The point x = 0 is a regular singular point of the given differential equation. Find the recursive relation for the series solution of the DE below. Show the substitution and all the steps to obtain the recursive relation. Do not solve the equation for y=y(x) xy" + 4y' - xy = 0,
The point x = 0 is a regular singular point of the given differential equation. Find the recursive relation for the series solution of the DE below. Show the substitution and all the steps to obtain the recursive relation. Do not solve the equation for y=y(x) xy" + 4y' - xy = 0,
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Differential Equations: Series Solutions at Regular Singular Points
The point \( x = 0 \) is a regular singular point of the given differential equation. In this exercise, we will **find the recursive relation for the series solution of the differential equation (DE) below**. Follow the steps to show the substitution and all the methods used to obtain the recursive relation. Note: We will not solve the equation for \( y = y(x) \) at this stage.
The given differential equation is:
\[ xy'' + 4y' - xy = 0, \]
To find the series solution, assume that the solution can be expressed in the form of a power series:
\[ y = \sum_{n=0}^{\infty} a_n x^n. \]
The first and second derivatives of \( y \) are:
\[ y' = \sum_{n=1}^{\infty} n a_n x^{n-1}, \]
\[ y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}. \]
Now, substitute \( y \), \( y' \), and \( y'' \) back into the original differential equation and obtain a recursive relation for the coefficients \( a_n \).
1. **Substitute \( y'' \) into the equation:**
\[ xy'' = x \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-1}. \]
2. **Substitute \( y' \) into the equation:**
\[ 4y' = 4 \sum_{n=1}^{\infty} n a_n x^{n-1}. \]
3. **Substitute \( y \) into the equation:**
\[ -xy = -x \sum_{n=0}^{\infty} a_n x^n = -\sum_{n=0}^{\infty} a_n x^{n+1}. \]
Combine these results to get:
\[ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-1} + 4 \sum_{n=1}^{\infty} n a_n x^{n-1} - \sum_{n](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff9cdbf72-d206-4fdd-bded-1942fceb9e15%2Ff1193bd9-bc50-45da-916c-1ecda4861552%2Fw47pakq_processed.png&w=3840&q=75)
Transcribed Image Text:### Differential Equations: Series Solutions at Regular Singular Points
The point \( x = 0 \) is a regular singular point of the given differential equation. In this exercise, we will **find the recursive relation for the series solution of the differential equation (DE) below**. Follow the steps to show the substitution and all the methods used to obtain the recursive relation. Note: We will not solve the equation for \( y = y(x) \) at this stage.
The given differential equation is:
\[ xy'' + 4y' - xy = 0, \]
To find the series solution, assume that the solution can be expressed in the form of a power series:
\[ y = \sum_{n=0}^{\infty} a_n x^n. \]
The first and second derivatives of \( y \) are:
\[ y' = \sum_{n=1}^{\infty} n a_n x^{n-1}, \]
\[ y'' = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}. \]
Now, substitute \( y \), \( y' \), and \( y'' \) back into the original differential equation and obtain a recursive relation for the coefficients \( a_n \).
1. **Substitute \( y'' \) into the equation:**
\[ xy'' = x \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-1}. \]
2. **Substitute \( y' \) into the equation:**
\[ 4y' = 4 \sum_{n=1}^{\infty} n a_n x^{n-1}. \]
3. **Substitute \( y \) into the equation:**
\[ -xy = -x \sum_{n=0}^{\infty} a_n x^n = -\sum_{n=0}^{\infty} a_n x^{n+1}. \]
Combine these results to get:
\[ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-1} + 4 \sum_{n=1}^{\infty} n a_n x^{n-1} - \sum_{n
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