Question

PLEASE HELP!!!!!!!!

Answer 1

\[ (6- \lambda)(4- \lambda)-(-2)(-4)=0\]

Answer 2

i did that and ended up with \[\lambda1=2 and \lambda2=8\]

Answer 3

yes these are your Eigenvalues.

Answer 4

im not sure what to do afterwards though, i solved for case 1 and a=b and a=b for that

Answer 5

but after that I'm lost

Answer 6

if you back-substitute the Eigenvalue you get the following

\[ 4x-4y=0 \\ \\-2x+2y=0\] One equation is redundant, both lead to x=y so you have an eigenvector of (1,1)

Answer 7

so v1=(1,1)^2?

Answer 8

Well that means that one solution is:

\[ x_1=c_1\left(\begin{matrix}1 \\ 1\end{matrix}\right)e^{3t} \]

Answer 9

oh ok

Answer 10

you get the next solution the same way, just different eigenvalue and eigenvector, and the most general solution by superposition.

Answer 11

isnt it supposed to be x1=c1(1,1)e^(2t)

Answer 12

above I used the Eigenvalue \(\lambda = 2 \) for back substitution, now you will have to do the same with \( \lambda =8 \)

Answer 13

yes sorry, I am sometimes really bad when it comes to LaTEX typesetting :D

Answer 14

its ok! :)

Answer 15

in the exponent there is your Eigenvalue, in the coefficient there is your eigenvector, but we agree that the values are 2 and 8.

Answer 16

and for x(0)=(1,5) what does that mean

Answer 17

for lamda=8 is a=-2b so a=-2 and b=1?

Answer 18

so x2=c2[-2 1]e^(8t)

Answer 19

well these are initial conditions, in the end you can split your equation up:

\[\left(\begin{matrix}x_1 \\ x_2\end{matrix}\right)=c_1\left(\begin{matrix}1 \\ 1\end{matrix}\right)e^{2t} + c_2\left(\begin{matrix}? \\ ?\end{matrix}\right)e^{8t}\]

this gives you two equations (I didn't calculate the second eigenvector yet, therefore the ??)

\[ \large x_1=c_1e^{2t}+c_2?e^{8t} \large \\x_2=c_1e^{2t}+c_2{e^{8t}}\]

Answer 20

there you can use initial conditions x(0) to evaluate c_1 and c_2

Answer 21

how do we solve for xp then?

Answer 22

for \(\lambda =8 \)

\[ -2x-4y=0\]

So I would say the Eigenvector is

\[\left(\begin{matrix}-4 \\ -2\end{matrix}\right)\]

Answer 23

oops I forgot to switch a sign. it should be 4/-2

Answer 24

oh ok thanks and how do we solve for xp after we solve for c1 and c2

Answer 25

I guess x_p means the particular solution?

Answer 26

because I haven't read that notation before.

Answer 27

yea xp means particular solution

Answer 28

but how do we solve for that after having x1 and x2

Answer 29

\[\Large \left(\begin{matrix}x_1 \\ x_2\end{matrix}\right)=c_1\left(\begin{matrix}1 \\ 1\end{matrix}\right)e^{2t}+c_2\left(\begin{matrix}4 \\ -2\end{matrix}\right)e^{8t}\]

You can split this up and read it line by line, this is only a vector notation for:

\[ \large x_1=c_1e^{2t}+c_2 4e^{8t} \\ \large x_2=c_1e^{2t}-c_22e^{8t} \]

Answer 30

two equations

Answer 31

yes i got that and now im solving for c1 and c2 but idk how to solve for the particular solution and thanks

Answer 32

well I would say that is the particular solution, just with the coefficients added, because the particular solutions requires two lines

Answer 33

I got c1=(11/3) and c2=-2/3

Answer 34

seems like your program wants you to add two lines in there

Answer 35

so the particular solution is the same as saying x1(t) and x2(t)?

Answer 36

in my intuition yes, but if that's not the answer then I am afraid that I can't help you there, you would have to ask someone whose native tongue is english, I haven't heard about such a thing before and here I would let the solution be in it's vector component form, and not in its split up form

Answer 37

ok i tried it and the first part is correct but i got the second part wrong. I think particular solution is x1 and x2 but did you get (11/3)e^(2t)-(8/3)e^(8t) for the first one and (11/3)e^(2t)-(4/3)e^(8t) for the second

Answer 38

the first one is right but second is wrong

Answer 39

let me see for the coefficients, I haven't calculated them yet.

Answer 40

c_2 = (-2/3) I got

Answer 41

never mind i caught my mistake! its positive 4/3 instead of negative for x2 thanks soo much

Answer 42

oh so you did it right, cool.

Answer 43

yes! I'm very happy! lol

Answer 44

very welcome, so the particular solution is them both added up? Or in the split up form?

Answer 45

its the split form

Answer 46

ahh, thanks for letting me know! Always fun to work with you! (-: I am AFK for a while now though, enjoy!

Answer 47

AFK?

Answer 48

Away from Keyboard

Answer 49

oh lol thanks!! and It's fun working with you too! I really appreciate it!

Answer 50

See you around then (-:

Answer 51

see you! :)