Solving a 2x2 matrix ode

Question

I am able to solve ordinary differential systems of the form

Find the solution of the ODE $$\frac{dX}{dt} = AX$$ where $X$ is a $2\times1$ matrix and $A$ is $2\times2$

However I am not able to find a solution where $X$ is $2\times2$

Can somebody please provide me hint on how to approach such a problem?

Take $A = \begin{pmatrix} 1 &-2\\ -2 &1\\ \end{pmatrix}$

Edit:

Write $U=P^{-1}X$ to obtain $U'=DU$ Thus $$ \begin{pmatrix} x_1' &x_2'\\ x_3' &x_4'\\ \end{pmatrix} = \begin{pmatrix} -1 &0\\ 0 &3\\ \end{pmatrix}\begin{pmatrix} x_1 &x_2\\ x_3 &x_4\\ \end{pmatrix}$$ This gives us

$$ \begin{pmatrix} x_1' &x_2'\\ x_3' &x_4'\\ \end{pmatrix} =\begin{pmatrix} -x_1 &-x_2\\ 3x_3 &3x_4\\ \end{pmatrix}$$ which gives us $$ \begin{pmatrix} x_1 &x_2\\ x_3 &x_4\\ \end{pmatrix} =\begin{pmatrix} e^{-t} &e^{-t}\\ e^{3t} &e^{3t}\\ \end{pmatrix}$$ Plugging it in $X=PU$ gives us $$X =\begin{pmatrix} 1 &1\\ -1 &1\\ \end{pmatrix} \begin{pmatrix} e^{-t} &e^{-t}\\ e^{3t} &e^{3t}\\ \end{pmatrix}$$

Answer 1

The matrix solution to $$\frac{dX}{dt} = AX$$

is the exponential $$ X= e^{tA}$$

where

$$A = \begin{pmatrix} 1 &-2\\ -2 &1\\ \end{pmatrix}$$

There are different methods to find $ e^{tA} $

One way is to find two linearly independent vectors satisfying your $$ \frac{dV}{dt} = AV$$ with initial values of $(1,0)$ and $(0,1)$

These vectors are then the columns of your matrix solution $X=e^{tA}$.

Answer 2

Hint: Write $A=PDP^{-1}$, where $D$ is diagonal, and set $U=P^{-1}X$. Then $$ \frac{dU}{dt} = P^{-1} \frac{dX}{dt} = P^{-1}AX = P^{-1}PDP^{-1}X = DU $$ which is easy to solve. Then $X=PU$.