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Separable Differential Equation (what's wrong?)

  1. Oct 7, 2009 #1
    So, here's the problem:

    [tex]\frac{dy}{dx} = \frac{8y}{5x}[/tex]

    To start off, I separate the integrals, which gives me:

    [tex]\frac{dy}{8y} = \frac{dy}{5x}[/tex]

    After that, I integrate both sides, which gives me:

    [tex]\frac{ln8y}{8} = \frac{ln5x}{5} + c [/tex]

    Now, the question says that it runs through (4, 1), so that is saying that y(4) = 1, correct?

    To solve for c, I just plug the 4 in where the Xs are and the 1 is where the Ys are?

    EDIT: Sorry, posted in the wrong forum! Please move this thread for me :\
     
  2. jcsd
  3. Oct 7, 2009 #2
    That's the correct approach.
     
  4. Oct 8, 2009 #3

    HallsofIvy

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    Notice by the way that
    [tex]\frac{ln(8y)}{8}= \frac{ln(5x)}{5}+ c[/tex]
    is the same as
    [tex]\frac{ln(y)+ ln(8)}{8}= \frac{ln(x)+ ln(5)}{5}+ c[/tex]
    [tex]\frac{ln(y)}{8}+ 1= \frac{ln(x)}{5}+ 1+ c[/tex]
    [tex]\frac{ln(y)}{8}= \frac{ln(x)}{5}+ c[/tex]

    And that is the same as
    [tex]ln(y)= \frac{8}{5} ln(x)+ 8c= ln(x^{\frac{8}{5}})+ 8c[/itex]
    so that
    [tex]y= e^{8c} x^{\frac{8}{4}}[/tex]
    and that can be written simply
    [tex]y= Cx^{\frac{8}{5}[/tex]
    where [itex]C= e^{8c}[/itex]
     
    Last edited: Oct 9, 2009
  5. Oct 8, 2009 #4
    Oh, okay! How did the 8c become only c in:

    [tex]\frac{8}{5} ln(x)+ 8c= ln(x^{\frac{8}{5}})+ c[/tex]

    Thanks for the help.
     
  6. Oct 8, 2009 #5
    8*c = c because 8, a constant, times a constant c, is still a constant.
     
  7. Oct 9, 2009 #6

    HallsofIvy

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    I have editted my post to make that clearer.
     
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