Exercise 7.2.24

As in the previous exercise, we write $Q^{-1}\mathit{AQ}=J$, where $J$ is the Jordan canonical form of $A$. Then solve $Y^{\prime }=\mathit{JY}$. Finally the answer should be $X=\mathit{QY}$.

1.

The coefficient matrix is

$$A=\left(\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right).$$

Compute

$$J=\left(\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right)$$

and

$$Q=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right).$$

Thus we know that

$$Y=e^{2t}\left(\begin{array}{c}\hfill \mathit{at}+b\hfill \\ \hfill a\hfill \\ \hfill 0\hfill \end{array}\right)+e^{3t}\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill c\hfill \end{array}\right).$$

and so the solution is

$$X=\mathit{QY}.$$

2.

The coefficient matrix is

$$A=\left(\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 2\hfill \end{array}\right).$$

So $J=A$ and $Q=I$. Thus we know that

$$Y=e^{2t}\left(\begin{array}{c}\hfill a{t}^2+\mathit{bt}+c\hfill \\ \hfill 2\mathit{at}+b\hfill \\ \hfill 2a\hfill \end{array}\right)$$

and so the solution is

$$X=\mathit{QY}.$$