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Substituting \(y=\sum_{n=0}^{\infty} c_{n} x^{n}\) into the differential equation we have $$\begin{aligned} (x+2) y^{\prime \prime}+x y^{\prime}-y &=\underbrace{\sum_{n=2}^{\infty} n(n-1) c_{n} x^{n-1}}_{k=n-1}+\underbrace{\sum_{n=2}^{\infty} 2 n(n-1) c_{n} x^{n-2}}_{k=n-2}+\underbrace{\sum_{n=1}^{\infty} n c_{n} x^{n}}_{k=n}-\sum_{n=0}^{\infty} c_{n} x^{n} \\ &=\sum_{k=1}^{\infty}(k+1) k c_{k+1} x^{k}+\sum_{k=0}^{\infty} 2(k+2)(k+1) c_{k+2} x^{k}+\sum_{k=1}^{\infty} k c_{k} x^{k}-\sum_{k=0}^{\infty} c_{k} x^{k} \\ &=4 c_{2}-c_{0}+\sum_{k=1}^{\infty}\left[(k+1) k c_{k+1}+2(k+2)(k+1) c_{k+2}+(k-1) c_{k}\right] x^{k}=0 \end{aligned}$$ Thus $$\begin{aligned} &4 c_{2}-c_{0}=0 (k+1) k c_{k+1}+2(k+2)(k+1) c_{k+2}+(k-1) c_{k}=0, \quad k=1,2,3, \dots \end{aligned}$$ and $$\begin{aligned} c_{2} &=\frac{1}{4} c_{0} \\ c_{k+2} &=-\frac{(k+1) k c_{k+1}+(k-1) c_{k}}{2(k+2)(k+1)}, \quad k=1,2,3, \ldots \end{aligned}$$ Choosing \(c_{0}=1\) and \(c_{1}=0\) we find $$c_{1}=0, \quad c_{2}=\frac{1}{4}, \quad c_{3}=-\frac{1}{24}, \quad c_{4}=0, \quad c_{5}=\frac{1}{480}$$ and so on. For \(c_{0}=0\) and \(c_{1}=1\) we obtain $$\begin{aligned} &c_{2}=0\\\ &c_{3}=0\\\ &c_{4}=c_{5}=c_{6}=\cdots=0 \end{aligned}$$ Thus, two solutions are $$y_{1}=c_{0}\left[1+\frac{1}{4} x^{2}-\frac{1}{24} x^{3}+\frac{1}{480} x^{5}+\cdots\right] \quad \text { and } \quad y_{2}=c_{1} x$$.

Step by step solution

01

Substitute the Series into the DE

Insert the series expression \( y = \sum_{n=0}^{\infty} c_{n} x^{n} \) for \( y \), and its derivatives \( y' \) and \( y'' \), into the given differential equation. The series for \( y'' \) is shifted by adjusting the indices so that all powers of \( x \) match.

02

Adjust Indices and Match Powers

Adjust indices for each series term in the differential equation, aligning them so that each term is expressed as a series in powers of \( x^k \). This requires recognizing the index shifts.

03

Combine Series

Combine similar series terms to create a unified series form. Each series should have the same power of \( x^k \), allowing the coefficients to be collected together.

04

Set Coefficients to Zero

Since the original differential equation must hold for all \( x \), set the coefficients of each power of \( x^k \) to zero. This provides a series of equations in terms of the coefficients \( c_n \).

05

Solve Homogeneous Equations

First, solve the equation for \( k=0 \), which gives \( 4c_2 - c_0 = 0 \) using \( c_0 = 1 \). Then solve the recurrence relation for \( c_{k+2} \) using the homogeneous equation set in Step 4.

06

Use Initial Conditions

With the initial conditions \( c_0 = 1 \) and \( c_1 = 0 \), solve for \( c_2 \) and subsequent coefficients recursively from the recurrence relation.

07

Construct Solution from Coefficients

Using the determined coefficients \( c_0, c_1, c_2, \ldots \), construct the series solution \( y_1 \). With \( c_0 = 0 \) and \( c_1 = 1 \), determine the second solution \( y_2 = x \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations

A differential equation involves a relationship between a function and its derivatives. These are often used to model physical systems, like the behavior of springs, circuits, or any system that changes over time. Understanding how to solve them is crucial because they help describe the real world in mathematical terms.

In the context of power series solutions, we substitute a series expression for the function into the differential equation. This approach differs from typical algebraic methods, as it relies on expressing the solution as an infinite sum. This infinite sum, or series, is composed of constant coefficients and powers of the variable for which we are finding the function. This method works particularly well for linear differential equations with variable coefficients.

Good examples of differential equations that are often solved using power series include Legendre's equation and Bessel's equation. They are prevalent in physics and engineering because they appear in wave equations, quantum mechanics, and other critical areas.

Recurrence Relations

Recurrence relations are equations that express each term of a sequence as a function of its preceding terms. In solving differential equations with series, we derive these relations based on the coefficients of the series terms.

When you plug a power series into a differential equation, the equation is converted into a recurrence relation. This relation tells you how to compute the terms of your series from one another. Often, this will result in a pattern that can be solved either explicitly (for select terms) or iteratively.

For instance, in our exercise, we end up with a recurrence relation expressed as \[ c_{k+2} = -\frac{(k+1)kc_{k+1} + (k-1)c_{k}}{2(k+2)(k+1)} \] which indicates how to find coefficients for higher-order terms based on the previous elements. These kinds of relations are foundational not just in solving power series problems but are also handy in computational mathematics methods like implementing algorithms.

Taylor Series Expansion

The Taylor Series is an essential concept in calculus, used to approximate functions. By expanding a function into an infinite sum of terms calculated from the derivatives of the function at a single point, this series can give us a precise depiction of the function around that point.

The power series solution method of solving differential equations is closely related to Taylor series expansions. We assume the solution can be written in this form: \[ y = \sum_{n=0}^{\infty} c_n x^n \] where each coefficient, \( c_n \), is derived based on the recurrence relations found when substituting the series into the differential equation.

This technique ensures that the power series converges to the true solution of the differential equation within a specific interval. It’s fascinating because it allows us to find solutions even when the function cannot be solved using elementary functions, demonstrating the power of calculus and analysis in understanding complex systems.