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Math Help - Bounded soloutions for an ode

  1. #1
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    Bounded soloutions for an ode

    Given this ode system:
    x' = 2x+y-7e^(-t) -3
    y'= -x+2y-1

    Find all the bounded soloution in [a,infinity) when a is a real number...

    I'm not realy sure what is a sufficient condition for bounded soloution in this question...Maybe there's something we can do and then we will not even need to solve the system...


    Help is Needed!!

    TNX a lot!
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  2. #2
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    Quote Originally Posted by WannaBe View Post
    Given this ode system:
    x' = 2x+y-7e^(-t) -3
    y'= -x+2y-1

    Find all the bounded soloution in [a,infinity) when a is a real number...

    I'm not realy sure what is a sufficient condition for bounded soloution in this question...Maybe there's something we can do and then we will not even need to solve the system...


    Help is Needed!!

    TNX a lot!
    This system can be written as
    \begin{bmatrix} x' \\ y'\end{bmatrix}= \begin{bmatrix}2 & 1 \\ -1 & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}+ \begin{bmatrix}-7e^{-t}- 3 \\ -1\end{bmatrix}

    Find the eigenvalues of that coefficient matrix. If there are any eigenvalues with positive real part, discard them. What is left will give the bounded solutions.
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  3. #3
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    Hey there,
    1. why should I discard the eignvalues with positive real part? How come it's a sufficient condition for bounded soloutions?
    2. " What is left will give the bounded solutions. " Why? The soloution I'll get by using the eignvalues will be a soloution for the homo. system only... I need a soloution for the whole system... How come the soloutions for the homo. system with a non-positive real part will give me the bounded soloution for the whole system?

    TNX!
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  4. #4
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    BTW -
    The eignvalues are 2+i, 2-i, and they both have positive real part..t
    Last edited by WannaBe; December 19th 2009 at 08:15 AM.
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  5. #5
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    Quote Originally Posted by WannaBe View Post
    Hey there,
    1. why should I discard the eignvalues with positive real part? How come it's a sufficient condition for bounded soloutions?
    2. " What is left will give the bounded solutions. " Why? The soloution I'll get by using the eignvalues will be a soloution for the homo. system only... I need a soloution for the whole system... How come the soloutions for the homo. system with a non-positive real part will give me the bounded soloution for the whole system?

    TNX!
    And what solution do you get for, say, eigenvalue 1+ 2i? e^x(Ccos(2x)+ Dsin(2x)), right? Don't you see that the positive real part becomes the coefficient of x in the exponential? That gets larger and larger without bound. If you have eigenvalue, say, -1+ 2i, you get solution e^{-x}(Ccos(2x)+ Dsin(2x)) which goes to 0 as x goes to infinity- that's definitely "bounded"! If an eigenvalue has 0 real part, say, 0+ 2i, then the solution is Ccos(2x)+ Dsin(2x) since cosine and sine are never larger than 1 or less than -1.

    Now, what are the eigenvalues of that matrix?
    Last edited by HallsofIvy; December 20th 2009 at 07:53 AM.
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  6. #6
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    I know everything you said, but tnx for the reminder...
    I've answered the question you've asked in the msg above you-
    The eignvalues are 2+i, 2-i, and they both have positive real part... In order to solve the entire system, we need to find fundemental basis of soloutions for the homo. system and one private soloution of the non-homogenic one...
    If we'll take 2+i, and find its eigenvectors, we'll get: (1,i) ... So we have a soloution:
    e^(2+i)t*(1,i)...Its real part is one soloution which is independant with the imag. part...So we'll get these two soloutions:
    x1= e^2t [ cost(1,0) -sint(0,1) ]
    x2 = e^2t[ sint(1,0) +cost(0,1) ]
    In order to solve the system, we now have to find a private soloution for the non-homogenic system, but it seems to be a very long process... Maybe from the two soloutions for the homo. system we will be able to find all the bounded soloutions for the entire system... As you can see, I realy need help here...

    TNX!
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  7. #7
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    Well NVM I've made it on my own

    TNX
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  8. #8
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    I'm glad. You seem to have misunderstood everything I said.
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