Solve Initial-Value Problem: Find Interval x=0

  • Thread starter KillerZ
  • Start date
In summary, the given initial-value problem has a unique solution for the interval centered about x=0, (-pi/2, pi/2), as tanx is continuous everywhere except at pi/2, 3pi/2, etc. and cosx cannot equal zero.
  • #1
KillerZ
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Homework Statement



Find an interval centered about x = 0 for which the given initial-value problem has a unique solution.

[tex]y^{''} + (tanx)y = e^{x}[/tex]

[tex]y(0) = 1[/tex] [tex]y^{'}(0) = 0[/tex]

Homework Equations



[tex]a_{i}(x), i=0,1,2,3,...,n[/tex] is continuous and

[tex]a_{n} \neq 0[/tex] for every x in I.

The Attempt at a Solution



[tex]a_{0} = tanx[/tex] is zero at x = 0

I am not sure if this is correct because tanx is continuous everywhere except at pi/2, 3pi/2, etc... so would interval be:

[tex]I = (0,\infty) or (-\infty , 0)[/tex]

or

[tex]I = (0,\pi/2) or (-\pi/2 , 0)[/tex]
 
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  • #2
KillerZ said:

Homework Statement



Find an interval centered about x = 0 for which the given initial-value problem has a unique solution.

[tex]y^{''} + (tanx)y = e^{x}[/tex]

[tex]y(0) = 1[/tex] [tex]y^{'}(0) = 0[/tex]

Homework Equations

How are what you have below relevant? What does ai(x) represent?
KillerZ said:
[tex]a_{i}(x), i=0,1,2,3,...,n[/tex] is continuous and

[tex]a_{n} \neq 0[/tex] for every x in I.

The Attempt at a Solution



[tex]a_{0} = tanx[/tex] is zero at x = 0

I am not sure if this is correct because tanx is continuous everywhere except at pi/2, 3pi/2, etc... so would interval be:

[tex]I = (0,\infty) or (-\infty , 0)[/tex]

or

[tex]I = (0,\pi/2) or (-\pi/2 , 0)[/tex]

Do you have a theorem that can be used for this problem? It might be titled Existence and Uniqueness Theorem.
 
  • #3
I found the interval:

as tanx = sinx/cosx

cosx can not equal zero

so the interval is:

[tex](-\frac{\pi}{2}, \frac{\pi}{2})[/tex]
 

1. What is an initial-value problem?

An initial-value problem is a type of differential equation that involves finding the function that satisfies a given set of initial conditions. These conditions usually involve the value of the function at a specific point, such as x=0.

2. How do you solve an initial-value problem?

To solve an initial-value problem, you first need to identify the type of differential equation and any given initial conditions. Then, you can use various methods such as separation of variables, substitution, or integrating factors to find the solution. Finally, you can check your solution by plugging in the initial conditions to ensure they are satisfied.

3. What is the interval x=0 in an initial-value problem?

The interval x=0 in an initial-value problem refers to the specific point on the x-axis where the initial conditions are given. It is often used as the starting point for finding the solution to the differential equation.

4. Why is it important to find the interval x=0 in an initial-value problem?

Finding the interval x=0 is important because it allows us to determine the starting point for finding the solution to the differential equation. It also helps to narrow down the range of values that need to be considered when solving the problem.

5. Can an initial-value problem have multiple solutions?

Yes, an initial-value problem can have multiple solutions. This can happen when the initial conditions are not specific enough to uniquely determine the solution. In such cases, additional constraints or information may be needed to find the unique solution to the problem.

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