Results 1 to 4 of 4

Math Help - Linear system--I have this 1/2 solved...

  1. #1
    Junior Member
    Joined
    May 2008
    Posts
    50

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    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 x_1

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

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2008
    Posts
    50
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    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 x_2 should be this

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

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

    So if we put these into E_2 we get...

    -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) +
     c_4\sin(t) + c_5e^{-2t} + 2e^{-t})=-3e^{-t}

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

    Now we have

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

    We need to solve these for c_3,c_4

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

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

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

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

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

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

    Now we finally can get the final solution

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

    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 c_1=5k_1 \\\ c_2=5k_2 in both equations you will get the same answer.

    Good luck.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear System
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 5th 2010, 04:04 AM
  2. System non-linear ODE's
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: January 24th 2010, 11:34 AM
  3. [SOLVED] System of Linear Equations
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: September 2nd 2009, 04:40 PM
  4. linear system
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 28th 2009, 09:17 AM
  5. linear system
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 28th 2008, 02:34 PM

Search Tags


/mathhelpforum @mathhelpforum