Problem 10 - A First Course in Differential Equations with Modeling Applications

Question

Find an interval centered about \(x=0\) for which the given initial-value problem has a unique solution. $$y^{\prime \prime}+(\tan x) y=e^{x}, \quad y(0)=1, \quad y^{\prime}(0)=0$$

Step-by-Step Solution

Answer

The interval is \((-\frac{\pi}{2}, \frac{\pi}{2})\).

Step 1: Identify the form of the differential equation

The given differential equation is \(y^{\prime \prime} + (\tan x) y = e^{x}\). This is a second-order linear differential equation.

Step 2: Assess the coefficients of the equation

The coefficient of \(y\) is \(\tan x\). The function \(\tan x\) is known to have vertical asymptotes (discontinuities) where \(x = \frac{\pi}{2} + n\pi\), with \(n\) being any integer. This impacts where any solution might be valid.

Step 3: Determine the point of interest

The initial conditions are provided at \(x = 0\). Thus, we need to find an interval centered around \(x = 0\) where there are no discontinuities in the coefficients.

Step 4: Define the continuity interval for \(\tan x\)

The function \(\tan x\) is continuous in the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\), because it has a vertical asymptote at \(x = \frac{\pi}{2}\) and \(x = -\frac{\pi}{2}\). This means \(y'' + (\tan x) y = e^x\) is defined and continuous in this interval.

Step 5: State the final interval of interest

We need to find an interval centered around \(x = 0\). Thus, a valid interval for the initial value problem that ensures uniqueness and existence of the solution is \((-\frac{\pi}{2}, \frac{\pi}{2})\).

Key Concepts

Initial Value Problem

An initial value problem involves solving a differential equation with some given conditions at a specific point. These conditions are called the initial conditions. In the given problem, we have the differential equation:

• \( y'' + ( an x) y = e^x \)

Along with the initial conditions:

• \( y(0) = 1 \)
• \( y'(0) = 0 \)

The purpose of the initial conditions is to find a unique solution that satisfies both the differential equation and these specified values at a given point, which, in this case, is \( x = 0 \). The initial value problem requires us to look closely not just at the equation itself, but also at the behavior and continuity of its coefficients.

Second-Order Linear Differential Equation

Second-order linear differential equations are a type of differential equation that involves the unknown function with its second derivatives. The general form of a second-order linear differential equation is:

• \( a(x)y'' + b(x)y' + c(x)y = f(x) \)

In our problem, the equation is:

• \( y'' + ( an x) y = e^x \)

Here, we can see that:

• \( a(x) = 1 \)
• \( b(x) = 0 \)
• \( c(x) = an x \)
• \( f(x) = e^x \)

One of the key characteristics of these equations is their linearity, which means the function and its derivatives appear with a power of one. Analyzing the coefficients and external term helps in understanding the behavior of the solution.

Continuity of Functions

Continuity plays a crucial role in solving differential equations. In our problem, the function \( an x \) acts as one of the coefficients in the equation. The continuity of \( an x \) is essential to determine where a solution might be valid.

\( an x \) has vertical asymptotes at \( x = \frac{\pi}{2} + n\pi \), where \( n \) is any integer. This means it is not defined at these points. Thus, it is continuous in the interval:

• \((-\frac{\pi}{2}, \frac{\pi}{2})\)

This continuity interval is important as it signifies where the equation remains well-defined and continuous. To ensure the validity and smoothness of solutions, we cannot have discontinuities in our coefficients within the targeted interval centered around \( x = 0 \).

Unique Solution Existence

The existence of a unique solution for a differential equation rests upon a combination of factors. The initial conditions and the continuity of the coefficients play critical roles in this context. According to the Existence and Uniqueness Theorem for solutions to differential equations, the equation must have continuous coefficients over a certain interval that includes the initial point.

In the exercise, because \( an x \) maintains continuity over the interval

• \((-\frac{\pi}{2}, \frac{\pi}{2})\)

and no discontinuities appear, a unique solution is ensured over this interval with initial conditions \( y(0) = 1 \) and \( y'(0) = 0 \). This means the specified solution will fit seamlessly into this interval, giving a predictable and consistent behavior from any point within. Having continuous coefficients and correctly applying the initial conditions makes the solution both unique and predictable within the specified interval.