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1st order diff eq.

  1. Oct 10, 2008 #1
    1. The problem statement, all variables and given/known data

    [itex]\frac{dy}{dx} = \frac{3x+2y}{3x+2y+2}[/itex]

    2. The attempt at a solution

    I thought I'd let u = 3x+2y+2, which would give me this,

    [itex]u' = 5 - \frac{4}{u}[/itex]. This is seperable, and we thus get

    [itex]\frac{1}{5}(3x + 2y +2) - \frac{4}{25}\ln|5(3x+2y+2)-4| = x + C[/itex]

    Can someone check this? I've been differentiating this beast and I keep getting different coefficients. Is what I got here correct?
     
    Last edited: Oct 10, 2008
  2. jcsd
  3. Oct 10, 2008 #2

    gabbagabbahey

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    I get

    [itex]\frac{1}{5}(3x + 2y +2) + \frac{4}{25}\ln|5(3x+2y+2)-4| = x + C[/itex]

    Perhaps you should show me your steps when integrating

    [itex]\int \frac{u}{5u-4} du[/itex]
     
  4. Oct 10, 2008 #3
    Gah, I actually missed TeXing the 5 inside the log function -- I got the same thing as you on paper. Thanks!
     
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