Good morning.

The question asks us to find the general solution to

$\displaystyle

\boldsymbol{x}'(t) = A\boldsymbol{x}(t) + \boldsymbol{b}(t)

$

whereby

$\displaystyle A = \begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix}$ , $\displaystyle \boldsymbol{b}(t) = \begin{bmatrix} 0 \\ 1 \end{bmatrix} $ .

Just in case this it is a source of confusion for you, I'll clarify that \boldsymbol{x} can be (and in this post, is) expressed as follows:

$\displaystyle

\boldsymbol{x}(t) = \begin{bmatrix} x \\ y \end{bmatrix}

$ .

Now I will generalize much less and try to spell out each step and computation for you. It will take a bit longer, but perhaps you need to see things this way.

We begin by plugging in what we know into $\displaystyle

\boldsymbol{x}'(t) = A\boldsymbol{x}(t) + \boldsymbol{b}(t)

$ .

We get

$\displaystyle \begin{bmatrix} x' \\ y' \end{bmatrix} =

\begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix}

\begin{bmatrix} x \\ y \end{bmatrix} +

\begin{bmatrix} 0 \\ 1 \end{bmatrix} =

$ $\displaystyle

\begin{bmatrix} x + 2y \\ -y \end{bmatrix} +

\begin{bmatrix} 0 \\ 1 \end{bmatrix} =

$ $\displaystyle

\begin{bmatrix} x+2y \\ -y +1 \end{bmatrix} $ .

So we have

$\displaystyle \begin{bmatrix} x' \\ y' \end{bmatrix} =

\begin{bmatrix} x+2y \\ -y +1 \end{bmatrix} $ .

Now we multiply both sides by the integrating factor $\displaystyle e^{-tA} $ .

$\displaystyle

e^{-tA} \begin{bmatrix} x' \\ y' \end{bmatrix} =

e^{-tA} \begin{bmatrix} x+2y \\ -y +1 \end{bmatrix}

$

Now, let's explain $\displaystyle e^{-tA}$ , which is called a matrix exponential. I'll give a general definition:

Let M be a an n x n matrix. $\displaystyle e^M$ is the matrix exponential of M , which is expressed by the power series

$\displaystyle e^{M} = \sum_{k=0}^\infty \frac{1}{k!} M^k $ .

Note that the exponential of M is well-defined, as the series always converges.

Are you following so far?