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Math Help - Particular solution of linear systems

  1. #1
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    Particular solution of linear systems

    Linear system of D.E's

    x' = -3x + 2y,
    y' = -3x + 4y,

    Initial conditions

    x(0) = 0;
    y(0) = 2;

    Using these equations, I derived x" - 4x' + 3x = 0

    And then found Yc = Ae^x + Be^(3x).

    Now what should I do?

    -WWB
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  2. #2
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    Quote Originally Posted by Whitewolfblue View Post
    Linear system of D.E's

    x' = -3x + 2y,
    y' = -3x + 4y,

    Initial conditions

    x(0) = 0;
    y(0) = 2;

    Using these equations, I derived x" - 4x' + 3x = 0

    And then found Yc = Ae^x + Be^(3x).

    Now what should I do?

    -WWB
    Please show how you got x" - 4x' + 3x = 0 (I get something different). Then you use x(0) = 0 to eliminate one of the arbitrary constants.

    Then you substitute x = x(t) into the first DE and solve for y. Then use y(0) = 2 to eliminate the other arbitrary constant.
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  3. #3
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    How I got the equation

    x' = -3x + 2y,
    y' = -3x + 4y,

    I eliminated the y's first in order to solve for x, I'm sure I have to eliminate x's next.

    I just realized what I did, I crossed my notes with the question I was working on.

    So looking at things I should have gotten:

    5.5x" + .5x' - 3x = 0 (that is if my new math is correct).

    What do I need to use the x(0) = 0 with? I need to find Yc again, correct?

    -WBB



    -EDIT-
    I got x(t) I think, not sure which one it is. In any case, when I eliminate y from the equations I get the equation:

    .5x" - .5x' - 3x = 0

    Thus producing Ae^(-2t) + Be^(3t).

    But, when I eliminate x from the equations I get:

    -(1/3)y" + (1/3)y' - y = 0

    I can't get the zero values for that equation because it never crosses the x axis. Did I do something wrong, or is there a step that I need to take in order to solve it?
    Last edited by Whitewolfblue; April 19th 2009 at 06:39 PM. Reason: Got farther
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  4. #4
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    Quote Originally Posted by Whitewolfblue View Post
    x' = -3x + 2y,
    y' = -3x + 4y,

    I eliminated the y's first in order to solve for x, I'm sure I have to eliminate x's next.

    I just realized what I did, I crossed my notes with the question I was working on.

    So looking at things I should have gotten:

    5.5x" + .5x' - 3x = 0 (that is if my new math is correct).

    What do I need to use the x(0) = 0 with? I need to find Yc again, correct?

    -WBB



    -EDIT-
    I got x(t) I think, not sure which one it is. In any case, when I eliminate y from the equations I get the equation:

    .5x" - .5x' - 3x = 0

    Thus producing Ae^(-2t) + Be^(3t). Mr F says: Now substitute x = 0 and t = 0 to get B = -A.

    But, when I eliminate x from the equations I get:

    -(1/3)y" + (1/3)y' - y = 0

    I can't get the zero values for that equation because it never crosses the x axis. Did I do something wrong, or is there a step that I need to take in order to solve it?
    Quote Originally Posted by mr fantastic View Post
    [snip]
    Then you use x(0) = 0 to eliminate one of the arbitrary constants. [See red above]

    Then you substitute x = x(t) into the first DE [that is, into x' = -3x + 2y] and solve for y. Then use y(0) = 2 to eliminate the other arbitrary constant.
    ..
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