Question

For each initial value problem, determine the largest \(t\)-interval on which Theorem \(3.1\) guarantees the existence of a unique solution. $$e^{t} y^{\prime \prime}+\frac{1}{t^{2}-1} y=\frac{4}{t}, \quad y(-2)=1, \quad y^{\prime}(-2)=2$$

Short answer

Answer: The largest t-interval on which the theorem guarantees the existence of a unique solution for the given initial value problem is \(-1 < t < 1\).

Step-by-step solution

Identify the differential equation and initial values

The given differential equation is: $$e^{t} y^{\prime \prime}+\frac{1}{t^{2}-1} y=\frac{4}{t},$$ and the initial values are: $$y(-2)=1, y^{\prime}(-2)=2$$

Determine the values of t for which the theorem is valid

To find the largest t-interval for which the theorem guarantees the existence of a unique solution, we need to find the domain of the differential equation, avoiding any singularities or discontinuities. Here, observe that the differential equation will have singularities when the denominators are equal to zero: - The first term \(e^{t} y^{\prime \prime}\) is continuous on all of \(\mathbb{R}\) - The second term involves the denominator \(t^2-1\), which is zero when \(t=\pm 1\) - The third term involves the denominator \(t\), which is zero when \(t=0\) Therefore, the equation has singularities at three points: \(t=1, -1, 0\) Since the initial condition is specified at \(t=-2\), and it must be inside the t-interval, the largest t-interval will be between the singularities to the left and right of \(t=-2\). Hence, the largest interval is between \(t=-1\) and \(t=1\), as these are the two closest singularities. Thus, the largest t-interval on which the theorem guarantees the existence of a unique solution for the given initial value problem is \(-1 < t < 1\).

Key concepts

Initial Value Problem

An initial value problem (IVP) in mathematics is a problem that involves finding a function that solves a differential equation and meets specific conditions at a given point. These conditions are known as "initial conditions," as they are generally specified at the beginning of the interval in which we are interested. In the given exercise, the initial value problem consists of the differential equation \(e^{t} y^{\prime \prime}+\frac{1}{t^{2}-1} y=\frac{4}{t}\) with initial conditions \(y(-2)=1\) and \(y^{\prime}(-2)=2\). You can think of this as finding a path that a function must follow, starting precisely from a given state. Initial conditions help anchor the solution, ensuring it behaves correctly from the point it begins.

Ordinary Differential Equations

Ordinary differential equations (ODEs) involve derivatives of functions concerning one independent variable. These are fundamental to describing various physical systems, from simple mechanical systems to complex biological processes. An ODE expresses the relationship between an unknown function and its derivatives. For instance, in our exercise, the equation \(e^{t} y^{\prime \prime}+\frac{1}{t^{2}-1} y=\frac{4}{t}\) links the unknown function \(y(t)\) and its second derivative \(y^{\prime \prime}(t)\). Solving an ODE typically involves finding a function \(y(t)\) that satisfies the given equation. The existence and uniqueness theorem play a crucial role here, as it assures us under certain conditions, such a solution not only exists but is unique. This is essential, as it means there isn't any ambiguity or multiple solutions satisfying the problem.

Singularities and Discontinuities

Singularities and discontinuities occur at points where a function or its derivative is undefined or behaves erratically, such as going towards infinity. In the context of differential equations, singularities can significantly affect the solutions. They indicate points where the assumptions of the existence and uniqueness theorem break down.In the given equation, singularities were identified by evaluating the denominators in each term. The singularities are found at points where these denominators become zero. For the equation \(e^{t} y^{\prime \prime}+\frac{1}{t^{2}-1} y=\frac{4}{t}\), these points are at \(t = 0, 1,\) and \(-1\). These are the values of \(t\) that must be avoided while finding the solution.Luckily, the initial condition of this problem is at \(t = -2\), which doesn't overlap with any of these singular points. Thus, the solution can be confidently found within the largest safe interval spanning between \(-1 < t < 1\), ensuring no interruptions in the behavior of the function within this interval.