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Elementary differential equations: transformations

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data

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

    a1= 1 b1=2 a2=2 b2=4
    a2/a1=b2/b1

    Therefore z=x+2y
    Here is where I get confused I understand that they must get a dz in the equations thus they take the derivative with respect to y of the following equation z=x+2y
    thus giving dz=dx+2 or dz-dx/2

    What I dont understand is the simplification process that occurs now

    (z+3)dx +(2z-1)(dz-dx/2)=0

    then the book goes on to transform it into

    7dx + (2z-1)dz=0

    I have no idea what they did to get to that step.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 16, 2012 #2

    vela

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    This isn't right. When you differentiate, you should get dz = dx + 2 dy. You can't have that lone 2 hanging around. It needs to be multiplied by dy. Then when you solve for dy, you get dy = (dz-dx)/2. Your second mistake was leaving out the parentheses.

    Does it make sense after you make those corrections? It's just algebra from here.

     
    Last edited: Sep 16, 2012
  4. Sep 16, 2012 #3
    It does clear up a few things but I am actually stuck on the algebra portion.

    What I am doing in that case is first multiplying both sides by 2 and I get

    2zdx+6dx+(2z-1)(dz-dx)=0
    2zdx+6dx+(2zdz-2zdx-dz+dx)=0

    7dx+2zdz-dz=0

    In my mind I want to make it 7dx+zdz=0

    But the book factors out the dz to make it 7dx+(2z-1)dz=0

    Why do they go that route?
     
  5. Sep 16, 2012 #4

    vela

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    To put it bluntly: because what you want to do in your mind is wrong. :smile:

    When you combine the dz terms, what you are doing is factoring. You pull out the common factor of dz to get (2z-1) dz. You can't do anything to simplify the expression from here.

    If you had, instead, 2z dz - z dz, you'd factor dz out to get (2z-z)dz. This time, you can simplify what's in the parentheses to end up with z dz. Or even better, you could pull out the common factor of z dz to get 2z dz - z dz = (2-1) z dz = z dz.
     
  6. Sep 16, 2012 #5
    Awesome thanks man! (: I love it when math makes sense.
     
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