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Differential Equation - Solve by substitution

  1. Feb 18, 2009 #1

    JJBladester

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    1. The problem statement, all variables and given/known data
    Solve the following differential equation by using substitution.


    2. Relevant equations

    (x + 2y - 1)dx + (2x + 4y - 3)dy = 0

    Let u = x + 2y

    3. The attempt at a solution

    u = x + 2y
    du/dx = 1 + 2(dy/dx)

    (x + 2y - 1)dx + (2x + 4y - 3)dy = 0

    (u - 1)dx + (2u - 3)dy = 0

    The solution the teacher gave is as follows but I didn't know how to go from my line above to the answer below. I believe it has something to do with getting the dx and dy out of the new equation involving u and then rearranging to solve by separation variables.

    K = x + (x + 2y)2 - 3(x + 2y)
     
  2. jcsd
  3. Feb 18, 2009 #2

    Tom Mattson

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    Use this...

    ...to eliminate dy in this

     
  4. Feb 18, 2009 #3

    JJBladester

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    If u = x + 2y

    then du/dx = 1 + 2(dy/dx)

    and dy/dx = (du/dx - 1)/2

    So...

    dy/dx + (x + 2y -1)/(2x + 4y - 2) = 0

    (du/dx - 1)/2 + (u - 1)/(2u - 3) = 0

    du/dx - 1 + (2u - 2)/(2u - 3) = 0

    If I divide (2u - 2) by (2u - 3), I get:

    1 + 1/(2u - 3)

    Then I have:

    du/dx - 1 + 1 + 1/(2u - 3) = 0

    du/dx + 1/(2u - 3) = 0

    (2u - 3)du + dx = 0

    The integration gives me:

    u2 - 3u + x = c

    So, my answer (after putting (x + 2y) back in for u) is:

    (x + 2y)2 -3(x + 2y) = c

    Got it, Tom. Thanks for the push in the right direction!
     
    Last edited: Feb 18, 2009
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