Question

Find the general solution of the given system of equations:

x' = (2, -1; 3,-2)x + (e^t; t)

(where ; indicates a new line of the matrix/vector)

Answer 1

\[x'=\left[\begin{matrix}2 & -1 \\ 3 & -2\end{matrix}\right]x+\left(\begin{matrix}e^t \\ t\end{matrix}\right)\]

Answer 2

Have you found your eigenvalues/vectors yet?

Answer 3

Yep! They are \[\left(\begin{matrix}1 \\ 1\end{matrix}\right)\]
and
\[\left(\begin{matrix}1 \\ 3\end{matrix}\right)\]

Answer 4

We're supposed to be using undetermined coefficients for this problem to find the particular solution though. I have the solution, but I have no idea how they found it. The final solution should be:

\[x=c1\left(\begin{matrix}1 \\ 1\end{matrix}\right)e^t+c2\left(\begin{matrix}1 \\ 3\end{matrix}\right)e^{-t}+\frac{ 3 }{ 2 }\left(\begin{matrix}1 \\ 1\end{matrix}\right)te^t-\frac{ 1 }{ 4 }\left(\begin{matrix}1 \\ 3\end{matrix}\right)e^t+\left(\begin{matrix}1 \\ 2\end{matrix}\right)t-\left(\begin{matrix}0 \\ 1\end{matrix}\right)\]

Answer 5

Okay, I assume you know how the homogeneous solution was found?

For the UC set up, first we split up the nonhomogeneous term like so:
\[\begin{pmatrix}e^t\\t\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}e^t+\begin{pmatrix}0\\1\end{pmatrix}t\]
and for the guess solution, we use
\[x_p=Ae^t+Bt+C\]
where \(A,B,C\) are unknown vectors.

Answer 6

Okay. Although, shouldn't there also be a vector with \[te^t\] in it?

Answer 7

Ah yes, that's right. I forgot about the homogeneous \(e^t\) solution.

Answer 8

So \(x_p=Ate^t+Be^t+Ct+D\).

Answer 9

Okay, sounds good so far

Answer 10

Then you have
\[{x_p}'=Ate^t+(A+B)e^t+C\]
Substituting into the original system,
\[Ate^t+(A+B)e^t+C=\begin{pmatrix}2&-1\\3&-2\end{pmatrix}(Ate^t+Be^t+Ct+D)+\begin{pmatrix}1\\0\end{pmatrix}e^t+\begin{pmatrix}0\\1\end{pmatrix}t\]

Answer 11

Then I assume I'd just equate like terms? That would give me \[A=\left[\begin{matrix}2 & -1 \\ 3 & -2\end{matrix}\right]A ;

A+B=\left[\begin{matrix}2 & -1 \\ 3 & -2\end{matrix}\right]B+\left(\begin{matrix}1 \\ 0\end{matrix}\right);

\left(\begin{matrix}0 \\ 0\end{matrix}\right)=\left[\begin{matrix}2 & -1 \\ 3 & -2\end{matrix}\right]C+\left(\begin{matrix}0 \\ 1\end{matrix}\right)\]

Answer 12

and \[C=\left[\begin{matrix}2 & -1 \\ 3 & -2\end{matrix}\right]D\]

Answer 13

I'm not really sure where to go from there though

Answer 14

That first equation is a bit problematic, though. It suggests \(A=\vec{0}\) :/

Answer 15

Or just that A is an eigenvector of the matrix corresponding to an eigenvalue of 1

Answer 16

But that wouldn't explain the \(\dfrac{3}{2}\) in the solution, would it?

Answer 17

Nope haha... dang.

Answer 18

Here's a slightly more detailed solution... but still doesn't show the work :\
http://www.math.ucsd.edu/~hus003/math20D/HW6.pdf

Answer 19

On page 10 of that

Answer 20

Well, given that \(A\) can be a scalar multiple of the first eigenvalue \(\left(A=\begin{pmatrix}k\\k\end{pmatrix}\text{ for real }k\right)\), I suppose you can use that to your advantage to solve for \(B\), then work backwards to find an appropriate \(k\).

Answer 21

I think k=3 works

Answer 22

I'm not really sure how to solve for B without finding k first :\

Answer 23

I would go ahead with \(A=\dbinom11\) and see what the solution ends up being. I don't see any particular reason to choosing \(k=\dfrac{3}{2}\).

Answer 24

I'll try that. Otherwise I'll probably just go to bed and ask my professor tomorrow haha... I'm exhausted XD