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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.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    What kind of techniques do you know? You can do this using undetermined coefficients.

  3. anonymous
    • one year ago
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    i know all those, but then, will i handle each differential equation on its own, or simulteneously?

  4. anonymous
    • one year ago
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    You 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 \[2\vec{a}t+\vec{b}=M\left(\vec{a}t^2+\vec{b}t+\vec{c}\right)+\begin{pmatrix}1\\1\end{pmatrix}t^2+\begin{pmatrix}4\\-6\end{pmatrix}t+\begin{pmatrix}-1\\5\end{pmatrix}\]

  5. anonymous
    • one year ago
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    Next 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 \(\vec{a},\vec{b},\vec{c}\).

  6. anonymous
    • one year ago
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    @SithsAndGiggles , YOU ARE A LIFE SAVER, thank you very Much!

  7. anonymous
    • one year ago
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    You're welcome!

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