So i'm given a differential equation
$$f'''(t)-2f''(t)-f'(t)+2f(t)=0$$ where $$t\in \mathbb{R}$$
and
$$C=\begin{pmatrix}0 & 1 & 0\\
0 & 0 & 1\\
-2 & 1 & 2\end{pmatrix}$$
the polynomial assosiated with $C$, $p(t)=t^3-2t^2-t+2$
Now assume that $f(t)$ is a solution of the differential equation and set that $\vec{x}(t)=(f(t),f'(t),f''(t))$
Question: How do I check that $\vec{x}(t)$ is a solution of the first order of systems?
Attempt:
Since this is disconnected we can solve it this way:
First we find the diag of $C$
which in this case is $$P=\begin{pmatrix}1 & \frac{1}{4}& 1\\ 1 & \frac{1}{2} & -1 \\ 1 & 1 & 1 \end{pmatrix},D=\begin{pmatrix}1 & 0& 0 \\0&2&0\\0&0&-1\end{pmatrix}$$
Then proceeed to solve the disconnected system $$\vec{y}'=D\vec{y}$$
$$\vec{y}(t)=\begin{pmatrix}C_1e^{t}\\C_2e^{2t}\\C_3e^{-t}\end{pmatrix}$$
where $$C_1,C_2,C_3\in \mathbb{R}$$
Which will help us solve $\vec{x}(t)=P\vec{y}(t)$
$$\vec{x}(t)=P\vec{y}(t)=\begin{pmatrix}C_1e^t+\frac{C_2e^{2t}}{4}+C_3e^{-t}\\ C_1e^t+\frac{C_2e^{2t}}{2}-C_3e^{-t}\\C_1e^t+C_2e^{2t}+C_3e^{-t}\end{pmatrix}$$
This is where i'm pretty much stuck… What am I supposed to do now? Thanks for any help in advance.
Best Answer
So far so good.
Now you have to show that $$x'=Cx$$
That is you have to take derivative of $$ \begin{pmatrix}C_1e^t+\frac{C_2e^{2t}}{4}+C_3e^{-t}\\ C_1e^t+\frac{C_2e^{2t}}{2}-C_3e^{-t}\\C_1e^t+C_2e^{2t}+C_3e^{-t}\end{pmatrix}$$
and show that it is the same as
$$\begin{pmatrix}0 & 1 & 0\\ 0 & 0 & 1\\ -2 & 1 & 2\end{pmatrix} \begin{pmatrix}C_1e^t+\frac{C_2e^{2t}}{4}+C_3e^{-t}\\ C_1e^t+\frac{C_2e^{2t}}{2}-C_3e^{-t}\\C_1e^t+C_2e^{2t}+C_3e^{-t}\end{pmatrix}$$