Question

For each initial value problem, determine the largest \(t\)-interval on which Theorem \(3.1\) guarantees the existence of a unique solution. $$y^{\prime \prime}+3 t^{2} y^{\prime}+2 y=\sin t, \quad y(1)=1, \quad y^{\prime}(1)=-1$$

Short answer

Answer: The largest t-interval is \((-\infty, \infty)\).

Step-by-step solution

Identify the coefficients and the forcing function

We are given the following second-order linear differential equation: \(y'' + 3t^2y' + 2y = \sin(t)\) The coefficients in this case are: \(a(t) = 3t^2\) \(b(t) = 2\) The forcing function is represented by: \(f(t) = \sin(t)\)

Check for continuity

To apply Theorem 3.1, we need to check if the coefficients and the forcing function are continuous on an open interval containing the initial condition point \((t, y) = (1, 1)\). The coefficient functions, \(a(t) = 3t^2\) and \(b(t) = 2\), are polynomial functions, and therefore, continuous on \(\mathbb{R}\), the entire real line. The forcing function \(f(t) = \sin(t)\) is a trigonometric function, which is also continuous on the entire real line. Since the coefficients and the forcing function are continuous on the entire real line, we can conclude that there is guaranteed to be a unique solution on an open interval containing the given initial condition point.

Determine the t-interval for the unique solution

As the coefficient functions and the forcing function are continuous on the entire real line, the largest t-interval on which Theorem 3.1 guarantees the existence of a unique solution is \((-\infty, \infty)\). Based on the analysis above and our knowledge of Theorem 3.1, we can conclude that there is a unique solution to the given initial value problem on the largest t-interval, which is \((-\infty, \infty)\).

Key concepts

Existence and Uniqueness Theorem

Understanding the existence and uniqueness theorem is essential when solving initial value problems involving differential equations. This theorem, often referred to as the Picard-Lindelöf theorem, provides the conditions under which a differential equation has a unique solution that passes through a given point in the plane.

The theorem states that if the function representing the rate of change and its partial derivative with respect to the unknown function are continuous within some interval around the initial condition, then there exists a unique solution to the initial value problem within this interval. In the case of the exercise provided, since the coefficients of the differential equation and the forcing function are continuous on the entire real line, the theorem assures us of the existence and uniqueness of the solution for all real numbers, constituting the interval \[(-\infty, \infty)\].

It's important to note that the applicability of this theorem hinges on the continuity of the differential equation's coefficients and the forcing functions, which serves as a prerequisite to our assurance of a solution's existence and uniqueness.

Second-order Linear Differential Equation

A second-order linear differential equation has the general form \(y'' + p(t)y' + q(t)y = g(t)\), where the coefficients \(p(t)\) and \(q(t)\) are functions of the independent variable \(t\), and \(g(t)\) is the forcing function. The solution to this kind of equation can describe a wide range of phenomena from mechanical vibrations to electrical circuits.

In our exercise, we are presented with the equation \(y'' + 3t^2y' + 2y = \sin(t)\), which is a clear example of a second-order differential equation. Here, \(p(t) = 3t^2\) and \(q(t) = 2\) are polynomial functions, which means they are particularly well-behaved – namely, continuous and differentiable everywhere, which simplifies the process of verifying the conditions needed for the existence and uniqueness theorem to apply.

Continuity of Coefficient Functions

In the realm of differential equations, the continuity of coefficient functions is a cornerstone for ensuring the applicability of many theorems, including the existence and uniqueness theorem. Coefficient functions are simply the functions that multiply the various derivatives of the unknown function in the differential equation.

In the exercise, the coefficient functions are given as polynomials: \(3t^2\) and a constant \(2\). Polynomials are continuous everywhere, which means these particular coefficient functions do not pose any restrictions on the interval where a solution might exist. Their continuity implies that no matter how \(t\) changes, the outcome of these coefficient functions will vary smoothly without any abrupt jumps or breaks, providing a stable structure to the behavior of the differential equation's solutions.

Trigonometric Forcing Function

The concept of a forcing function in differential equations refers to an external influence that drives the system represented by the equation. When the forcing function is a trigonometric function, as in \(f(t) = \sin(t)\) from our exercise, it often models phenomena that have periodic behavior, such as oscillations.

Trigonometric functions like \(\sin(t)\) and \(\cos(t)\) are continuous and periodic, making them ideal candidates for modeling systems subject to repetitive external forces. In the context of the exercise, the \(\sin(t)\) term on the right side of the equation introduces a sinusoidal oscillation as an input to the system. Due to the properties of trigonometric functions, this does not affect the interval of existence for solutions to the differential equation; specifically, \(\sin(t)\) is continuous for all \(t\), which reinforces the conclusion that a unique solution exists for the entirety of the real number line.