Question

Finding solutions for a system from the fundamental matrix.

See Below

Answer 1

\[V(t)=\left[\begin{matrix}0 & 4te^-t & e^-t\\ 1 & e^-t & 0 \\ 1 & 0 & 0\end{matrix}\right]\]

Answer 2

Verify that v is a fundamental matrix for the system

\[x'=\left[\begin{matrix}-1 & 4 & 4 \\ 0 & -1 & 1\\ 0 & 0 & 0\end{matrix}\right]\]

Answer 3

so i found the eigenvalues and then the vectors, but I do not get what they got for V(t).

the eigenvalues are 0 and -1, with -1 being a double root. I got the v1(t), when lambda is 0 but i do not get what I was supposed to for when lambda is -1, what am i doing wrong

Answer 4

The third column, the 4 is supposed to be -4.

Answer 5

@wio

Answer 6

Hmmm, I'm not completely sure what is being asked.

Answer 7

Is it just a diffeq but with repeated roots?

Answer 8

well, if I was given the x'=Ax matrix, and I solved for the eigenvalues and vectors, just like we did last night, and put together the matrix from the three solutions, I should get V(t), but I dont, I get the correct eigenvalues, and I get the correct v1(t) , when lambda is zero, i get
(0e^0t, 1e^0t, 1e^0t), which is (0, 1, 1) which is what I am supposed to get for the first column of V(t), but when lambda is -1, I do not get either of those column vectors in V(t).

Answer 9

what is your eigen vector for \(-1\)?

Answer 10

i get (0, 0, 1)^T

Answer 11

By the way `\begin{bmatrix}` will auto add brackets.\[
\begin{bmatrix}-2 & 4 & 4 \\ 0 & -2 & 1\\ 0 & 0 & -1\end{bmatrix}
\]

Answer 12

We gotta find the null space in this...

Answer 13

i thought the diagonal is -1 - (-1), so wouldnt both the -2 be 0's

Answer 14

Wait! I messed it up...

Answer 15

\[
\begin{bmatrix}0 & 4 & 4 \\ 0 & 0 & 1\\ 0 & 0 & 1\end{bmatrix}
\]

Answer 16

\[\left[\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1\\ 0 & 0 & 0\end{matrix}\right]\]

Answer 17

It seems that there are multiple ways of doing this one...

Answer 18

and thats what i end with, x2=0 x3=0, x3=x3

Answer 19

Hmm, basically it comes down to how you have to deal with repeated eigen values in diffeqs.

Answer 20

how do you do that

Answer 21

The middle row is the only row that is a bit strange.

Answer 22

that is what i was thinking to... because i have no idea how they get that t in there.

Answer 23

well if you are given all the solutions like in the V(t), how would we get the A matrix in x'=Ax if it wasnt given already to us?

Answer 24

Okay, so from what I've seen, once you find a vector \(v_1\) for a repeated eigen vector, you need to find a new vector \(b\), such that \[
(A-\lambda_1I)b=v_1
\]Then the solution is \[
v_1te^{\lambda_1t}+be^{\lambda_1t}
\]

Answer 25

Or so it says in this book...

Answer 26

So we find \(b\)\[
\begin{bmatrix}0 & 4 & 4 \\ 0 & 0 & 1\\ 0 & 0 & 1\end{bmatrix}
b
=
\begin{bmatrix}1 \\ 0\\ 0 \end{bmatrix}
\]

Answer 27

where did you get (1, 0, 0) from? I thought the vector was (0, 0, 1) or did I do that wrong

Answer 28

Because \([c,0,0]^T\) is the null space, the eigenvector.

Answer 29

maybe I am doing this wrong but I am getting (1, 4, 0)^T

Answer 30

How are you getting stuff like that?

Answer 31

I'm getting \([0,1/4,0]^T\)

Answer 32

ohhh yes you are right, i see what I did... but I just multiplied by 4, but yes, i got c2 is 1/4, so it would be what you got.

Answer 33

\[
\begin{bmatrix}0 & 4 & 0 \\ 0 & 0 & 1\\ 0 & 0 &0\end{bmatrix}
b
=
\begin{bmatrix}1 \\ 0\\ 0 \end{bmatrix}
\]It row reduces to this, so I guess there are many solutions...
this is confusing as hell.

Answer 34

yes it sure is. And that didn't quite work either... Im wondering if there is a to start from V(t), and find x' .

Answer 35

I have to run to the grocery store, if you find anything please let me know, otherwise I will be right back.

Answer 36

Anyway, my full solution is: \[
\begin{bmatrix}0 \\ 1\\ 1 \end{bmatrix}e^{0t}+
\left(\begin{bmatrix}1 \\ 0\\ 0 \end{bmatrix}te^{-t}+
\begin{bmatrix}0 \\ 1/4\\ 0 \end{bmatrix}e^{-t}\right) +
\begin{bmatrix}1 \\ 0\\ 0 \end{bmatrix}e^{-t}
\]This simplifies to: \[
\begin{bmatrix}0 \\ 1\\ 1 \end{bmatrix}+
\begin{bmatrix}1 \\ 0\\ 0 \end{bmatrix}te^{-t}+
\begin{bmatrix}1 \\ 1/4\\ 0 \end{bmatrix}e^{-t}
\]

Answer 37

which just isn't consistent with their solution.. so I'm confused.

Answer 38

well it would be if the last two vectors were switched.

Answer 39

It's strange, because I'm getting:\[
V(t)=
\begin{bmatrix}
0 & te^{-t} & 4e^{-t}\\
1 & 0 & e^{-t} \\
1 & 0 & 0\end{bmatrix}
\]

Answer 40

It's not the order of vectors that is a problem, it's the fact that they have a \(t\) in the wrong column.

Answer 41

well i agree with you, i do not see any reason why it wouldnt be what you have there.

Answer 42

http://www.youtube.com/watch?v=V_L8RELShNI


the guy's voice is kinda annoying, but the video explains it decently

Answer 43

ohhhhhhkaaaayyyy, haha, it is kind of annoying, but in this problem the k vector, would be (4, 1, 0), but not sure how to get that, but it was easy to understand, thank you again for your time and help.