How to solve this linear homogeneous diff equation when A is not constant I know how to solve the differential equation $\dot{x} = A x$ when $A$ is a constant $n\times n$ matrix. However, I cannot solve the problem when $A$ also depends on $t$. 
To be more specific, $A (t)=\left(\begin{array}{cc}
1 & -1/t\\
1+t & -1
\end{array}\right)$, where $t>0$. 
I can verify that $x(t)=\left(\begin{array}{c}
1\\
t
\end{array}\right)$ is a solution. However, in order to find the fundamental solution,
I need another linearly independent solution. I tried to set the other solution to be $\left(\begin{array}{c}
y_{1}(t)\\
y_{2}(t)
\end{array}\right)$ and plugged it into the equation. Then I got the following:
$y_{1}t-y_{2}=\dot{y}_{1}t$ and $y_{1}+y_{1}t-y_{2}=\dot{y}_{2}$. Then, I was stuck.
I also tried $(\det\Phi)^{\prime}=trA\det\Phi$ and got nowhere. 
 A: You have an ODE of the form 
$$\dot{x} (t) = A (t) \, x(t)$$
which a control theorist would call a "linear time-varying (LTV) unforced system" (unforced because there is no control input). The general solution of this ODE is the following
$$x (t) = \Phi (t,t_0)\, x(t_0)$$
where the transition matrix $\Phi (t,t_0)$ is given by the Peano-Baker series (take a look at lecture 5 on Hespanha's book for more details). Since the Peano-Baker series is scary, an alternative would be to differentiate both sides of $x (t) = \Phi (t,t_0)\, x(t_0)$ with respect to time to obtain $\dot{x} (t) = \dot{\Phi} (t,t_0)\, x(t_0)$. Since we have $\dot{x} (t) = A (t) \, x(t) = A \, \Phi (t,t_0)\, x(t_0)$, we finally obtain the matrix differential equation
$$\dot{\Phi} (t,t_0) = A (t) \, \Phi (t,t_0)$$
Let $\phi_1 (t)$ and $\phi_2 (t)$ be the 1st and 2nd columns of $\Phi (t,t_0)$, respectively. Since $\Phi (t_0,t_0) = I_{2 \times 2}$, we have the initial conditions for the ODEs on $\phi_1 (t)$ and $\phi_2 (t)$.
