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Separation of Variables

  1. Jan 31, 2008 #1
    1. The problem statement, all variables and given/known data
    Find an implicit and explicit solution for the given initial-value problem

    [tex]\frac{dx}{dt}=4(x^2+1)[/tex] for [tex]x(\frac{\pi}{4})=1[/tex]

    [tex]\frac{dx}{dt}=4(x^2+1)[/tex]

    [tex]\Rightarrow \frac{dx}{x^2+1}=4dt[/tex]

    [tex]\Rightarrow \tan^{-1}x+C=4t[/tex]

    Now I am a little stuck. Usually I just plug in my values. I am thinking of taking the tan of both sides, eh?
     
  2. jcsd
  3. Jan 31, 2008 #2
    x = tan(4t-C) looks like a good idea to me :)
     
  4. Feb 1, 2008 #3

    HallsofIvy

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    Yes, x= tan(4t- C) is sensible. But you don't HAVE to do that to "plug in the values": your equation, as it is, says [itex]tan^{-1} + C= \pi[/itex]. What is arctan 1?
     
  5. Feb 1, 2008 #4
    Okay, I see now. But I have a new question, or rather just need some clarification.

    In Calculus, we always integrated one dependent variable wrt one independent, which is what I am doing above here, except that something new arises in these separation of variables problems.

    Since I am integrating TWO functions that are said to be equivalent, I end up with TWO arbitrary constants. Now I have been shown that they could just be moved to one side of the equation, and since one is just as arbitrary as the other, I can just write them as C (just one). I am fine with this.

    But in the above example, suppose I had chose to put +C on the RHS instead of the left.
    Now solving for this arbitrary constant I would have gotten that [itex]C=-\frac{3}{4}\pi[/itex] instead of [itex]+\frac{3}{4}\pi[/itex]

    What does this all mean?
    Sorry if it is obvious!

    Casey
     
  6. Feb 1, 2008 #5
    It just means that

    [tex]\tan^{-1}x + \frac{3\pi}{4} = 4t[/tex]

    is equivalent to

    [tex]\tan^{-1}x = 4t - \frac{3\pi}{4}[/tex]

    :smile:
     
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