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Homework Help: Differential equations, qualitative solution

  1. Mar 9, 2010 #1
    dy/dt = -ty^2 y(0) = 1
    What can you say about the solution over the interval 0<=t <=2?

    The text says that the solution is decreasing and never zero because y(t) = 0 for all t is an equilibrium solution.

    I can see why the solution can never cross the x-axis, but why can't the solution become zero?
     
  2. jcsd
  3. Mar 9, 2010 #2

    rock.freak667

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    What did you get for the solution y(t)?
     
  4. Mar 9, 2010 #3
    I got y = [tex]\frac{1}{t^2 + 1}[/tex], but I got the impression that looking at the derivative was enough to make the conclusion
     
  5. Mar 14, 2010 #4
    Anyone?
     
  6. Mar 14, 2010 #5

    rock.freak667

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    if it crosses the t-axis, then shouldn't y=0 ? and what does that mean?
     
  7. Mar 14, 2010 #6
    Yes, but why can't it become zero and stay there?
     
  8. Mar 14, 2010 #7

    rock.freak667

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    How exactly would it become zero? That would mean that at some value of 't' the corresponding value of 'y' would be zero.

    [tex]0 = \frac{1}{t^2+1}[/tex]


    You would end up with a false equality, which can only mean that y(t) is never zero for all values of t.
     
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