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Dx/dy particular solution

  1. May 15, 2014 #1

    gingermom

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    1. The problem statement, all variables and given/known data



    Find the particular solution of the differential equation that satisfies the initial condition.

    dx/dy = e^x+y x(1) = 0

    2. Relevant equations

    it does specifically state dx/dy not dy/dx


    3. The attempt at a solution

    dx/dy= e^x *e^y
    1/e^x dx=e^y dy

    integrate both sides
    -e^-x =e^y +C
    e^-x=-e^y+C
    -x=ln(-e^y +c)
    x=-ln (-e^y+c) but I can't take a ln of -e so how do I find c?

    0=-ln(-e^1+c)
    0=ln(C-e)
    1=c-e Can I do this? Since 0 raised to any power is still zero, I am not sure about whether I can get rid of the Ln this way.
    then c= 1+C
    If so the particular solution would be
    x(y)=-ln(-e^y+1+e)


    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 15, 2014 #2

    strangerep

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    This is different from your original PDE. Did you write it incorrectly?
     
  4. May 15, 2014 #3

    haruspex

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    Looks like gingermom meant e^(x+y).

    gingermom, does your solution satisfy the conditions set out in the question?
     
  5. May 15, 2014 #4

    gingermom

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    Yes it starts with e^(x+y) but is that not equal to e^x *e^y? that is what I used to separate the variables. If you mean does x(1) =0 fit the equation then yes. If you mean is the derivative of that equal to e^(x+y) - not meaning to be flip, but if I was 100 percent sure I wouldn't have asked the question.
     
  6. May 15, 2014 #5

    gingermom

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    Did you mean that I found the specific solution in terms of x(y) and the way the question is worded I need to go back and solve that for y(x)? I am not clear on what you are asking when you say does my solution satisfy the question?
     
  7. May 15, 2014 #6

    HallsofIvy

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    What you originally wrote would normally be interpreted as [tex]dx/dy= e^x+ y[/tex].

    But haruspex's question was related to "x(0)= 1".

    With "x(y)=-ln(-e^y+1+e)" x(0)= -ln(-1+ 1 + e)= -ln(e)= -1, not 1.
     
  8. May 15, 2014 #7

    gingermom

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    Oh I see what you mean - sorry for the wrong input. I multiplied both sides by -1 to get
    0=ln(e^1 +c)
    0=ln1
    ln1=ln (e^1 +c)
    1=e^1 +c
    c= 1-e^1 so I should not have multiplied by -1 because both sides are not in ln duh!
    so it would be
    0=- ln(e^1 +c)
    0=ln1
    ln1=-1 ln (e^1 +c) -r can I multiple
    1=(e^1+c)^-1
    c=e-1

    so the particular solution would be
    x(y) = -ln(-e^y+e-1)
     
  9. May 15, 2014 #8

    haruspex

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    No, the given condition is x(1) = 0, not x(0) = 1.
    But that does not satisfy x(1) = 0, whereas what you originally posted does!
    The point of my question in post #3 is that your solution in the OP does satisfy all the given conditions and is therefore the right answer.
    Your mistake in post #7 is here:
     
  10. May 15, 2014 #9

    gingermom

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    Well I guess I will know soon - since I turned in the corrected answer rather than my original post - guess that is what I guess for asking.
     
  11. May 15, 2014 #10

    gingermom

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    Sorry for my comment - it is what I get for not paying more attention to the details - everyone's help is greatly appreciated
     
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