Problem 10 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The unique solution exists in the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\).

Step 1: Identify the Differential Equation

The given differential equation is a second-order linear differential equation: \[ y'' + ( an x) y = e^x. \]We need to find an interval around \( x = 0 \) where it has a unique solution.

Step 2: Analyze the Coefficient Functions

The differential equation can be expressed in standard form: \[ y'' + p(x) y' + q(x) y = g(x). \] Here, \( p(x) = 0 \), \( q(x) = \tan x \), and \( g(x) = e^x \). The terms \( p(x) \), \( q(x) \), and \( g(x) \) need to be continuous in the interval for the existence of a unique solution.

Step 3: Determine the Continuity of Coefficient Functions

The function \( an x \) is continuous wherever it is defined, except where it has vertical asymptotes. \( an x\) has vertical asymptotes at \( x = \frac{\pi}{2} + k\pi \), where \( k \) is an integer. We need to consider the largest possible interval around \( x = 0 \) that doesn't include these asymptotes.

Step 4: Find the Proper Interval

Since the first vertical asymptote of \( \tan x \) occurring closest to \( x = 0 \) is at \( x = \frac{\pi}{2} \), the largest open interval without discontinuities is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). This ensures \( q(x) \) is continuous.

Step 5: Verify the Existence of Unique Solution

For the initial value problem \[ y(0) = 1, \quad y'(0) = 0, \]to have a unique solution, the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\) with continuous \( p(x) \), \( q(x) \), and \( g(x) \) guarantees the existence and uniqueness of the solution according to the theory of linear differential equations.

Key Concepts

Initial-Value Problems

Initial-value problems are a specific type of problem commonly found in the study of differential equations. In these problems, we are often given a differential equation along with specific values for the unknown function and its derivatives at a particular point. For example, in the exercise provided, we have the equation \[ y'' + \tan(x) y = e^x, \]along with the initial values \( y(0) = 1 \) and \( y'(0) = 0 \).These initial conditions help determine a specific solution among the infinitely many that could satisfy the differential equation in general. The key is to use these conditions to ensure that the solution not only fits the form of the differential equation but also passes through the specified initial values.

• Initial-value problems specify the value of the function and some of its derivatives at a particular point.
• This helps in finding a unique solution specific to the given conditions.

Understanding these initial conditions is critical, as they define the particular path or curve that our solution will follow in its domain.

Existence and Uniqueness Theorem

One crucial aspect of solving differential equations is determining whether a solution exists and whether it is unique. This is where the Existence and Uniqueness Theorem comes into play. For a second-order linear differential equation like the one in the problem, \[ y'' + \tan(x) y = e^x, \]the theorem offers a framework to verify these conditions based on the properties of the equation's coefficients.The theorem assumes that given \[ y'' + p(x) y' + q(x) y = g(x), \]if the functions \( p(x) \), \( q(x) \), and \( g(x) \) are continuous over an interval containing the initial value point, then there is a guarantee of both the existence and uniqueness of the solution.

• The continuity of these functions plays a pivotal role.
• Vertically asymptotic points in functions like \( \tan(x) \) can affect the interval of solution.

In our case, \( \tan(x) \) creates vertical asymptotes at points like \( x = \frac{\pi}{2} \), which limits our interval for a continuous solution to \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). This interval choice ensures all conditions for a unique solution are met.

Differential Equation Solutions

Solving a differential equation is all about finding a function that satisfies the equation. For second-order linear differential equations, this often involves looking for solutions of the form \( y = Ce^{rx} \), where \( C \) and \( r \) are constants, which describe the behavior of the solution over its interval.However, with the presence of a non-homogeneous term such as \( e^x \) in our equation \[ y'' + \tan(x) y = e^x, \]the task becomes finding a particular solution that accounts for this term along with the general solution of the associated homogeneous equation.

• The general solution consists of the solution to the homogeneous part plus a particular solution.
• Methods like Variation of Parameters or the Method of Undetermined Coefficients might be applied, depending on the nature of \( g(x) \).

Each unique initial-value problem can, therefore, lead to a solution that not only fulfills the differential equation's requirements but also respects the initial conditions set by the problem constraints. This dual satisfaction highlights the intricate balance required when solving such equations.