Question

For each initial value problem, determine the largest \(t\)-interval on which Theorem \(3.1\) guarantees the existence of a unique solution. $$t y^{\prime \prime}+\frac{\sin 2 t}{t^{2}-9} y^{\prime}+2 y=0, \quad y(1)=0, \quad y^{\prime}(1)=1$$

Short answer

The given initial value problem contains a second-order linear ordinary differential equation with non-constant coefficients. After analyzing the coefficients and their continuous domains, we found that the largest t-interval in which Theorem 3.1 guarantees the existence of a unique solution to the given initial value problem is \(t \in (-3, 3)\).

Step-by-step solution

Identify the coefficients of the ODE

The given second-order linear ODE is $$t y^{\prime \prime}+\frac{\sin 2 t}{t^{2}-9} y^{\prime}+2 y=0.$$ The coefficients are: 1. \(p(t) = t\) 2. \(q(t) = \frac{\sin 2t}{t^2 - 9}\) 3. \(r(t) = 2\)

Identify the continuous domains of the coefficients

The coefficient \(p(t) = t\) is continuous for all real values of \(t\). The coefficient \(r(t) = 2\) is also continuous for all real values of \(t\). The coefficient \(q(t) = \frac{\sin 2t}{t^2 - 9}\) needs careful consideration. The denominator \(t^2 - 9\) introduces potential discontinuities at \(t = -3, 3\). We need to find the largest open interval that contains \(t = 1\) and doesn't include the discontinuities.

Determine the t-interval for Theorem 3.1 to hold

As we determined in the last step, \(p(t)\) and \(r(t)\) are continuous for all real values of \(t\). The coefficient \(q(t)\) is continuous for all real values except \(t = -3, 3\). Therefore, we need to find the largest open interval containing the initial point \(t = 1\) that doesn't include the discontinuities. This interval is given by \((-3, 3)\). Thus, the largest \(t\)-interval on which Theorem 3.1 guarantees the existence of a unique solution to the given initial value problem is \(t \in (-3, 3)\).

Key concepts

Existence of a Unique Solution

Understanding the existence of a unique solution for initial value problems is crucial in differential equations. The theory behind this concept ensures that, under certain conditions, given a differential equation and an initial condition, there will be exactly one function that satisfies both.

In the context of our exercise, Theorem 3.1 refers to what's often known as the Existence and Uniqueness Theorem. This theorem states that if the functions defining the differential equation are continuous on an open interval containing the initial value point, then there exists a unique solution that goes through that point. In simpler terms, if the mathematical 'terrain' is smooth at and around the starting position, you can be confident that there will be one and only one path that starts there and follows the rules set out by the ODE.

In the given problem, we need to ensure that the coefficients of the ODE are continuous within the interval containing the initial condition, which is at \( t = 1 \). Any discontinuities would prevent us from assuring the existence of a unique solution, thus the identification and avoidance of these points are essential.

Second-Order Linear ODE

A second-order linear ODE is an equation that involves the unknown function \( y(t) \), its first derivative \( y'(t) \), and its second derivative \( y''(t) \), and it can be generally expressed in the form \( a(t) y'' + b(t) y' + c(t) y = g(t) \), where \( a(t) \), \( b(t) \), and \( c(t) \) are the coefficients and \( g(t) \) is a known function.

These types of ODEs are called 'linear' because the unknown function and its derivatives only appear to the first power and are not multiplied together nor appear in a function (like a sine or exponential function). The linearity has a profound consequence on the solutions: they can be added together or multiplied by constants to produce new solutions, a property known as superposition.

In the given exercise, the equation given is in the standard format without a \( g(t) \) term, implying that it is homogeneous. Identifying the type of ODE is the first step since it guides us on what theorems or methods to apply when seeking a solution.

Continuous Coefficients

Our equation's continuous coefficients are functions that do not have any jumps, breaks, or points of discontinuity within an interval of interest. In differential equations, continuous coefficients are important conditions for many theorems that guarantee the existence and uniqueness of solutions.

In our exercise, we have the coefficients \( p(t) = t \) and \( r(t) = 2 \), which are clearly continuous everywhere because they either represent a straight line (in the case of \( r(t) \)) or the identity function (in the case of \( p(t) \)). However, the coefficient \( q(t) = \frac{\sin 2t}{t^2 - 9} \) requires careful attention since it becomes discontinuous where its denominator is zero, specifically at \( t = 3 \) and \( t = -3 \). The interval around our initial condition \( t = 1 \) must avoid these points of discontinuity.

Therefore, as we apply Theorem 3.1, we conclude that our largest interval for guaranteeing a unique solution that includes the initial condition but excludes the discontinuity points is \( (-3, 3) \). Continuous coefficients are the foundation that supports the entire structure of existence and uniqueness theorems, making them an essential concept in studying differential equations.