Question

Find all the solutions to: xdot = [1 2; 0 1] * x

Answer 1

can you tell me what this means

Answer 2

Sure! I'm asking how to solve a system of linear differential equations of the form:

\[x \prime = \left[\begin{matrix}1 & 2 \\ 0 & 1\end{matrix}\right] * \left(\begin{matrix}x1 \\ x2\end{matrix}\right)\]

Answer 3

oh thats x prime?

Answer 4

do you know the steps?

Answer 5

So from what I understand, I first have to find the eigenvalues of the matrix \[\left[\begin{matrix}1 & 2 \\ 0 & 1\end{matrix}\right]\] , which gives me
det(\[\left[\begin{matrix}1-\lambda & 2 \\ 0 & 1-\lambda\end{matrix}\right]\]) = 0. Solving this gives me \[\lambda = 1\], with a multiplicity of two. I'm confused as to how to capture the effect of multiplicity.

Answer 6

i'm looking for an example

Answer 7

Do you have a Cramster account? I've found a good example there if you can't find one. The problem is, it's not helping me as much as I would've hoped.

Answer 8

ok it looks like we also need to find eigenvectors

Answer 9

Right, and when I solve for the eigenvectors I get v1 = \[\left(\begin{matrix}0 \\ 1\end{matrix}\right)\]

And because there is a double root, it's the only eigenvector found. From what I've read, the multiplicity adds other terms but that's where my confusion lies.

Answer 10

why is it so hard to find an example lol

Answer 11

Haha, try this one!

http://www.cramster.com/solution/solution/1085373

Answer 12

okay i think this is good
i'm gonna try to work your problem

Answer 13

Thank you! I'd be grateful if you could get me out of this rut.

Answer 14

hmmm... they were about to find another vector that wasn't linear dependent

Answer 15

i don't see how we can find a vector that is linear independent from the first one

Answer 16

I think the multiplication by 't' makes it linearly independent, right?

Answer 17

thats really the only thing thats holding us back

Answer 18

Would you happen to know anyone else that help us? I'm not sure this should be this difficult.

Answer 19

the eigenvectors are [1 0] and [0 1]

Answer 20

Hi phi! Would you mind telling me how you got that?

Answer 21

i don't know how he got that
but i do have something but i don't know if it is right

Answer 22

i will rewrite it

Answer 23

Alright!

Answer 24

i got these two vectors for lambda=1
[1 0] and [1 1/2]
i will show you how in a second

Answer 25

Alright that would be great, I'm not sure how you got those.

Answer 26

oops I lied, the evecs are [1 0] and [-1 0]

Answer 27

but aren't those linear dependent phi?

Answer 28

Hmm, alright I think I get it. I'm just going to go through your work once more really quickly. Thank you again for taking the time to answer it!

Answer 29

well like i said i don't know if it is right

Answer 30

so I guess I'm confused by how the first one you deduced that p1 could be anything?

Answer 31

Oh because it is multiplied by zero?

Answer 32

yep

Answer 33

If you guys find a second eigenvector please let me know!

Answer 34

Yeah I understand this method, I hope it's right! I'm going to think about it some more.

Answer 35

I get

\[c_1\left[\begin{matrix}1 \\ 0\end{matrix}\right]e^t+c_2\left(\left[\begin{matrix}1 \\ 0\end{matrix}\right]t+\left[\begin{matrix}0 \\ 1/2\end{matrix}\right]\right)e^t\]

Answer 36

Hi Zarkon, could you possibly explain how you got that?

Answer 37

I just followed the example