Find fundamental matrix for the equation $\dot{y}=A(t)y$.

Question

So I want to find the fundamental matrix for the equation \begin{equation}\begin{split}\dot{y}=A(t)y,\quad \text{where }A(t)=\begin{pmatrix} t & 4\\ -1 & t\end{pmatrix}.\qquad (1)\end{split}\end{equation}So far I have computed the fundamental matrix for $\dot{y}=By$ where $B=\begin{pmatrix} 0&4 \\ -1 &0\end{pmatrix}$ and found it to be $$\begin{pmatrix} \cos(2t)&2\sin(2t)\\ -\frac{\sin(2t)}{2}&\cos(2t)\end{pmatrix}.$$ I have also found the fundamental matrix for $\dot{y}=ty$ which is $e^{\frac{t^2}{2}}$ (EDIT: I suppose the fundamental matrix for this equation is $\begin{pmatrix} e^{\frac{t^2}{2}}& 0\\ 0&e^{\frac{t^2}{2}}\end{pmatrix}$). As a hint I been told to use these two results to find the fundamental matrix of (1), but I'm not sure how to do it and could use some help/hints.

Answer 1

You have to use the following result to obtain the fundamental matrix. Let a solution of $x'=Bx$ be $x(t)$ and a solution of $y'=tI_2y$ be $y$. Then $x(t)y(t)$ is a solution of $z'=(B+tI_2)z$. In fact $$ (x(t)y(t))'=x'(t)y(t)+x(t)y'(t)=Bx(t)y(t)+tI_2x(t)y(t)=(B+tI_2)x(t)y(t).$$

Answer 2

Any fundamental matrix is of the form $U(t)X$ where $U(t)$ is

$$ U(t) =\operatorname{Texp} \left ( \int_0^t ds A(s) \right) $$

But in our case $A(s)$ commutes with itself at different times so the time ordered exponential is just a standard exponential and the result is

$$ U(t) =\exp \left ( \int_0^t ds A(s) \right) $$

Now you can use the fact that $A(s)$ is a sum of two commuting objects and you obtain the answer that you gave in the comments.