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Math Help - Linear System with Constant Coefficients

  1. #1
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    Linear System with Constant Coefficients

    Dx1 = -x1 - 3e^(-2t)
    Dx2 = -2x1 - x2 - 6^(e^-2t)

    Find the general solution of the system.

    Can a calc guru help walk me through how to solve this system? Thanks,

    Kim
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  2. #2
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    Quote Originally Posted by Kim Nu View Post
    Dx1 = -x1 - 3e^(-2t)
    Dx2 = -2x1 - x2 - 6^(e^-2t)

    Find the general solution of the system.

    Can a calc guru help walk me through how to solve this system? Thanks,

    Kim
    Lets start by writing each equation in operator form

    E_1: \\\ (D+1)x_1=-3e^{-2t}

    E_2: \\\ (D+1)x_2+2x_1=-6e^{-2t}

    Now multiply equation 2 by -\frac{1}{2}(D+1) and add it 2 equation 1 to get

    -\frac{1}{2}(D+1)^2x_2=3(D+1)e^{-2t}-3e^{-2t} \iff -\frac{1}{2}(D+1)^2x_2=-6e^{-2t}

    (D+1)^2x_2=12e^{-2t}

    So x_{2c}=c_1e^{-t}+c_2te^{-t}

    The particular solution is of the form x_{2p}=Ae^{-2t}

    so we get

    4Ae^{-2t}+2(-2)Ae^{-2t}+Ae^{-2t}=12e^{-2t} \iff A=12

    So x_{2p}=12e^{-2t} and the solution is

    x_2= c_1e^{-t}+c_2te^{-t}+12e^{-2t}

    I did this very quickly so check my work.

    You should be able to finish from here.

    I hope this helps. Good luck.
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  3. #3
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    Hi Empty Set,

    Thanks a bunch for the response!

    One question on this step:



    How do you make the D on the right hand side dissappear? Shouldn't it be:

    3(D + 1)e^(-2t) - 3e^(-2t) = 3e^(-2t) [(D + 1) -1] = (3e^(-2t))(D)
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  4. #4
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    Quote Originally Posted by Kim Nu View Post
    Hi Empty Set,

    Thanks a bunch for the response!

    One question on this step:



    How do you make the D on the right hand side dissappear? Shouldn't it be:

    3(D + 1)e^(-2t) - 3e^(-2t) = 3e^(-2t) [(D + 1) -1] = (3e^(-2t))(D)
    Remember that D =\frac{d}{dt}

    So it we distribute the operator we get

    3(D+1)e^{-2t}-3e^{-2t}=3[De^{-2t}+e^{-2t}]-3e^{-2t}

    Now we do what D tells us we take a derivative

    3[De^{-2t}+e^{-2t}]-3e^{-2t}=3[-2e^{-2t}+e^{-2t}]-3e^{-2t}=-6e^{-2t}+3e^{-2t}-3e^{-2t}=-6e^{-2t}

    I hope this clears it up.

    Good luck.
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  5. #5
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    Hi Empty Set,

    Thanks for your teaching. I reworked the problem and arrived at the same point you did for x2. However the answer in the back of the book for x2 is:

    x2 = -2c1te^(-t) + c2e^(-t) + 12e^(-2t)

    How do you suppose that -2 snuck its way in front of the c1 for the x2c solution? Thanks,

    Kim
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  6. #6
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    Quote Originally Posted by Kim Nu View Post
    Hi Empty Set,

    Thanks for your teaching. I reworked the problem and arrived at the same point you did for x2. However the answer in the back of the book for x2 is:

    x2 = -2c1te^(-t) + c2e^(-t) + 12e^(-2t)

    How do you suppose that -2 snuck its way in front of the c1 for the x2c solution? Thanks,

    Kim
    When you solve for x_1 you will get two more constants

    c_3, c_4

    But since each of these systems are only first order and linear you can only have ONE constant per equation.

    You will need to plug both of your solutions into the original system and solve for a relationship between c_1,c_2,c_3,c_4 and eliminate two of them from your solution. Notice that both solutions only have c_1,c_2 in them.

    I hope this gives you somewhere to start...

    I'm leaving for work now, but will try to look back later.

    Good luck.
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  7. #7
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    Thanks Empty Set. I'll look forward to some more input on this one. I have no problem findin the correct answer for x1; its the x2 that is confusing me.
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    Here is the rest...

    Since you have solved for x_1 I will just use my solution

    x_1=c_3e^{-t}+3e^{-2t}

    Note that this is a different constant then from x_2

    Now we know that the constants c_1,c_2,c_3 are related by what I called E_2 before. So we plug in x_1,x_2 and start simplifying.

    (D+1)x_2+2x_1=-6e^{-2t}

    -c_1e^{-t}+c_2e^{-t}-c_2te^{-t}-24e^{-2t}+c_1e^{-t}+c_2te^{-t}+12e^{-2t}+2c_3e^{-t}+6e^{-2t}=-6e^{-2t}

    Collecting like terms are reducing we get

    (-c_1+c_2+c_1+2c_3)e^{-t}=0 \iff (c_2+2c_3)e^{-t}=0

    so c_2=-2c_3

    So now we get the solutions

    x_1=c_3e^{-t}+3e^{-2t} and

    x_2=c_1e^{-t}\underbrace{-2c_3te^{-t}}_{c_2=-2c_3}+12e^{-2t}

    I hope this helps.

    Good luck.
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  9. #9
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    You are the master!
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