Find the particular solution of a differential equation

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Homework Statement



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

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

Homework Equations



N/A

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.
 
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Answers and Replies

  • #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?
 
  • #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.
 
  • #4
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Yeah I realized that, but the 1/2 is still throwing me off. I'm sure it's something simple that I've forgotten.
 
  • #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.
 
  • #6
Dick
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Yeah I realized that, but the 1/2 is still throwing me off. I'm sure it's something simple that I've forgotten.

I'm not sure what's throwing you off. C isn't 20. It's something else. Solve for it!
 
  • #7
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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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