Question

Find three linearly independent solutions of ........

see below.

Answer 1

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

Answer 2

\[\left(\begin{matrix}1 \\ 1\\1\end{matrix}\right)\]

Answer 3

Hmm, I'm not sure what the question is asking. The only one variable and a solution..... Do you mean to say this: \[
x'=\left[\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 4 & 4 & -1\end{matrix}\right]x
\]

Answer 4

@zonazoo This diffeq?

Answer 5

yes and yes. sorry i left out the x

Answer 6

Then can you find the eigen vectors and eigen values?

Answer 7

yes. umm, give me a sec. let me see what i remember.

Answer 8

The answer is going to be \[
x = e^{\lambda_1 t}v_1+e^{\lambda_2 t}v_2+e^{\lambda_3 t}v_3
\]Where \(\lambda_i\) is some eigen value, and \(v_1\) is its corresponding eigen vector.

Answer 9

where does e^t come from?

Answer 10

The reason for this is that: \[
x'=Ax=\lambda x \implies x=e^{\lambda t}
\]So we have general solutions for each eigen value, and when you have multiple solutions to a diffeq, you add them to get the full solution.

Answer 11

\[
x'=kx \implies x=e^{kt}
\]Is a simple diffeq learned in Calc 1

Answer 12

There are some things I'm glossing over here, like constants of integration and such, but I frankly don't know enough to explain it rigorously.

Answer 13

oh, yeah calc was 15 years ago, so alot of this im trying to get back.
so my e.v. are -2, -1, 2

and then after i solve for each eigenvector i plug that into v1, v2, and v3 and i will get my solutions?

Answer 14

Yep.

Answer 15

and the fundamental set of solutions would be just
e^-2t, e^-t, e^2t

and fundamental meaning that all solutions are scalar multiplies of that set?

Answer 16

Well, each one needs an eigen vector next to it, because \(x\) is a vector and it certainly isn't going to be equal to a scalar.

Answer 17

and to get the eigenvector i just plug my eigenvalue back into the matrix and row reduce?

Answer 18

For each eigenvector, subtract the eigen value from the diagonal, then solve for the null space.

Answer 19

for lambda = -2

\[\left[\begin{matrix}2 & 1 & 0 \\ 0 & 2 & 1 \\ 4 & 4 &1\end{matrix}\right]\]

and solving for the null space.. which is just row reduction right. Ax=0

Answer 20

v1=(1/4, -1/2, 0)^T?

Answer 21

Okay for finding the null spae there are two ways to do it: eye it, or let \(Ax=b\) and keep track of the \(b\) entries as you row reduce.

Answer 22

In this case, the easiest thing to do is eye it. You can show that to make the top row 0, you want \(v_1 = [-1,2,\_]\)

Answer 23

Looking at the second row, you can tell that if the middle variable is \(2\) you want the last one to be \(-4\)

Answer 24

So you get \(v_1=[-1,2,-4]\)

Answer 25

ohhh okay, v1 = (1/4, -1/2, 1) and multiply by 4, to get (1, -2, 4).

so then the first solution would be

x1(t)= (e^-2t, -2e^2t, 4e^2t) ???

Answer 26

wait, why are our signs different?

Answer 27

The eigen values do NOT correspond to different components of \(x\)

Answer 28

If you multiplied by \(-4\) the signs would be the same.

Answer 29

right... okay, i meant the solution was (e^-2t, -2e^-2t, 4e^-2t)

what do you mean? values do not correspond to components of x?

Answer 30

To get the full solution, you need to find every eigen vector.

Answer 31

Getting just one eigen vector/value pair does not give you a full solution for every component. If gives you a partial solution for every component.

Answer 32

right, i would to the same process, for lambda = -1, and =2, and i would replace the -2t with the lambda i used to find the eigenvector, and get 3 total solutions.

Answer 33

Yeah.

Answer 34

Remember you originally had the following system: \[
\begin{array}{rcl}
x'_1 &=&&&x_2&& \\
x'_2 &=& &&&&x_3 \\
x'_3 &=& 4x_1&+&4x_2&+&x_3
\end{array}
\]

Answer 35

i am curious though, when i solved for det(A-lambdaI) i got

\[\lambda ^{3}+\lambda ^{2}-4\lambda-4\]

and the book says that is actually my third order equation for this problem

y'''+y''-4y'-4y = 0

is that always the case that the determinant will for the equation?

Answer 36

Because you are multiply \((c_1-\lambda)(c_2-\lambda)(c_3-\lambda)...(c_n-\lambda)\) you will get an \(n\) degree polynomial for an \(n\times n\) matrix. This means you will have \(n\) roots. This comes from the fundamental theorem of algebra.

Answer 37

It does get hairy when you have repeated roots or imaginary roots though.

Answer 38

well i thank you for all your help, the book doesnt even mention to use eigenvalues or vectors in this chapter, i guess it is implied that we should of known that. but i understand now. you are awesome.