-2.4.2 interval of initial value problem

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Discussion Overview

The discussion revolves around determining an interval in which the solution of a given initial value problem exists. The problem involves a differential equation and a specific initial condition, with a focus on the implications of the interval specified in the problem statement.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant isolates y' to express it as $y'=-\dfrac{y}{t(t-4)}$, indicating a concern about the interval of existence.
  • Another participant notes that the problem specifies "0< t< 4", highlighting that y' does not exist at t=0 and t=4, which may affect the solution interval.
  • There is a question about how the initial condition y(2)=2 fits into determining the interval, suggesting it may provide specific constraints on y.
  • A participant suggests a "brute force" approach to solve the differential equation to find y(t) and deduce the solution intervals.
  • Another participant emphasizes that even without the specified interval, the points t=0 and t=4 lead to a non-existence of y', indicating that solutions cannot extend across these points, but solutions could exist outside the specified interval.
  • It is reiterated that the initial condition y(2)=2, which lies within the interval 0< t< 4, confirms that the correct interval for the solution is indeed 0< t< 4.

Areas of Agreement / Disagreement

Participants generally agree that the interval for the solution is 0< t< 4 based on the initial condition and the behavior of y' at the endpoints. However, there are discussions about the implications of the initial condition and the nature of the solution outside the specified interval, indicating some unresolved nuances.

Contextual Notes

There are limitations regarding the assumptions about the behavior of the solution at the endpoints t=0 and t=4, as well as the implications of the initial condition on the solution's existence.

karush
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Determine an interval in which the solution of the given initial value problem is certain to exist

$t(t-4)y'+y=0 \quad y(2)=2\quad 0<t<4$

ok my first step was isolate y'
s
$y'=-\dfrac{y}{t(t-4)}$

not sure what direction to go since we are concerned about an interval
 
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Well, the problem specifically says "0< t< 4" and y' does not exist at t= 0 and t= 4.
 
how does y(2)=2 fit into this
doesn't that give us specific y interval
 
karush said:
how does y(2)=2 fit into this
doesn't that give us specific y interval
You could always do "brute force" if you can't figure out a work around. Boundary conditions get rid of integration constants. Solve the differential equation. What is y(t)? What does that tell you about the solution interval(s)?

-Dan
 
IF the "0< t< 4" were not there you would still be able to observe that if x= 0 or x= 4, the denominator would be 0 so y' would not exist. That would tell you that a solution cannot be extended across x= 0 or x= 4. But you could have solutions for x< 0 or x> 4. The fact that you are given y at x= 2, which is between 0 and 2 tells you that the correct interval is 0< x< 4, not x< 0 or x> 4. But, as I said, that is redundant since the problem itself says "0< x< 4".
 

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