Question

Find the general solution of x'=(2, 1, -5, -2)x+(-cos t, sin t). ( 2 and 1 on left, -5 and -2 on right. -cos t on top, sin t on bottom. how do I find the particular solution for the 2x1 matrix using undetermined coefficients?)

Answer 1

\[x'=\left[\begin{matrix}2&-5\\1&-2\end{matrix}\right]x +\left[\begin{matrix}-cost\\sintt\end{matrix}\right]\]

Answer 2

@SithsAndGiggles

Answer 3

@Idealist @SithsAndGiggles is here, we are fine

Answer 4

Which method is easier to use for the particular solution in this case? Undetermined coefficients or variation of parameters?

Answer 5

hey, why I don't see any thing pop up? I saw Sith wrote something

Answer 6

how to find the general solution for 2x1 matrix?

Answer 7

Homogeneous solution:
\[\begin{align*}\begin{vmatrix}2-\lambda&-5\\1&-2-\lambda\end{vmatrix}=-(4-\lambda^2)+5&=0\\
4-\lambda^2&=5\\
\lambda^2&=-1\\
\lambda&=\pm i
\end{align*}\]
Finding the eigenvectors:
\[\lambda_1=i:\\
\begin{pmatrix}2-i&-5\\1&-2-i\end{pmatrix}\begin{pmatrix}\eta_1\\\eta_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}~~\Rightarrow~~\vec{\eta_1}=\begin{pmatrix}2+i\\1\end{pmatrix}\]
So the second eigenvector must be \(\vec{\eta_2}=\begin{pmatrix}2-i\\1\end{pmatrix}\).

Answer 8

I'll have to review how to write up the final general solution using the eigenvectors.

In any case, I would suggest continuing with undetermined coefficients, but only because I don't remember how to use variation of parameters.

Answer 9

can you show me how to find it using undetermined coefficients?

Answer 10

but for the first eigenvector, I got a=i, b=2/5i+1/5 and for the second eigenvector, I got a=-i, b=-2/5i+1/5.

Answer 11

For the homogeneous part, you should have
\[\vec{x_c}=C_1\begin{pmatrix}2\cos t-\sin t\\\cos t\end{pmatrix}+C_2\begin{pmatrix}2\sin t+\cos t\\\sin t\end{pmatrix}\]
I'll go over the details of finding the first eigenvector:
\[\lambda_1=i:\\
\begin{pmatrix}2-i&-5\\1&-2-i\end{pmatrix}\begin{pmatrix}\eta_1\\\eta_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\]
Multiply the top row by \(2+i\) and the bottom row by -5:
\[\begin{pmatrix}(2-i)(2+i)&-5(2+i)\\-5&5(2+i)\end{pmatrix}\begin{pmatrix}\eta_1\\\eta_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\\
\begin{pmatrix}5&-5(2+i)\\-5&5(2+i)\end{pmatrix}\begin{pmatrix}\eta_1\\\eta_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\\
\begin{pmatrix}5&-5(2+i)\\0&0\end{pmatrix}\begin{pmatrix}\eta_1\\\eta_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\]
So you have
\[5\eta_1=5(2+i)\eta_2\\
\eta_1=(2+i)\eta_2\]
So, \(\eta_2=1~~\Rightarrow~~\eta_1=2+i\).

Answer 12

wait a minute.

Answer 13

how did you get 0, 0 on the bottom?

Answer 14

Sorry, that didn't answer your question... The step was adding the rows together. I thought you were talking about the zero vector's zeros.

Answer 15

I get it now. how about the second eigenvector?

Answer 16

The second eigenvector isn't important in the long run, but since you know one you also know the other. They come in conjugate pairs, so you don't have to show the same work for it.

Answer 17

I got the same 2 eigenvectors, now what?

Answer 18

Now for the U.C. part of the answer. I'm still not sure how to approach it, so you'll have to give me a moment.

Answer 19

okay.