finding solutions for homogeneous linear systems of ode with non-constant coefficients For the matrix $A: \mathbb R \rightarrow \mathbb R^{2 \times 2}$
$ A(t) := \begin{pmatrix} 0 & 1\\ 0 & 2t \end{pmatrix}$
I want to find two linearly independent solutions of $\dot x(t) = A(t)x(t)$, but I can only know and find methods for systems with constant coeffiecients. 
Can anybody give me instructions on how to deal with the non-constant ones?
 A: A closed-form solution for systems with time-varying $A$ is difficult to find in general.
However, in this specific case a solution is easier to obtain since the $x_2$ dynamics is decoupled from $x_1$. From the form of $A(t)$ we have
$$\dot{x}_2=2tx_2$$
or equivalently
$$\frac{dx_2}{x_2}=2tdt$$
If we now integrate over $[0,t]$ we obtain
$$\ln(x_2(t))-\ln(x_2(0))=t^2$$
i.e.
$$x_2(t)=\exp(t^2)x_2(0)$$
Since $\dot{x}_1=x_2$ we can integrate to calculate $x_1$
$$x_1(t)=x_1(0)+x_2(0)\int_0^t{exp(s^2)ds}=x_1(0)+x_2(0)\exp(t^2)D_{+}(t)$$
where $D_{+}(t)$ is the Dawson function.
Since the transition matrix 
$$\Phi(t)=\left[\matrix{ 1 & \exp(t^2)D_{+}(t) \\ 0 & \exp(t^2)}\right]$$
is always invertible, choosing a set of independent initial conditions will result in linear independent solutions. For example if we select initial conditions
$$x_{01}=\left[\matrix{1 \\0}\right], x_{02}=\left[\matrix{0 \\1}\right]$$  we will get the linear independent solutions
$$x(t,x_{01})=\left[\matrix{1 \\0}\right], \qquad x(t,x_{02})=\left[\matrix{\exp(t^2)D_{+}(t) \\ \exp(t^2)}\right]=\exp(t^2)\left[\matrix{D_{+}(t) \\ 1}\right]$$
