Problem 13 - Advanced Engineering Mathematics

Question

Use an appropriate infinite series method about \(x=0\) to find two solutions of the given differential equation. $$ x y^{\prime \prime}-(x+2) y^{\prime}+2 y=0 $$

Step-by-Step Solution

Answer

Two solutions are found by using a power series expansion around \(x=0\) and solving the recurrence relation.

Step 1: Introduction to Power Series Solution

We start by assuming a power series solution about \(x=0\), which takes the form \( y(x) = \sum_{n=0}^{\infty} a_n x^n \). The derivatives are \( y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \) and \( y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \).

Step 2: Substitute into Differential Equation

Substitute \(y(x)\), \(y'(x)\), and \(y''(x)\) into the differential equation: \[ x \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} -(x+2) \sum_{n=1}^{\infty} n a_n x^{n-1} + 2 \sum_{n=0}^{\infty} a_n x^n = 0 \]

Step 3: Simplify and Change Indices

Simplify and adjust indices to have each term as a power series with \(x^n\) terms:- For the second derivative term: \( \sum_{n=2}^{\infty} n(n-1) a_n x^{n-1} \).- For the first derivative term: \( \sum_{n=1}^{\infty} n a_n x^n + 2 \sum_{n=1}^{\infty} n a_n x^{n-1}\).Re-index terms so each power of \(x\) matches.

Step 4: Combine Terms

Combine the terms to get a single power series: \[ \sum_{n=0}^{\infty} \left[ n(n-1)a_n + 2(n+1)a_{n+1} - 2a_n \right] x^n = 0 \].For consistency, this gives the recurrence relation.

Step 5: Solve Recurrence Relation

For the series to equal zero for all \(x\), the coefficients must each be zero:\[ n(n-1) a_n + 2(n+1) a_{n+1} - 2 a_n = 0 \] Solving for \(a_{n+1}\), we get:\[ a_{n+1} = \frac{n(n-1) - 2}{2(n+1)} a_n \].

Step 6: Identify Specific Solutions

Calculate terms starting with initial conditions. Typically, assume \(a_0\) and \(a_1\) (like usual cases, let \(a_0 = 1, a_1 = 0\) and \(a_0 = 0, a_1 = 1\)) to find two series solutions.- First solution: \( y_1(x) \) with \(a_0 = 1, a_1 = 0\).- Second solution: \( y_2(x) \) with \(a_0 = 0, a_1 = 1\).

Step 7: Explicit Calculation

Explicitly compute the first few coefficients using the recurrence relation for both initial conditions to form the series representation of solutions.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In simpler terms, they involve an unknown function along with its various derivatives and can be used to model a wide variety of real-world systems. For instance, they can describe how populations grow, how heat dissipates in a material, or how a mechanical system responds to external forces.
Differential equations can be categorized into several types, such as ordinary differential equations (ODEs) or partial differential equations (PDEs), with the former involving functions of one variable and the latter involving functions of multiple variables. In the given exercise, we are dealing with an ODE.
To solve these equations analytically, especially when they have variable coefficients or nonlinearities, techniques like the power series method can be applied. This involves expressing the solution as an infinite series of powers of the independent variable, which allows for manipulation and substitution into the original equation for simplicity and further analysis.

Recurrence Relation

A recurrence relation is a mathematical formula that expresses each term of a sequence as a function of the preceding terms. It essentially sets the blueprint for generating all terms of a sequence from its initial values.

In the context of solving differential equations using a power series, a recurrence relation naturally emerges. When the power series is substituted into the differential equation, each power of the variable acts as a separate equation, leading to conditions that establish the relations between the coefficients of the series. For instance, in this exercise, the recurrence relation was given by:

• \[ a_{n+1} = \frac{n(n-1) - 2}{2(n+1)} a_n \]

This relation effectively tells us how to compute each term of the series based on the previous terms. By systematically applying this relation, we can derive terms of the series and thereby construct the solution to the differential equation.

Initial Conditions

Initial conditions are specific values for the unknown function and possibly its derivatives at a certain point, usually at the start or boundary of a problem domain. They are crucial for completely determining the solution to a differential equation.

For linear differential equations, like the one in this exercise, multiple solutions exist and initial conditions help in selecting one particular solution among them. Here, the problem utilizes initial conditions to generate two unique power series solutions by setting initial coefficients differently:

• For the first solution, the initial coefficients are set as \(a_0 = 1, a_1 = 0\), which begins the sequence and tailors the series uniquely.
• For the second solution, \(a_0 = 0, a_1 = 1\) is used, altering the starting point and thereby the entire series solution.

Choosing different initial conditions changes the path of the solution, allowing exploration of the variety of potential solutions provided by the power series method.