Question

In each of Problems I through 6 determine (without solving the problem) an interval in which the solution of the given initial value problem is certain to exist. $$ y^{\prime}+(\tan t) y=\sin t, \quad y(\pi)=0 $$

Short answer

Question: Determine an interval in which the solution of the following initial value problem is certain to exist without actually solving the problem: \(y' + (\tan t)y = \sin t, y(\pi) = 0\). Answer: The solution of the given initial value problem is certain to exist on the interval \((\frac{\pi}{2}, \frac{3\pi}{2})\).

Step-by-step solution

Identify the given ODE and initial condition

We are given the first-order linear ODE: $$ y' + (\tan t)y = \sin t $$ and the initial condition: $$ y(\pi) = 0 $$

Analyze the continuity of the functions p(t) and q(t)

We have the functions \(p(t) = \tan t\) and \(q(t) = \sin t\). Both of these functions have points of discontinuity. In this case, \(\tan t\) is discontinuous at its asymptotes, which occur at \((2n + 1)\frac{\pi}{2}\) for all integers \(n\). \(\sin t\) is continuous for all real t.

Determine the open interval containing the initial value t = π

We need to find an open interval containing the initial value \(t = \pi\) in which both \(p(t)\) and \(q(t)\) are continuous. The discontinuities of the function \(p(t) = \tan t\) are of the form \((2n + 1)\frac{\pi}{2}\). Specifically, to the left of \(t = \pi\), the closest asymptote is \(t = \frac{3\pi}{2}\), and to the right, the closest asymptote is \(t = \frac{\pi}{2}\). Therefore, the open interval \((\frac{\pi}{2}, \frac{3\pi}{2})\) contains the initial value \(t = \pi\) and is a region where both \(p(t)\) and \(q(t)\) are continuous.

Conclusion

We have found an interval, \((\frac{\pi}{2}, \frac{3\pi}{2})\), in which the solution of the given initial value problem is certain to exist. This is because the functions \(p(t) = \tan t\) and \(q(t) = \sin t\) are continuous on this interval, which contains the initial value of \(t = \pi\). The Existence and Uniqueness Theorem guarantees a unique solution on this interval.

Key concepts

Initial Value Problem

An initial value problem consists of a differential equation accompanied by a specific value, known as the initial condition. This condition typically is in the form of a point on the solution curve that the resulting function must satisfy.
For example, in our exercise, the differential equation is given by: - \( y' + (\tan t)y = \sin t \) and the initial condition by: - \( y(\pi) = 0 \)
In essence, solving an initial value problem involves finding the function \( y(t) \) that not only satisfies the differential equation but also passes through the given point \((\pi, 0)\).
Initial value problems are crucial in many scientific and engineering fields as they often describe systems at a given starting point and help predict future behaviors.

First-order Linear ODE

A first-order linear ordinary differential equation (ODE) is a type of equation used to model various real-world scenarios, from physics to finance. It is called "first-order" because it involves only the first derivative of the unknown function. An ODE of this kind typically takes the form:
\[ y' + p(t)y = q(t) \]
In our case,
- \( p(t) = \tan(t) \) - \( q(t) = \sin(t) \)
The task is to find the function \( y(t) \) that satisfies this equation. First-order linear ODEs have various methods of solutions, such as the integrating factor method or matrix exponentials for systems, making them highly versatile in application.
The presence of functions \( p(t) \) and \( q(t) \) determines the path of solution and the techniques used. This type of ODE is foundational in understanding dynamic systems and processes.

Continuity of Functions

Continuity is a fundamental concept when discussing differential equations, particularly due to the Existence and Uniqueness Theorem. For the theorem to apply, specific functions within the equation must be continuous over a certain interval.
In our exercise, we have: - \( p(t) = \tan t \) - \( q(t) = \sin t \)
The function \( \sin t \) is continuous everywhere on the real line. However, \( \tan t \) is discontinuous at points \( (2n + 1)\frac{\pi}{2} \), where \( n \) is any integer.
This discontinuity caused by \( \tan t \) impacts where we can assure the existence of a unique solution. Thus, we focus on finding intervals where both functions are continuous. Understanding continuity helps in setting the stage for solving differential equations and is essential for ensuring predictions or models are reliable over specified domains.

Interval of Existence

The interval of existence refers to the region on the real number line where a unique solution to a differential equation is guaranteed by the Existence and Uniqueness Theorem.
In our particular problem, to find such an interval, both functions \( p(t) = \tan t \) and \( q(t) = \sin t \) must be continuous. We've identified that \( \tan t \) has discontinuities at every \( (2n + 1)\frac{\pi}{2} \).
Given the initial value \( t = \pi \), we determine the closest points of discontinuity are \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \). Thus, an open interval where both functions are continuous, and that includes \( t = \pi \), is \( (\frac{\pi}{2}, \frac{3\pi}{2}) \).
In this interval, a unique solution is assured by the theorem, indicating where the system precisely behaves according to our constraints. This step is critical in ensuring proper predictive capabilities in modeling scenarios via differential equations.