Question

Find a particular solution to x'=x+2y+2t, y'=3x+2y-4. a)using integrating factor method, b) using eigenvector decomposition, c)using undetermined coeffiecients

Answer 1

I don't know about the first two, so I'll just stick with the last method.
\[\left(\begin{matrix}x\\y\end{matrix}\right)'=\left(\begin{matrix}1&2\\3&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)+t\left(\begin{matrix}2\\0\end{matrix}\right)+\left(\begin{matrix}0\\-4\end{matrix}\right)\]
Find the corresponding characteristic solution using the usual eigenvalue method:
\[\left(\begin{matrix}x\\y\end{matrix}\right)'=\left(\begin{matrix}1&2\\3&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)_c\]
As a guess for the particular solution, try
\[\left(\begin{matrix}x\\y\end{matrix}\right)_p=t\vec{a}+\vec{b}\]
This gives you
\[\left(\begin{matrix}x\\y\end{matrix}\right)_p'=\color{red}{\vec{a}=

\left(
\begin{matrix}
1&2\\
3&2
\end{matrix}
\right)

\left(t\vec{a}+\vec{b}\right)

+t

\left(
\begin{matrix}
2\\
0
\end{matrix}
\right)

+

\left(
\begin{matrix}
0\\
-4
\end{matrix}
\right)}\]
Let \(A=\left(\begin{matrix}1&2\\3&2\end{matrix}\right)\).

This gives you
\[\vec{a}=At\vec{a}+A\vec{b}+t\left(\begin{matrix}2\\0\end{matrix}\right)+\left(\begin{matrix}0\\-4\end{matrix}\right)\]
Matching up "coefficients," you get
\[\begin{cases}
A\vec{a}+\left(\begin{matrix}2\\0\end{matrix}\right)=\vec{0}\\
A\vec{b}+\left(\begin{matrix}0\\-4\end{matrix}\right)=\vec{a}
\end{cases}\]
Once you figure out \(\vec{a}\) and \(\vec{b}\), you should get the right solution. I'd check with WolframAlpha just to be sure.