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System of Differential Equations

  1. Apr 17, 2016 #1
    1. The problem statement, all variables and given/known data
    dx/dt + dy/dt = 2x + 2y + 1

    dx/dt + 2(dy/dt) = y + 3
    2. Relevant equations

    3. The attempt at a solution
    Dx + Dy = 2x + 2y + 1

    Dx + 2Dy = y + 3

    --Rearranging----------------------------
    Dx - 2x = -Dy + 2y + 1
    Dx = -2Dy + y + 3
    ---Factoring-----------------------------
    (D - 2)x = (-D + 2)y + 1
    Dx = (-2D + 1)y + 3
    --Eliminating x--
    (D)(Dx-2x) = ((-D+2)y + 1) (D) <--- multiply by D
    -(D-2)(Dx) = ((-2D + 1)y + 3) -(D-2) <---- multiply by -(D-2)

    I get to this point and try to cancel out terms but it becomes a mess and attempting to find the solution usually leaves my with polynomials I can't factor, I don't know what to do..
     
  2. jcsd
  3. Apr 17, 2016 #2

    Ray Vickson

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    Why not do it the easy way? Solve for Dx and Dy in terms of x and y, so you have a standard linear system of the form
    [tex] \pmatrix{x'(t)\\y'(t)} = \pmatrix{a_1 & b_1 \\ a_2 & b_2} \pmatrix{x\\y} + \pmatrix{c_1\\c_2},[/tex]
    where the ##a_i, b_i, c_i## are constants. Then use standard methods.
     
  4. Apr 17, 2016 #3
    I have very little Linear Algebra knowledge, I don't know if i could set up the
    linear system correctly
    |Dx| = |-Dy + 2x + 2y|
    |Dy| = |-2Dy + y + 3 |
     
  5. Apr 17, 2016 #4

    Ray Vickson

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    I don't understand what you are writing here. Anyway, you do not need to know any linear algebra; you just have two linear equations in the two unknowns Dx and Dy, and you can solve them the way you learned back in school.
     
  6. Apr 17, 2016 #5
    I guess i'm confused because I have, derivatives on both sides when solving for Dx and Dy
     
  7. Apr 18, 2016 #6

    vela

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    Here are the original differential equations:

    dx/dt + dy/dt = 2x + 2y + 1
    dx/dt + 2(dy/dt) = y + 3

    Try subtracting the first equation from the second.
     
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