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Elementary Differential equations : seperable making it explicit

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data

    (y+2)dx +y(x+4)dy=0


    2. Relevant equations



    3. The attempt at a solution

    ∫(1/x+4)dx + ∫(y/y+2)dy=0

    ln(x+4)-2ln(y+2)+y=lnc

    Here is where I get confused how do I make this into an explicit solution,

    that +y really bothers me.

    I was thinking ln [ (x+4)/(y+2)^2] +y = lnc

    raising both sides to e^x

    (x+4)/(y+2)^2+e^y=c

    but the answer is

    x+4 = [(y+2)^2]*e^(-y+1)
     
  2. jcsd
  3. Sep 15, 2012 #2
    That's not correct. When you raise both the sides to e, you get this:
    [tex]e^{\ln((x+4)/(y+2)^2)+y}=c[/tex]
    which is equal to
    [tex]\frac{x+4}{(y+2)^2} \cdot e^y=c[/tex]

    And can you check the answer you posted? I get (-y+c) instead of (-y+1).
     
  4. Sep 15, 2012 #3
    They hold that the answer is -y+c. I forgot to mention that it is a boundary problem y(-3)=-1

    Would that have anything to do with it?

    Also from what you got I do not quite understand how you got -y+c

    if you multiply both sides by [(y+2)^2]/e^y

    wouldnt it be x+4= c(y+2)^2/e^y ?
     
  5. Sep 15, 2012 #4
    Yep, it would be that, but 1/e^y=e^(-y).
    c is a arbitrary constant, that is we can write it like this, e^c which is again a constant or we can even write it as ln(c) which is also equal to a constant. To get the answer you posted, i wrote c as e^c.
     
  6. Sep 15, 2012 #5
    Excellent, I appreciate your help on these problems.
     
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