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Diff. equation help

  1. Oct 29, 2007 #1
    1. The problem statement, all variables and given/known data
    (3x-2y+1)dx+(3x-2y+3)dy=0

    this a piece of my solution: (pls correct if i'm wrong)

    I thought of solving it using miscellaneous substitution using (3x-2y+1) as u...
    du= 3dx- 2dy ,dx=(du+2dy)/3 so,

    >u[(du+2dy)/3) + udy +2dy =0

    >u(du +2dy) + 3udy + 6dy =0

    >udu +(5u+6)dy=0

    >[u/ (5u+6)] du + dy =0
    integrating it:

    > (5u+6)/25 + 6/25 [ln (5u+6)] + y = 0

    >(5u+6) + 6[ln(5u+6) + 25y =0

    >(15x-10y+11) + 6[ln(15x-10y+11)] +25y =0


    but this answer is way too different from the answer on the book the answer there is:
    5(x+y+c) = 2ln[15x-10y+11] ... what could be my mistake?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 29, 2007 #2

    mjsd

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    Homework Helper

    you ans is same as the book's....simplify it!
    note you are basically doing the reverse of this
    f(x,y)=K where K is some constant then
    [tex]\frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} =\frac{df}{dt}=\frac{d(K)}{dt}=0[/tex]
    where K is 5c in your book I think
     
  4. Oct 29, 2007 #3
    thanks.. but uhm.. sorry I do not understand it well.. could you please explain further how can i simplify so that i could come out to the same answer as the book.. please.
     
  5. Oct 29, 2007 #4
    Okay, you made a mistake in integration:
    (u/ (5u+6))du + dy = 0
    ((1 - (6/5u+6))/5) du +dy = 0
    On integrating:
    15x-10y+11 - 6ln(15x-10y+11) +25y = k
    15x+15y+11+k = 6ln(15x-10y+11)
    15(x+y+c) = 6ln(15x-10y+11)
    5(x+y+c) = 2ln(15x-10y+11)

    Done!
     
  6. Oct 29, 2007 #5
    a.. ok.. wow, never noticed that.. thanks a lot!
     
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