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Simple ODE

  1. Apr 5, 2006 #1
    Can anyone walk me through solving this?

    [tex]\frac{dy}{dx} = \frac{2y}{x}[/tex]
    Seperate veriables...
    [tex]\frac{dy}{2y} = \frac{dx}{x}[/tex]
    ...and integrate from here, but the answer is [itex]y = cx^2[/itex] and I don't see how they got that.
     
    Last edited: Apr 5, 2006
  2. jcsd
  3. Apr 5, 2006 #2

    TD

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    Homework Helper

    Show us how you integrated, it shouldn't be so hard :smile:
     
  4. Apr 5, 2006 #3
    Ya, I know :uhh:

    Can't I say that it's equal to
    [tex]\int \frac{1}{2}y^{-1}dx =\int \frac{1}{x} dx[/tex]
    ...isn't the integral of [itex]\frac{1}{x} = ln (|x|)[/itex]?
     
  5. Apr 5, 2006 #4
    Yes, that looks right so far.
     
  6. Apr 5, 2006 #5
    the left side of your equation should be inegrated with respect to y not x
     
  7. Apr 6, 2006 #6
    Sorry, that was a typo.

    So I should get 1/2 ln(|y|) + C' = ln(|x|) + C after I integrate? Hows that equal to y =cx^2? :\
     
  8. Apr 6, 2006 #7
    combine C' and C into one constant, D, and refer to the properties of natural log
     
  9. Apr 6, 2006 #8
    C - C' = D

    [tex]\ln y = \ln x^2 + D[/tex]

    and I can just remove ln, even with D there?
     
    Last edited: Apr 6, 2006
  10. Apr 6, 2006 #9
    no, use the properties of ln to simplify your equation.

    hint:
    [tex]\ln a - \ln b = \ \ ?? [/tex]

    [tex]e^{\ln a} = \ \ ??[/tex]

    [tex]b \ln a = \ \ ??[/tex]
     
  11. Apr 6, 2006 #10
    So,

    [tex]\ln \frac{y}{x^2} = D[/tex]
    [tex]e^{\ln y/x^2} = e^D[/tex]

    But then wouldn't I end up with

    [tex]y = e^Dx^2[/tex]
     
  12. Apr 6, 2006 #11

    TD

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    Yes, but since e is a constant and D is too, e^d is just another constant! Name it c and you have what you want :smile:
     
  13. Apr 6, 2006 #12
    Oh, well that makes sense. Thanks everyone. :smile:
     
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