-2.4.2 interval of initial value problem

In summary, the given initial value problem for $t(t-4)y'+y=0$, with $y(2)=2$, is certain to exist for the interval $0<t<4$, as both the boundary conditions and the differential equation indicate that the solution cannot be extended beyond this interval.
  • #1
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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  • #2
Well, the problem specifically says "0< t< 4" and y' does not exist at t= 0 and t= 4.
 
  • #3
how does y(2)=2 fit into this
doesn't that give us specific y interval
 
  • #4
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
 
  • #5
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".
 

Related to -2.4.2 interval of initial value problem

1. What is an initial value problem?

An initial value problem is a type of differential equation that involves finding a function that satisfies a given set of conditions. These conditions typically include an initial value, such as the value of the function at a certain point, and a differential equation that describes the behavior of the function.

2. What does the "-2.4.2 interval" refer to in an initial value problem?

The "-2.4.2 interval" refers to the specific range of values for the independent variable (usually denoted as t) in which the initial value problem is being solved. In this case, it means that the initial value problem is being solved over the interval from -2 to 4, with a step size of 0.2.

3. How is the solution to an initial value problem typically represented?

The solution to an initial value problem is typically represented as a function, with the independent variable (usually denoted as t) on the horizontal axis and the dependent variable (usually denoted as y) on the vertical axis. The function can be graphed to show the behavior of the dependent variable over the given interval.

4. What is the significance of the initial value in an initial value problem?

The initial value is the starting point for the solution to an initial value problem. It represents the value of the dependent variable (usually denoted as y) at a specific point in the given interval. This value is used in conjunction with the differential equation to determine the behavior of the dependent variable over the entire interval.

5. How is the "-2.4.2 interval" chosen for an initial value problem?

The "-2.4.2 interval" is typically chosen based on the specific problem being solved. It may be given as part of the problem or it may be chosen by the scientist based on the desired range of values for the independent variable. The step size (in this case, 0.2) is also chosen based on the desired level of precision for the solution.

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