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Differential Equation

  1. Jan 11, 2009 #1
    I'm trying to solve this firrst order diff. equation, where I'm given the initial value, x(0)=2


    [tex]\frac{dx}{dt}=\frac{3x+4}{\sqrt{t}}[/tex]

    [tex]\frac{dx}{3x+4}=\frac{1}{\sqrt{t}}dt[/tex]

    [tex]\int\frac{1}{3x+4}dx=\int\frac{1}{\sqrt{t}}dt[/tex]

    [tex]ln(3x+4)=ln(t^{\frac{1}{2}})[/tex]

    this is as far as I got, do I sub in x(0)=2 into the LHS? if not, could i have some pointers to help carry on?

    cheers.
     
    Last edited: Jan 11, 2009
  2. jcsd
  3. Jan 11, 2009 #2

    Hootenanny

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    Your solution looks good up until the final line. You may want to re-check your integrals.
     
  4. Jan 11, 2009 #3
    [edit] - there should be 'dt' s on the RHS in the 2nd and 3rd line of work
     
  5. Jan 11, 2009 #4

    HallsofIvy

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    [tex]\int \frac{dt}{\sqrt{t}}= \int t^{-1/2}dt[/tex]
    is NOT [itex]ln(t^{1/2})[/itex]!
     
  6. Jan 11, 2009 #5
    also dont forget your constant that comes from the integration
     
  7. Jan 12, 2009 #6
    oh right, my mistake lol

    so, the last line should be

    [tex]ln(3x+4)=\frac{1}{2}ln(t^{\frac{1}{2}})[/tex]

    I have no idea where to go from here with the initial value that I was given x(0)=2! would I sub in x=2 into the equation?
     
  8. Jan 12, 2009 #7

    Dick

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    As other people have been trying to tell you, the integral of t^(-1/2) DOES NOT involve a log. It's just power law. And I still don't see a constant of integration.
     
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