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anonymous
 one year ago
I have tried the method of finding equilibrium point, and then say X=x(t)x*, where x* is the EP, yet I can't figure out the question below, how do they get the particular solution? The first part on eigenvalues, eigenvectors, the complimentary solution, I am all sorted, just the part on the particular solution. Please help.
anonymous
 one year ago
I have tried the method of finding equilibrium point, and then say X=x(t)x*, where x* is the EP, yet I can't figure out the question below, how do they get the particular solution? The first part on eigenvalues, eigenvectors, the complimentary solution, I am all sorted, just the part on the particular solution. Please help.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What kind of techniques do you know? You can do this using undetermined coefficients.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i know all those, but then, will i handle each differential equation on its own, or simulteneously?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can work with the particular solutions either way. Individually, it might be easier, but we're given a quadratic expression, so we might as well work with them simultaneously. Let \(M=\begin{pmatrix}1&1\\1&1\end{pmatrix}\). We'll use \({\bf x}_p=\vec{a}t^2+\vec{b}t+\vec{c}\) as our trial solution. \[{{\bf x}_p}'=2\vec{a}t+\vec{b}\] Substituting into the original system, we have \[2\vec{a}t+\vec{b}=M\left(\vec{a}t^2+\vec{b}t+\vec{c}\right)+\begin{pmatrix}1\\1\end{pmatrix}t^2+\begin{pmatrix}4\\6\end{pmatrix}t+\begin{pmatrix}1\\5\end{pmatrix}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Next we match up the corresponding power of \(t\). \[\begin{cases} \vec{0}=M\vec{a}+\begin{pmatrix}1\\1\end{pmatrix}\\ 2\vec{a}=M\vec{b}+\begin{pmatrix}4\\6\end{pmatrix}\\ \vec{b}=M\vec{c}+\begin{pmatrix}1\\5\end{pmatrix} \end{cases}\] then solve for \(\vec{a},\vec{b},\vec{c}\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles , YOU ARE A LIFE SAVER, thank you very Much!
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