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Math Help - solve by systematic elimination...

  1. #1
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    solve by systematic elimination...

    dx/dt = -5x - y

    dy/dt = 4x-y

    x(1) = 0 and y(1)= 1

    ok so rewriting this problem....

    Dx + 5x + y = 0

    Dy - 4x + y = 0

    rewriting it again I get...

    x(D+5) + y = 0
    -4x + y(D+1) = 0

    so i chose to eliminate the x first and got my y to be...

    y = C1e^-3t + C2te^-3t

    and this is where i get confused, i thought i was doing the problem right but my book has a different answer. i know the initial values will replace the C1 and C2 values...it seems that the e^-3t should be e^-3t+3 which i dont see how they got...

    any help would be appreciated..thanks in advance!
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  2. #2
    Behold, the power of SARDINES!
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    So your start seem good, but I do have to comment you really can put the x and y in front of the D terms. so we have

    \begin{array}{ccccc}<br />
(D+5)x & + & y & =& 0 \\<br />
-4x & + & (D+1)y & = & 0<br />
 \end{array}

    Now you want to eliminate x so we multiply the top equation by 4 and let (D+5) act on the bottom equation to get

    \begin{array}{ccccc}<br />
4(D+5)x & + & 4y & =& 0 \\<br />
-4(D+5)x & + & (D+5)(D+1)y & = & 0<br />
 \end{array}

    Now if we add the two equations we get

    4y+(D+5)(D+1)y=0 \iff [4+(D+5)(D+1)]y=0

    [D^2+6D+9]y=0 \iff (D+3)^2y=0

    This has complimentary solution

    y(t)=c_1e^{-3t}+c_2te^{-3t}

    Now if we plug this into the 2nd equation we get

    \displastyle x=\frac{y+y'}{4}

    y'(t)=-3c_1e^{-3t}+c_2(e^{-3t}-3te^{-3t})

    This gives

    x(t)=\frac{1}{4}\left( -3c_1e^{-3t}+c_2(e^{-3t}-3te^{-3t})+c_1e^{-3t}+c_2te^{-3t}\right)

    \displaystyle x(t)=\frac{1}{4}[(-2c_1+c_2)e^{-3t}-2c_2te^{-3t}]

    0=(-2c_1+c_2)e^{-3}-c_2e^{-3} \iff -2c_1-c_2=0

    1=c_1e^{-3}+c_2e^{-3} \iff e^{3}=c_1+c_2

    This gives c_1=-e^{3} \quad c_2=2e^{3}

    Now we can answer your question

    If you put these into the two equations above you will get what you want

    y(t)=-e^{3}e^{-3t}+2e^{3}te^{-3t}=-e^{-3t+3}+2e^{-3t+3}
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  3. #3
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    ok let me go back and make my corrections..thanks alot!
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