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.)
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.