## 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.

1. anonymous

2. anonymous

What kind of techniques do you know? You can do this using undetermined coefficients.

3. anonymous

i know all those, but then, will i handle each differential equation on its own, or simulteneously?

4. anonymous

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

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

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7. anonymous

You're welcome!