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Square-root and differential equations

  1. Nov 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi all

    I have the following expression

    [tex]
    \frac{dy}{dt}= \sqrt{C-y(t)^2},
    [/tex]

    where C is a constant, and y is the variable, which depends on t. What I need to do is to solve this differential equation, but my problem is that this is a math-class (and not a physics-class), so I need to be very rigorous.

    Now it is almost mandatory to comment of the argument of a squareroot. What I know is that the solution y is real.

    Thus the term C-y(t) cannot be less than zero. But my question is: Does it make sense to talk about the term "C-y(t)" being less than zero, when y(t) is time-dependent?
     
    Last edited: Nov 26, 2009
  2. jcsd
  3. Nov 26, 2009 #2

    LCKurtz

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    Sure, you can insist [itex]y^2(t) \le C[/itex] or [itex]|y(t)| \le \sqrt{C}[/itex] for all t. Obviously C will have to be positive. If you just go ahead and solve it by separation of variables you will find that restriction automatically holds for your solution.
     
  4. Nov 26, 2009 #3
    Thanks. I must admit, I am not entirely sure what you mean when you say "restriction automatically holds for your solution.".

    I have a final question, which is in no way related to this, but perhaps you can answer it anyway. Do you know if the set R2 (i.e. all of R2) can be considered as an open set?
     
  5. Nov 26, 2009 #4

    LCKurtz

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    Just go ahead and solve the DE. The y(t) that you get will work under the square root sign with no problems about a negative under the square root.
    Yes. R2 is open. And closed.
     
  6. Nov 26, 2009 #5
    Thanks! That was very kind of you.
     
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