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Find the particular solution of a differential equation

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the particular solution of the differential equation dy/dx = (x-4)e^(-2y)

    Satisfying initial condition y(4)=ln(4)

    2. Relevant equations

    N/A

    3. The attempt at a solution

    I separated this into dy/e^(-2y) = (x-4)dx

    I then integrated it to get e^(2y)/2 = x^2/2 - 4x

    I then tried to isolate y and got y= 1/2ln(x^2-8x+C)

    Plugging in the values, I get ln(4)=1/2ln(-16+C)

    What do I do with that 1/2 out in front? If it wasn't a part of the equation, C would equal 20. That is, if I'm correct so far.
     
    Last edited: Jan 30, 2012
  2. jcsd
  3. Jan 30, 2012 #2

    Simon Bridge

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    Put the everything inside the ln - it will make you feel happier.

    [tex]e^{2y} = x^2-8x+c \Rightarrow y=\ln \sqrt{x^2-8x+c}[/tex]... now the initial condition makes sense?
     
  4. Jan 30, 2012 #3

    Dick

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    You add the +C after you integrate. Not at the very end. So you should have gotten e^(2y)/2 = x^2/2 - 4x + C. You can actually solve for C at that step as well. Then isolate y.
     
  5. Jan 30, 2012 #4
    Yeah I realized that, but the 1/2 is still throwing me off. I'm sure it's something simple that I've forgotten.
     
  6. Jan 30, 2012 #5

    Simon Bridge

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    [tex]a\ln(x)=ln(x^a)[/tex]
    But solve for c while y is still in the exponential like Dick says also works.
     
  7. Jan 30, 2012 #6

    Dick

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    I'm not sure what's throwing you off. C isn't 20. It's something else. Solve for it!
     
  8. Jan 30, 2012 #7
    I forgot the rule that Simon posted. It makes complete sense now. I will look at it the other way as well. Thank you both!
     
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