Question

where i am wrong? http://i.imgur.com/j2v9F.jpg

Answer 1

im sorry old link

Answer 2

I know, I was looking at that too.
I can't see any errors in your work, so it must be a notation thing :/

Answer 3

i have a new challenge appeared

Answer 4

you can try me, but I'm not the master of ODE's or anything...

Answer 5

http://i.imgur.com/fGuVp.png

Answer 6

do you have an idea?

Answer 7

I am getting only one value for lambda: 3
is that what you got?

Answer 8

no

Answer 9

i got -4 and -2

Answer 10

rather I get lambda=-3\[(A-\lambda I)=\left[\begin{matrix}-2-\lambda & 1 \\ -1 & -4-\lambda\end{matrix}\right]\]so\[\det(A-\lambda I)=(-2-\lambda)(-4-\lambda)+1=8+6\lambda+\lambda^2+1\]\[\lambda^2+6\lambda+9=(\lambda+3)^2=0\to\lambda=-3\]where did I/you make a mistake?

Answer 11

haha,yes you are right.i forgot adding 1 to the equation

Answer 12

I love it when it's something simple :)

Answer 13

Though this is where the problem get's hard for me, having to find rho...

Answer 14

i will try it now with the new lambda

Answer 15

You have a root of multiplicity two, and only one eigenvector.
As you said, it's getting more difficult.

Answer 16

i think i don't know how to solve this kind of problem, i checked for the solved examples in my textbook but there isn't any...

Answer 17

what changes when there is only one eigenvector?

Answer 18

Have you worked on the D operator methods?

Answer 19

you have to find another vector to go along with\[\overrightarrow\eta=\overrightarrow\rho(A-\lambda I)\]I got\[\overrightarrow{\eta}=\left(\begin{matrix}1 \\ -1\end{matrix}\right)\]which is giving me problems getting rho.

Plus you have to add a \[te^{\lambda t}\]to one of the solutions to keep them linearly independent.

This is a good page on the subject:
http://tutorial.math.lamar.edu/Classes/DE/RepeatedEigenvalues.aspx

Answer 20

I'm not familiar with the D operator, and that's the second time I've heard that term today. So I must ask: what is it?
can you send me a link?

Answer 21

This means that the solution would be:
[x,y] = [1,-1](C1+tC2)e^3t
Do a differentiation to check that there is no errors in the solution.

Answer 22

i am checking

Answer 23

The D operator is basically a replacement of d/dt by the operator D, so
d2/dt2 would be come D^2, etc.
This allows us to solve systems of differential equations like we solve algebraic systems.
Try the link: http://en.wikipedia.org/wiki/Differential_operator

Answer 24

I was not suggesting that eta=[1,-1] was the only vector we need, so that can't be right, can it?
we need vector rho as I defined it above, but with eta=[1,-1] I can't solve the system :/

Answer 25

[x,y] = [1,-1](C1+tC2)e^3t , is it 3t or -3t?

Answer 26

Unfortunately that's the only one we have! So we can safely go along with your suggestion.

Answer 27

Sorry, you're right. The eigenvalue is -3, so it's e^-3t.

Answer 28

well\[(A-\lambda I)=(A+3I)=\left[\begin{matrix}1 & 1 \\ -1 & -1\end{matrix}\right]\]so\[\overrightarrow{\eta}=\left(\begin{matrix}1 \\ -1\end{matrix}\right)\]right?

Answer 29

yes

Answer 30

Right, when we do the echelon form, we usually put the linearly dependent rows as zero, so it would look like:\[\left[\begin{matrix}1 & 1 \\ 0 & 0\end{matrix}\right]\]
But the other form is perfectly understandable, because the second line is clearly linearly dependent on the first.

Answer 31

btw, i understand your solution but i am having difficulties at placing them into system because it is divided by A and B

Answer 32

i wrote e^(-3t) to the first rows and t*e^(-3t) to the second

Answer 33

so then how do I solve\[\left[\begin{matrix}1 & 1 \\ -1 & -1\end{matrix}\right]\overrightarrow{\rho}=\left(\begin{matrix}1 \\ -1\end{matrix}\right)\]???

Answer 34

With the second row as zero, we then know that we need one "free variable", one which we can assign any value.
The first row reduces to
x1+x2=0
So if we put x2=-1, then x1=1, hence [1,-1].
We could equally well put x2=1, then x1=-1, and [-1,1] is an equally valid eigenvector, since they are linearly dependent of each other.

Answer 35

For ruhmeshalle, that's what you do.
I lumped C1 and C2 together, instead of separately like A and B.

Answer 36

so then we just need\[\overrightarrow{x}=Ae^{\lambda t}\overrightarrow{\eta}+B(te^{\lambda t}\overrightarrow{\eta}+e^{\lambda t}\overrightarrow{\rho})\]where A and B are the constants mathmate was referring to, and all those vector have now been found.

Answer 37

Since we don't have a second eigenvector, you can substitute 0 for rho and end up with something like
[x,y] = e^(-3t)(A+Bt) [1,-1] (or C1 and C2 as I called them)

Answer 38

ruhmeshalle, do you have the initial conditions?

Answer 39

But don't we need the other eigenvector?

no I think no initial conditions, it says "most general solution"

Answer 40

http://i.imgur.com/J5sJW.png

Answer 41

it is going to kill me

Answer 42

yes, because you only used the eta vector and we never found rho

Answer 43

still haven't dealt with the problem I posted above with rho...

Answer 44

If there are no initial conditions, you can then substitution the solution into the original equations to see if it works.

Answer 45

Are eta and rho supposed to be different eigenvectors (in a normal case)?

Answer 46

mathmate,system says i am half correct. where i am wrong? i did exactly what you said

Answer 47

@mathmate
that was my understanding.
look at the beginning of this
http://tutorial.math.lamar.edu/Classes/DE/RepeatedEigenvalues.aspx

Answer 48

It is due to the double root.

Answer 49

You're right, we ARE missing rho.
We need to get a second vector out of the first AND A, let me get back to you.

Answer 50

so where have I gone wrong? It seems certain that eta is [1,-1] which and that \[(A-\lambda I)=\left[\begin{matrix}1 & 1 \\ -1 & -1\end{matrix}\right]\]which is making this unsolvable for me.

Answer 51

why do we need A ?
or which A are you referring to, the constant you originally called C1 ?

Answer 52

Oh wait, I think I got the idea...

Answer 53

I am working on it, I think rho would be [1,1], but I have to work out the details.
How many lives do you have?

Answer 54

infinite

Answer 55

Good, one thing less to worry about. I'll be back.

Answer 56

The formula I posted above leads to \[\rho_1+\rho_2=1\]\[-\rho_1-\rho_2=-1\]so we can write\[\rho_2=1-\rho_1\]and let\[\rho_1=0\]so\[\overrightarrow{\rho}=\left(\begin{matrix}1 \\ 0\end{matrix}\right)\]will work I think...

Answer 57

sorry, other way around\[\overrightarrow{\rho}=\left(\begin{matrix}0 \\ 1\end{matrix}\right)\]

Answer 58

sorry, don't know why I made lambda negative on accident...
I meant,
so try that with\[\overrightarrow{\eta}=\left(\begin{matrix}1 \\ -1\end{matrix}\right)\]in\[\overrightarrow{x}=Ae^{\lambda t}\overrightarrow{\eta}+B(te^{\lambda t}\overrightarrow{\eta}+e^{\lambda t}\overrightarrow{\rho})\]

Answer 59

still didn't work

Answer 60

nuts...

Answer 61

t*e^(-3t)+e^(-3t)

t*e^(-3t)-e^(-3t)

Answer 62

i hope i wrote it truly...

Answer 63

no, you are missing a term I think

Answer 64

http://i.imgur.com/7Gzbl.png

Answer 65

It turns out that the solution should be:
x=e^(-3t)(C1+C2(1+t)), and
y=e^((-3)t)(-C1-C2 t)
I have checked them against the original equations, and both are satisfied.

Answer 66

yes, but in terms of the way he has to formulate his answers it's hard to see how to enter that.

Answer 67

it says it is wrong again

Answer 68

oh, the bottom should be -te^(-3t)+e^(-3t) I think

Answer 69

on the B part only I mean

Answer 70

We need to formulate the result into the format requiring two vectors, one of which is [1,-1] and the other one [1,0].

Answer 71

that's what I got :) well, I made the other [0,1] but that should work as well, right?

Answer 72

te^(3t)
-te^(3t)+e^(-3t)

should be what goes into B I think...

Answer 73

It is possible, because C1 and C2 are undetermined.
You can differentiate and substitute into the original equations and see if they work as well.

Answer 74

yes!

Answer 75

now it is true

Answer 76

Hooray!
The first time you multiplied the one with the t by rho, it is the other way around.

Answer 77

Look again at what I got above for rho and eta and the form I posted above for the general solution and you will see that is the solution I had earlier.

Answer 78

i will

Answer 79

thank you very much for dealing with this problem for very long time

Answer 80

It's ok, this was fun, I let the boring ones go by.

Answer 81

L style

Answer 82

recognize...

Answer 83

;)

Answer 84

ruhmeshalle, could you post the screen shot?

Answer 85

ok, just a second

Answer 86

http://i.imgur.com/P340t.jpg

Answer 87

here it is

Answer 88

i like this green color..

Answer 89

Thank you!

Answer 90

thank you too!