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Differential equations, qualitative solution

  • Thread starter MaxManus
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  • #1
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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?
 

Answers and Replies

  • #2
rock.freak667
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What did you get for the solution y(t)?
 
  • #3
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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
 
  • #4
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Anyone?
 
  • #5
rock.freak667
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Anyone?
if it crosses the t-axis, then shouldn't y=0 ? and what does that mean?
 
  • #6
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if it crosses the t-axis, then shouldn't y=0 ? and what does that mean?
Yes, but why can't it become zero and stay there?
 
  • #7
rock.freak667
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Yes, but why can't it become zero and stay there?
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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