Problem 30 - Advanced Engineering Mathematics

Question

Use the power series method to solve the given initial-value problem. $$ (x+1) y^{\prime \prime}-(2-x) y^{\prime}+y=0, y(0)=2, y^{\prime}(0)=-1 $$

Step-by-Step Solution

Answer

The solution in power series form is derived using the recurrence relation and initial conditions.

Step 1: Assume a Power Series Solution

Assume that the solution to the differential equation can be expressed as a power series centered at 0:\[ y(x) = \sum_{n=0}^{\infty} a_n x^n \]This implies:\[ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} \]\[ y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} \]

Step 2: Substitute Power Series into Differential Equation

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

Step 3: Organize Terms and Adjust Indices

Adjust indices and group similar terms in the power series:- For the \((x+1) y''\) term, multiply and separately consider the series for the constant 1 and the series multiplied by x.- For the \((2-x)y'\) term, multiply and consider the effect of multiplying by -x.After combining, you'll have powers of x indexed from 0 upwards with corresponding expressions for \( a_n \).

Step 4: Derive the Recurrence Relation

By comparing coefficients for each power of x, derive a recurrence relation for the coefficients \( a_n \). This will involve setting the sums of coefficients for each power of x to zero.

Step 5: Solve the Recurrence Relation

Use the initial conditions \( y(0) = 2 \) and \( y'(0) = -1 \) to find specific values for some of the \( a_n \).\(a_0 = 2\) from \(y(0) = 2\).\(a_1 = -1\) from \(y'(0) = -1\).Plug these values into the recurrence relation to determine other coefficients.

Step 6: Write the Power Series Solution

Express the solution as a power series using the calculated coefficients. The series provides the solution to the differential equation in terms of x.\[ y(x) = a_0 + a_1 x + a_2 x^2 + \ldots \]

Step 7: Verify the Solution

Double-check the expressions to ensure all contributions align with the initial-value problem, ensuring that solutions for corresponding coefficients (\( a_0 \), \( a_1 \), etc.) satisfy the conditions set by the problem.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In simpler terms, they express the relationship between a function and the rate at which it changes. The primary goal is to find the unknown function that satisfies the equation. In our exercise, the function we are trying to find is denoted by \( y \), and it's related to its first and second derivatives, \( y' \) and \( y'' \), respectively.
Key characteristics of differential equations include:

• Order: Determined by the highest derivative in the equation. Our given differential equation is of the second order because it involves \( y'' \).
• Linearity: A differential equation is linear if it can be written in the form \( a_n(x)y^(n) + a_{n-1}(x)y^(n-1) + \ldots + a_1(x)y' + a_0(x)y = f(x) \). The equation from the exercise is linear because all terms involving the function or its derivatives are linear.

Understanding the structure and type of differential equation helps us determine the appropriate solution method, such as the power series method used here.

Initial-Value Problems

In differential equations, initial-value problems (IVPs) provide specific information that help pin down the exact solution from many possible solutions.Typically, initial values specify the function and its derivatives at a particular point.
In the problem we're solving, the initial conditions are given by \( y(0) = 2 \) and \( y'(0) = -1 \). This means that at \( x = 0 \), the function \( y \) has a value of 2, and its first derivative has a value of -1.
Initial conditions are crucial because:

• They allow us to determine the specific constants (or coefficients) in our solution. Without them, we can only determine a family of solutions.
• They help tailor the general solution to satisfy real-world constraints or exact requirements.

When using the power series method, these initial conditions assist in finding the values of the initial coefficients of our series.

Recurrence Relation

A recurrence relation is an equation that recursively defines a sequence; each term of the sequence is defined as a function of preceding terms.In this exercise, we derive the recurrence relation for the coefficients of the power series that represent the solution.
The basic steps to establish a recurrence relation include:

• Substituting the power series expressions into the differential equation.
• Grouping the terms by powers of \( x \).
• Setting the coefficients of each power of \( x \) equal to zero.

By solving these equations, we derive relations between the coefficients, such as \( a_{n+1} \) in terms of \( a_n \) and possibly other prior coefficients.
For our exercise, we used our initial conditions \( y(0)=2 \) and \( y'(0)=-1 \) to find the values of specific coefficients. These coefficients are used to generate the entire solution via the recurrence relation.