_1 A(s)\ ds = M(t) + I
\end{align}
which means we need to find the exponetial matrix
\begin{align}
\exp\left( M(t)+I \right) = \exp [M(t)] \exp(I)=\exp[M(t)]. 
\end{align}
Before we compute anything, we shall look at the eigen-decomposition of $M$. Observe since
\begin{align}
M(t) = \begin{pmatrix}
a(t) & b(t)\\
b(t) & a(t)
\end{pmatrix}  
\end{align}
which means the characteristic polynomial is given by
\begin{align}
p(z) = (z-a)^2-b^2 = z^2-2az+(a^2-b^2).
\end{align}
Hence the roots of $M(t)$ for any fixed time is given by $a(t)-b(t)$ and $a(t)+b(t)$. Thus, it follows 
\begin{align}
M(t) = 
\frac{1}{2}
\begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}
\begin{pmatrix}
a(t)+b(t) & 0\\
0 & a(t)-b(t)
\end{pmatrix}
\begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}
\end{align}
which means
\begin{align}
\exp[M(t)] = \frac{1}{2}
\begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}
\begin{pmatrix}
e^{a(t)+b(t)} & 0\\
0 & e^{a(t)-b(t)}
\end{pmatrix}
\begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}.
\end{align}
The rest of the computation is straightforward. 
