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Thread: Linear system--I have this 1/2 solved...

  1. #1
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    Linear system--I have this 1/2 solved...

    E1: (2D + 1)x1 + (D^2 - 4)x2 = -7e^(-t)
    E2: Dx1 - (D + 2)x2 = -3e^(-t)

    Here's how I solved for x1:

    1. Multiply E2 by (D - 2) to get:

    (Dx1)(D-2) - (D^2 - 4)x2 = -3e^(-t) * (D-2)

    2. Simplify the above and then add to E1 to get:

    (D^2 + 1)x1 = 2e^(-t)

    3. x1c = c1cos (t) + c2sin (t)

    4. Then using Ae^(-t), x1p = e^(-t)

    5. x1 = c1cos (t) + c2sin (t) + e^(-t)


    How do I solve for x2? I'm lost again on this one! Thanks,

    Kim
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Kim Nu View Post
    E1: (2D + 1)x1 + (D^2 - 4)x2 = -7e^(-t)
    E2: Dx1 - (D + 2)x2 = -3e^(-t)

    Here's how I solved for x1:

    1. Multiply E2 by (D - 2) to get:

    (Dx1)(D-2) - (D^2 - 4)x2 = -3e^(-t) * (D-2)

    2. Simplify the above and then add to E1 to get:

    (D^2 + 1)x1 = 2e^(-t)

    3. x1c = c1cos (t) + c2sin (t)

    4. Then using Ae^(-t), x1p = e^(-t)

    5. x1 = c1cos (t) + c2sin (t) + e^(-t)


    How do I solve for x2? I'm lost again on this one! Thanks,

    Kim
    Go back to your original equations and try to eliminate $\displaystyle x_1$

    Hint try $\displaystyle D \cdot E_1+[-(2D+1)]E_2$ and see what happens.

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  3. #3
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    Hey thanks Empty Set,

    Now my posts are definitely getting annoying but:

    I took your suggestion, which is the correct course of action; I end up with

    x2 = c3cos(t) + c4sin(t) + c3e^(-2t) + 2e^(-t)

    Then I take the before mentioned solved x1 value and x2 value and plug them into E2: Dx1 - (D + 2)x2 = -3e^(-t)

    I work this all out and end up with:
    sin (t)[-c1 + c3 -2c4] + cos (t)[c2 - c4 - 2c3) = -8e^(-t)

    The answer in the back of the book is:

    x1 = 5c1cost(t) + 5c2sin(t) + e^(-t)
    x2 = (c1 + 2c2)cost + (-2c1 + c2)sin(t) + c3e^(-2t) + 2e^(-t)

    So my answers are close to being correct, I just don't understand how those coefficient were arrived at. Do you have any suggestions? Thanks,

    Kim
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by Kim Nu View Post
    Hey thanks Empty Set,

    Now my posts are definitely getting annoying but:

    I took your suggestion, which is the correct course of action; I end up with

    x2 = c3cos(t) + c4sin(t) + c3e^(-2t) + 2e^(-t)

    Then I take the before mentioned solved x1 value and x2 value and plug them into E2: Dx1 - (D + 2)x2 = -3e^(-t)

    I work this all out and end up with:
    sin (t)[-c1 + c3 -2c4] + cos (t)[c2 - c4 - 2c3) = -8e^(-t)

    The answer in the back of the book is:

    x1 = 5c1cost(t) + 5c2sin(t) + e^(-t)
    x2 = (c1 + 2c2)cost + (-2c1 + c2)sin(t) + c3e^(-2t) + 2e^(-t)

    So my answers are close to being correct, I just don't understand how those coefficient were arrived at. Do you have any suggestions? Thanks,

    Kim
    I think $\displaystyle x_2$ should be this

    $\displaystyle x_2 = c_3\cos(t) + c_4\sin(t) + c_5e^{-2t} + 2e^{-t}$

    And your $\displaystyle x_1 $ from your 1st post is
    $\displaystyle x_1 = c_1\cos (t) + c_2\sin (t) + e^{-t}$

    So if we put these into $\displaystyle E_2$ we get...

    $\displaystyle -c_1\sin(t)+c_2\cos(t)-e^{-t}-(-c_3\sin(t)+c_4\cos(t)-2c_5e^{-2t}-2e^{-t})-2(c_3\cos(t) +$
    $\displaystyle c_4\sin(t) + c_5e^{-2t} + 2e^{-t})=-3e^{-t}$

    $\displaystyle (-c_1+c_3-2c_4)\sin(t)+(c_2-c_4-2c_3)\cos(t)=0$

    Now we have

    $\displaystyle -c_1+c_3-2c_4=0 \\\ c_2-c_4-2c_3=0$

    We need to solve these for $\displaystyle c_3,c_4$

    To get $\displaystyle c_3$ so we want to eliminate $\displaystyle c_4$ so we multiply the 2nd by -2 and add it to the first to get

    $\displaystyle -2(c_2-c_4-2c_3)-c_1+c_3-2c_4=0 \iff -c_1-2c_2+5c_3=0$

    $\displaystyle 5c_3=c_1+2c_2 \iff c_3=\frac{c_1+c_2}{5}$

    Now we need to find $\displaystyle c_4$ so we want to dliminate $\displaystyle c_3$ so now we multiply the first by 2 and add it to the 2nd

    $\displaystyle 2(-c_1+c_3-2c_4)+c_2-c_4-2c_3=0 \iff -2c_1+c_2-5c_4=0$

    $\displaystyle c_4=\frac{-2c_1+c_2}{5}$

    Now we finally can get the final solution

    $\displaystyle x_1 = c_1\cos (t) + c_2\sin (t) + e^{-t}$

    $\displaystyle x_2 = \frac{c_1+c_2}{5}\cos(t) + \frac{-2c_1+c_2}{5}\sin(t) + c_5e^{-2t} + 2e^{-t}$

    The answers are equivalent if you replace $\displaystyle c_1=5k_1 \\\ c_2=5k_2$ in both equations you will get the same answer.

    Good luck.
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