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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.

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  • $\begingroup$ First you find $e^{Ct}.$ If you have $C = Q^{-1}D Q$ for diagonal $D,$ then $e^{Dt}$ is also diagonal with those exponential terms, after which $e^{Ct} = Q^{-1}e^{Dt} Q.$ At the very, very end, given an $x(0) = x_0,$ you get $x = e^{Ct} x_0$ $\endgroup$
    – Will Jagy
    Oct 17, 2018 at 20:47
  • $\begingroup$ I guess I can tell what you mean by disconnected. You need to figure out both my matrices $Q$ and $Q^{-1}.$ I suppose your $P$ is one of them, not sure which. $\endgroup$
    – Will Jagy
    Oct 17, 2018 at 20:49
  • $\begingroup$ $P$ is your $Q$ I guess. I think I get it now. Thank you. $\endgroup$ Oct 17, 2018 at 21:00

1 Answer 1

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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

$$\begin{pmatrix}0 & 1 & 0\\ 0 & 0 & 1\\ -2 & 1 & 2\end{pmatrix} \begin{pmatrix}C_1e^t+\frac{C_2e^{2t}}{4}+C_3e^{-t}\\ C_1e^t+\frac{C_2e^{2t}}{2}-C_3e^{-t}\\C_1e^t+C_2e^{2t}+C_3e^{-t}\end{pmatrix}$$

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  • $\begingroup$ They don't seem pretty identical $\endgroup$ Oct 17, 2018 at 21:34
  • $\begingroup$ for $x'=$$\begin{pmatrix}C_1e^t& C_2e^{2t}/2 & -C_3e^{-t}\\ C_1e^t & C_2e^{2t} & C_3e^{-t}\\ C_1e^t& 2C_2e^{2t} & -C_3e^{-t}\end{pmatrix}$$ Which is not the same as $$\begin{pmatrix}0 & 1 & 0\\ 0 & 0 & 1\\ -2 & 1 & 2\end{pmatrix} \begin{pmatrix}C_1e^t+\frac{C_2e^{2t}}{4}+C_3e^{-t}\\ C_1e^t+\frac{C_2e^{2t}}{2}-C_3e^{-t}\\C_1e^t+C_2e^{2t}+C_3e^{-t}\end{pmatrix}$$ $\endgroup$ Oct 17, 2018 at 21:36
  • $\begingroup$ Check your $ x'$ again, it has to be a vector not a matrix. $\endgroup$ Oct 17, 2018 at 21:48
  • $\begingroup$ Ah, i got it now. Just an arithmetic error. Thanks! $\endgroup$ Oct 18, 2018 at 13:43
  • $\begingroup$ Good, thanks for your attention . $\endgroup$ Oct 18, 2018 at 13:59

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