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The solution of a differential equation is: $$\left \{ \begin{array}{lcl} x_1(t) & = & e^{-2t}-e^{-5t} \\ x_2(t) & = & e^{-2t}+e^{-3t}+e^{-5t} \\ x_3(t) & = & e^{-3t} + e^{-5t} \end{array}\right .$$ with initial conditions $$\left \{ \begin{array}{lcl} x_1(0) & = & 0 \\ x_2(0) & = & 3 \\ x_3(0) & = & 2 \end{array}\right .$$ What is the corresponding differential equation?

I had worked another linear algebra question using back substitution and I know that is what I am going to need to do here. In the previous problem I was substituting $A=QR$ and finding $\mathcal A$. Just not sure which approach to use in this situation.

$Ax=b$

$A=x^{-1}b$

Any help is greatly appreciated. :)

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  • $\begingroup$ I slightly changed your post. If I messed up somewhere please feel free to roll back. $\endgroup$ – Kaster Jun 26 '13 at 19:51
  • $\begingroup$ @Kaster~ Thanks! Looks great. I will look at the edits to see how you did that. I'm still learning. :) $\endgroup$ – Lanae Jun 26 '13 at 19:55
  • $\begingroup$ Can you solve direct problem? I mean, for example if I provide you with some sort of solution and say that the system I found is the one you're looking for, will you be able to check it? In other word to solve it and make sure that solution you find is the one you specified in your post. $\endgroup$ – Kaster Jun 26 '13 at 20:00
  • $\begingroup$ @Kaster~ I should be able to check that by $Ax=b$ I have my $\mathcal x$ and my $\mathcal b$. Just looking for $\mathcal A$. So I should be able to back substitute somehow to get my $\mathcal A$. Correct? $\endgroup$ – Lanae Jun 26 '13 at 20:03
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Writing

$$ \begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix}=\begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}\begin{pmatrix}e^{-2t} \\ e^{-3t} \\ e^{-5t}\end{pmatrix},$$

we have

$$\begin{array}{ll} \color{DarkOrange}{\frac{d}{dt}\begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix}} & = \begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}\begin{pmatrix}-2e^{-2t} \\ -3e^{-3t} \\ -5e^{-5t}\end{pmatrix} \\ & = \color{Purple}{\begin{pmatrix}-2 & 0 & 5 \\ -2 & -3 & -5 \\ 0 & -3 & -5\end{pmatrix}} \begin{pmatrix}e^{-2t} \\ e^{-3t} \\ e^{-5t}\end{pmatrix} \\ & = \color{Blue}{\begin{pmatrix}? & ? & ? \\ ? & ? & ? \\ ? & ? & ?\end{pmatrix}}\color{Green}{\begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}}\begin{pmatrix}e^{-2t} \\ e^{-3t} \\ e^{-5t}\end{pmatrix} \\ & \color{DarkOrange}{= \begin{pmatrix}? & ? & ? \\ ? & ? & ? \\ ? & ? & ?\end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3\end{pmatrix}.} \end{array}$$

One may solve for the unknown matrix via

$$\begin{array}{ll} & \color{Purple}{\begin{pmatrix}-2 & 0 & 5 \\ -2 & -3 & -5 \\ 0 & -3 & -5\end{pmatrix}}=\color{Blue}{\begin{pmatrix}? & ? & ? \\ ? & ? & ? \\ ? & ? & ?\end{pmatrix}}\color{Green}{\begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}} \\ \iff & \begin{pmatrix}-2 & 0 & 5 \\ -2 & -3 & -5 \\ 0 & -3 & -5\end{pmatrix}\begin{pmatrix} 1 & 0 & -1 \\ 1 & 1 & 1 \\ 0 & 1 & 1\end{pmatrix}^{-1}=\begin{pmatrix}? & ? & ? \\ ? & ? & ? \\ ? & ? & ?\end{pmatrix}. \end{array}$$

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  • $\begingroup$ ~ so my $$\mathcal A=\pmatrix{-5&3&-3\\2&-4&1\\2&-2&-1}$$ $\endgroup$ – Lanae Jun 26 '13 at 20:31
  • $\begingroup$ ~ How do I turn that $\mathcal A$ matrix into the differential equation that I'm looking for? I'm sure its something so simple that I'm not seeing. Thanks! $\endgroup$ – Lanae Jun 27 '13 at 21:07
  • $\begingroup$ @Lanae The differential equation you are looking for is already written in my answer. It's a linear system with $x'=Ax$, where $A$ is the matrix we found. I will use more color. $\endgroup$ – anon Jun 27 '13 at 21:43
  • $\begingroup$ ~ Thanks a bunch! I knew I was thinking too much. I have another question posted about finding $\mathcal S$ and $\Lambda$ from a diagonizable matrix, $\mathcal A$, would you tell me what you think? $\endgroup$ – Lanae Jun 27 '13 at 22:15

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