Some Review [Chapter 7.9]

First, in the previous post, I made an error. It has been corrected. The correction is also shown here:
Compute the inverse Laplace transform of
$\frac{e^{-2s}}{s^{1/2}}$
[CORRECTION: It should be -2s instead of 2s]

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When solving a nonhomogeneous linear system, a reasonable question
to ask is which method to use. Given the choice, use what you think
is easiest. But you might not be given a choice and so you should
be familiar with all the techniques.

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I find the book explained the first method fairly well. Let me know
if you feel otherwise, please specify what is giving you trouble.

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Let's go through some details on Example 2 page 435.

Going from (18) to (19):

$x^{\prime}=ae^{-t}-ate^{-t}-be^{-t}+c$

$Ax+g=Aate^{-t}+Abe^{-t}+Act+Ad+\left(\begin{array}{c}
2\\
0\end{array}\right)e^{-t}+\left(\begin{array}{c}
0\\
3\end{array}\right)t$

Thus $a-b=Ab+\left(\begin{array}{c}
2\\
0\end{array}\right)$, $Aa=-a$, $c=Ad$, and $0=Ac+\left(\begin{array}{c}
0\\
3\end{array}\right)$

To start solving for the unknowns. Let's ignore the terminology about
eigenvectors and just do what we would do with a system of equations,
start with the equations with the least variables: $Aa=-a$ and $Ac=-\left(\begin{array}{c}
0\\
3\end{array}\right)$. Let's start with the first. This is the same as $(A+I)a=0$.

$A+I=\left(\begin{array}{cc}
-1 & 1\\
1 & -1\end{array}\right)\to\left(\begin{array}{cc}
-1 & 1\\
0 & 0\end{array}\right)$. Rank one, so one free variable, let $x_{1}=\alpha$. Then $x_{2}=\alpha$.
Thus we have $a=\left(\begin{array}{c}
\alpha\\
\alpha\end{array}\right)$

Great and the second: $\left(\begin{array}{cccc}
-2 & 1 & | & 0\\
1 & -2 & | & -3\end{array}\right)\to\left(\begin{array}{cccc}
0 & -3 & | & -6\\
1 & -2 & | & -3\end{array}\right)\to\left(\begin{array}{cccc}
0 & 1 & | & 2\\
1 & -2 & | & -3\end{array}\right)\to\left(\begin{array}{cccc}
0 & 1 & | & 2\\
1 & 0 & | & 1\end{array}\right)$. So $c=\left(\begin{array}{c}
2\\
1\end{array}\right)$. (We could have also found $A^{-1}$ which is handy for all these
computations)

Okay so next $Ad=c$ is simple.

Finally we have the fourth equation which I have rearranged to be
$Ab+b=a-\left(\begin{array}{c}
2\\
0\end{array}\right)$. Then we have $(A+I)b=\left(\begin{array}{c}
\alpha-2\\
\alpha\end{array}\right)$. Write $b=\left(\begin{array}{c}
b_{1}\\
b_{2}\end{array}\right)$

Thus $\left(\begin{array}{cccc}
-1 & 1 & | & \alpha-2\\
1 & -1 & | & \alpha\end{array}\right)\to\left(\begin{array}{cccc}
-1 & 1 & | & \alpha-2\\
0 & 0 & | & 2\alpha-2\end{array}\right)$

At this point, we need to have $2\alpha-2=0$ in order for the system
of equations to be consistent. So $\alpha=1$.

$\to\left(\begin{array}{cccc}
-1 & 1 & | & -1\\
0 & 0 & | & 0\end{array}\right)\to-b_{1}+b_{2}=-1$ . Let $b_{1}=\beta$. Then $b_{2}=-1+\beta$. Thus $b=\left(\begin{array}{c}
\beta\\
-1+\beta\end{array}\right)=\left(\begin{array}{c}
0\\
-1\end{array}\right)+\beta\left(\begin{array}{c}
1\\
1\end{array}\right)$.

Note: There are many correct ways to express $b$. For example, if
I choose $b_{2}=\gamma$, then $b_{1}=\gamma+1$ so that $b=\gamma\left(\begin{array}{c}
1\\
1\end{array}\right)+\left(\begin{array}{c}
1\\
0\end{array}\right)$. To see that this and the above are the same, let $\gamma=\beta-1$.

At this point we are done and we are free to choose any remaining
constants as we please, because we are looking for any particular
solution. In this case, we just have $\beta$ to choose, so let $\beta=0$.
(In the second case, we could let $\gamma=0$ and get a different
particular solution, but that's fine.)

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I like method of undertermined coefficients because then I don't have
to memorize a formula. But variations of parameters is nice if you
like integrating and memorizing formulas. I refer you to the book.
Let me know if there are difficulties.

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Just to apply the Laplace Transform component by component. Just like
with Section 6.2, you'll need the initial value to use this method.
It's unlikely you'll be forced to solve a nonhomogenous system of
equations with this method, since there will be other places on the
exam to test your knowledge of the Laplace Transform.