Question

need help differential equations matrice problem

http://i.imgur.com/kDW5j.png

Answer 1

i found eigenvalues and vectors for the first matrice but for the second one?

Answer 2

i don't know how to organize them together

Answer 3

http://www.jirka.org/diffyqs/htmlver/diffyqsse24.html

Answer 4

at example 3.9.1 there is a related question.

Answer 5

As a start, for
\[\left[\begin{matrix}-1 & -3 \\ 1 & -5\end{matrix}\right]\]
The eigenvalues are -2 and -4.
You can almost find them by inspection because the sum of the eigenvalues is the trace of the matrix (-1+-5)=-6, and the product is the determinant (-1*-5-(-3)*1)=+8.
The eigenvectors are <3,1> and <1,1>, which are real and distinct, and you should be able to solve for the general solution.
The first equation is not homogeneous, so you need a particular solution.
Hope to be able to come back for that some time later, or if TuringTest could help.

Answer 6

i will try to solve it by taking help from the link i sent, they are very related. thanks though

Answer 7

Yes, it is very related, and it looks very detailed. You need to solve a non-homogeneous equation before attempting a system, so go for it.

Answer 8

Well, mathmate has given us the complimentary solution:\[\overrightarrow{y_c}=c_1e^{-2t}\left(\begin{matrix}1 \\ 3\end{matrix}\right)+c_2e^{-4t}\left(\begin{matrix}1 \\ 1\end{matrix}\right)\]So now we must find the particular solution. I'm thinking that it can be done best with variation of parameters.

Answer 9

Here's what I've got so far:
We're going to make use of the formula of variation of parameters for a system\[\overrightarrow{x_p}=X\int X^{-1}\overrightarrow{g}dt\]for a proof of that formula and all it means check here: http://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousSystems.aspx
For now suffice to say that if you read that you will understand that X is simply the our answer for the complimentary solution written as a matrix.\[\overrightarrow{y_c}=c_1e^{-2t}\left(\begin{matrix}1 \\ 3\end{matrix}\right)+c_2e^{-4t}\left(\begin{matrix}1 \\ 1\end{matrix}\right)\]so\[X=\left[\begin{matrix}e^{-2t} & e^{-4t} \\ 3e^{-2t} & e^{-4t}\end{matrix}\right]\]we need the inverse of this matrix, which I found to be\[X^{-1}=\left[\begin{matrix}-\frac{1}{2}e^{2t} & \frac{1}{2}e^{2t} \\ \frac{3}{2}e^{4t} & -\frac{1}{2}e^{4t}\end{matrix}\right]\]and the other vector in the formula above, g, is just the part of vector in our equation that is making it non-homogeneous:\[\overrightarrow{g}=\left(\begin{matrix}e^{-t} \\ 0\end{matrix}\right)\]Now it's just a matter of plugging in this into the formula and solving.

I'll see what I get...

Answer 10

Ok, I got\[X^{-1}\overrightarrow{g}=\left(\begin{matrix}-\frac{1}{2}e^t \\ \frac{3}{2}e^{3t}\end{matrix}\right)\]so\[\int X^{-1}\overrightarrow{g}=\int\left(\begin{matrix}-\frac{1}{2}e^t \\ \frac{3}{2}e^{3t}\end{matrix}\right)dt=\left(\begin{matrix}-\frac{1}{2}e^t \\ \frac{1}{2}e^{3t}\end{matrix}\right)\]which means\[X\int\limits X^{-1}\overrightarrow{g}=\left[\begin{matrix}e^{-2t} & e^{-4t} \\ 3e^{-2t} & e^{-4t}\end{matrix}\right]\left(\begin{matrix}-\frac{1}{2}e^t \\ \frac{1}{2}e^{3t}\end{matrix}\right)=\left(\begin{matrix}0 \\ -e^{-t}\end{matrix}\right)=e^{-t}\left(\begin{matrix}0 \\ -1\end{matrix}\right)\]should be our particular solution.

Answer 11

i was afk

Answer 12

Putting that all together before the initial conditions gives\[\overrightarrow{y}(t)=c_1e^{-2t}\left(\begin{matrix}1 \\ 3\end{matrix}\right)+c_2e^{-4t}\left(\begin{matrix}1 \\ 1\end{matrix}\right)+e^{-t}\left(\begin{matrix}0 \\ -1\end{matrix}\right)\]using the initial condition\[\overrightarrow{y}(0)=c_1\left(\begin{matrix}1 \\ 3\end{matrix}\right)+c_2\left(\begin{matrix}1 \\ 1\end{matrix}\right)+\left(\begin{matrix}0 \\ -1\end{matrix}\right)=\left(\begin{matrix}1 \\ 0\end{matrix}\right)\]so it seems that we have to solve\[c_1+c_2=1\]\[3c_1+c_2-1=0\]which is giving me a headache since that gives\[c_1=0\]\[c_2=1\]and I'm not sure if that's okay, having C1 be zero and wiping out the first term.

However it's what I'm getting and so I will post what it becomes:\[\overrightarrow{y}(t)=e^{-4t}\left(\begin{matrix}1 \\ 1\end{matrix}\right)+e^{-t}\left(\begin{matrix}0 \\ -1\end{matrix}\right)\]I hope that is the answer, but I will have to find a way to check.

Answer 13

i am checking your solution now

Answer 14

*finger crossed
but it is my first attempt...

Answer 15

The vector on the left is correct.!

Answer 16

now i am checking the other one

Answer 17

sweet, that was the one I was worried about.

Answer 18

oh but i can't say the same thing for the other one :/

Answer 19

No!!!
screenshot please?

Answer 20

aye aye captain, just a second

Answer 21

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

Answer 22

You have the final solution in the place where the complimentary one goes. Enter\[\overrightarrow{y_c}=c_1e^{-2t}\left(\begin{matrix}1 \\ 3\end{matrix}\right)+c_2e^{-4t}\left(\begin{matrix}1 \\ 1\end{matrix}\right)\]for that one.

Answer 23

That HAS to be right.

Answer 24

i am sorry but it is not :/

Answer 25

but mathmate and I have the same answer o-0

Answer 26

http://i.imgur.com/gFlI2.png

Answer 27

ok wait, I think it's just a matter of the one on the left being the other way round...\[\overrightarrow{y_c}=c_1e^{-2t}\left(\begin{matrix}3 \\ 1\end{matrix}\right)+c_2e^{-4t}\left(\begin{matrix}1 \\ 1\end{matrix}\right)\]but then I have to do the rest of it over agian....

Answer 28

but when e^-4t was on the left side it was also true

Answer 29

oh you mean

Answer 30

3

Answer 31

that's because c1 and c2 are arbitrary, it doesn't matter which is which

Answer 32

yes the three

Answer 33

i didn't see

Answer 34

I'm thinking that is right, which means I have to do the rest over again...

Answer 35

let me do it this time because you have already worked so much on this problem. i feel bad when people spend their time for me for such a long time.. you should have a rest bro. it's christmas time, thank you

Answer 36

But now I want to know the answer :/

Answer 37

did switching the 3 change anything?

Answer 38

i feel relieved, now it is all up to you :D

Answer 39

is that a yes? it worked as\[\overrightarrow{y_c}=c_1e^{-2t}\left(\begin{matrix}3 \\ 1\end{matrix}\right)+c_2e^{-4t}\left(\begin{matrix}1 \\ 1\end{matrix}\right)\]
???

Answer 40

i am checking

Answer 41

YES!

Answer 42

the green color appeared

Answer 43

the color of victory

Answer 44

Okay, then I am doing it right, but it is going to be a while before I have the correct particular.

Give it some time, and don't worry, if I wanted to work on another problem I would.

Answer 45

you really remind me of L :D

Answer 46

and mathmate is watari

Answer 47

I wish :P
lol

Answer 48

this is going much faster now that I know what to do...

Answer 49

i think again c1=0 and c2=1

Answer 50

→=c1e−2t(31)+c2e−4t(11) so this is the one which i should put the constant values in

Answer 51

i did and the answer is (e^(-4t) , e^(-4t)-e^-t)

Answer 52

not quite there yet, I'm about to look for the constants.

Answer 53

yes I got \[c_1=0\]\[c_2=-\frac{1}{3}\]and a final answer of\[\overrightarrow{y}(t)=-\frac{1}{3}e^{-4t}\left(\begin{matrix}3 \\ 1\end{matrix}\right)+e^{-t}\left(\begin{matrix}\frac{4}{3} \\ \frac{1}{3}\end{matrix}\right)\]Let's hope...

Answer 54

i don't know if i am writing it wrong but it didn't accept the first one

Answer 55

i will send you screenshot

Answer 56

great,second one is true

Answer 57

so\[y_1(t)=-e^{-4t}+\frac{4}{3}e^{-t}\]\[y_2(t)=-\frac{1}{3}e^{-4t}+\frac{1}{3}e^{-t}\]is what should be in your lines I think.

Answer 58

yes i wrote it exactly the same

Answer 59

and which part is wrong?

Answer 60

but first part is wrong

Answer 61

y1(t)=−e−4t+43e−t this

Answer 62

http://i.imgur.com/9mg4O.png

Answer 63

Now how could that possibly be...?
I'll have to think about that...

Answer 64

hey

Answer 65

are you there

Answer 66

It's gotta be the 4/3...
yeah I'm here

Answer 67

i just wrote -1/3 front of e^(-4t) and it is true now

Answer 68

(-1/3)e^(-4t)+(4/3)e^-t

Answer 69

it really is interesting

Answer 70

well that seems strange, how could that be necessary?

Answer 71

i don't have any applicable suggestion at this point because changing one of these values makes the second equation false

Answer 72

if\[\overrightarrow{y}(t)=-\frac{1}{3}e^{-4t}\left(\begin{matrix}3 \\ 1\end{matrix}\right)+e^{-t}\left(\begin{matrix}\frac{4}{3} \\ \frac{1}{3}\end{matrix}\right)\]led to that the issue must be the only way to get a -1/3 in there is if that first vector was [1,1] which we know it isn't.

Answer 73

so it makes very little sense to me...

Answer 74

there may be some disturbance in the system though

Answer 75

computers are not trustable

Answer 76

let's blame the system and get away with it :p

Answer 77

I would actually bet on it in this case.
Do you mind if I type what I did and you can evaluate it later?
I just like using LaTeX and have most of it saved already, just need to change some things...

Answer 78

sure it is ok

Answer 79

i just have 1 question left about this chapter, same type of question

Answer 80

i will try solving it by following your steps

Answer 81

Our complimentary solution is:\[\overrightarrow{y_c}=c_1e^{-2t}\left(\begin{matrix}3 \\ 1\end{matrix}\right)+c_2e^{-4t}\left(\begin{matrix}1 \\ 1\end{matrix}\right)\]Variation of parameters:\[\overrightarrow{x_p}=X\int X^{-1}\overrightarrow{g}dt\]Where X is the matrix made up of our solutions for the complimentary solution, and g is the vector in the equation that makes it non-homogeneous. For us\[X=\left[\begin{matrix}3e^{-2t} & e^{-4t} \\ e^{-2t} & e^{-4t}\end{matrix}\right]\]and\[\overrightarrow{g}=\left(\begin{matrix}e^{-t} \\ 0\end{matrix}\right)\]We need the inverse of X to use this formula which I found to be\[X^{-1}=\left[\begin{matrix}\frac{1}{2}e^{2t} & -\frac{1}{2}e^{2t} \\ -\frac{1}{2}e^{4t} & \frac{3}{2}e^{4t}\end{matrix}\right]\]so\[X^{-1}\overrightarrow{g}=\left[\begin{matrix}\frac{1}{2}e^{2t} & -\frac{1}{2}e^{2t} \\ -\frac{1}{2}e^{4t} & \frac{3}{2}e^{4t}\end{matrix}\right]\left(\begin{matrix}e^{-t} \\ 0\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}e^{t} \\ -\frac{1}{2}e^{3t}\end{matrix}\right)\]then\[\int X^{-1}\overrightarrow{g}dt=\int\left(\begin{matrix}\frac{1}{2}e^{t} \\ -\frac{1}{2}e^{3t}\end{matrix}\right)dt=\left(\begin{matrix}\frac{1}{2}e^{t} \\ -\frac{1}{6}e^{3t}\end{matrix}\right)\]so\[X\int X^{-1}\overrightarrow{g}dt=\left[\begin{matrix}3e^{-2t} & e^{-4t} \\ e^{-2t} & e^{-4t}\end{matrix}\right]\left(\begin{matrix}\frac{1}{2}e^{t} \\ -\frac{1}{6}e^{3t}\end{matrix}\right)=\left(\begin{matrix}\frac{4}{3}e^{-t} \\ \frac{1}{3}e^{-t}\end{matrix}\right)\]Which means our particular solution is\[\overrightarrow{y_p}=e^{-t}\left(\begin{matrix}\frac{4}{3} \\ \frac{1}{3}\end{matrix}\right)\]...

Answer 82

So putting that all together gives\[\overrightarrow{y}(t)=c_1e^{-2t}\left(\begin{matrix}3 \\ 1\end{matrix}\right)+c_2e^{-4t}\left(\begin{matrix}1 \\ 1\end{matrix}\right)+e^{-t}\left(\begin{matrix}\frac{4}{3} \\ \frac{1}{3}\end{matrix}\right)\]and the initial condition gives\[\overrightarrow{y}(0)=c_1\left(\begin{matrix}3 \\ 1\end{matrix}\right)+c_2\left(\begin{matrix}1 \\ 1\end{matrix}\right)+\left(\begin{matrix}\frac{4}{3} \\ \frac{1}{3}\end{matrix}\right)=\left(\begin{matrix}1 \\ 0\end{matrix}\right)\]leading to the system\[3c_1+c_2+\frac{4}{3}=1\]\[c_1+c_2+\frac{1}{3}=0\]which solves to\[c_1=0\]\[c_2=-\frac{1}{3}\]so our final solution SHOULD be\[\overrightarrow{y}(t)=-\frac{1}{3}e^{-4t}\left(\begin{matrix}3 \\ 1\end{matrix}\right)+e^{-t}\left(\begin{matrix}\frac{4}{3} \\ \frac{1}{3}\end{matrix}\right)\]

Answer 83

Somebody tell me why it's not quite right, please.

Answer 84

i can repost this question if you want to

Answer 85

thanks for such great work

Answer 86

should i ?

Answer 87

because it seem to be true to me

Answer 88

or i can ask it to one of my professors tomorrow and give you feedback

Answer 89

Sure, link people to it or let me know somehow if you find out what I did wrong.

Answer 90

Let me work it out.

Answer 91

Yeah, probable better to just work from scratch.

Answer 92

Homogenous I get:\[\vec{y}_H(t)=c_1 e^{2t}(1,-1)^T+c_2 e^{4t} (-3,5)^T\]
Where T denotes the transpose (keeps me from typing "matrices"

Answer 93

no, that can't be right. mathmate has the same answer for that I do, and it was accepted by the computer.

Answer 94

I get\[\overrightarrow{y_c}=c_1e^{-2t}\left(\begin{matrix}3 \\ 1\end{matrix}\right)+c_2e^{-4t}\left(\begin{matrix}1 \\ 1\end{matrix}\right)\]

Answer 95

Oh, I wrote the matrix down wrong. x.x

Answer 96

The problem only seems to occur at the end, so it will be a while before you get there I think.

Answer 97

Yeah, that negative sign made a huge difference. I got the same thing you did, let me do the rest.

Answer 98

Almost done