i, I tried to solve this differential equation involving series methods, but I think the a2, a3, a4 factors might be wrong because I ended up with an "x" in the answer and there should just be number digits. Could I get some help please? (x+1)y" + xy = 0 where initial conditions are y(0) = 1, y'(0) = 2.
i, I tried to solve this differential equation involving series methods, but I think the a2, a3, a4 factors might be wrong because I ended up with an "x" in the answer and there should just be number digits. Could I get some help please? (x+1)y" + xy = 0 where initial conditions are y(0) = 1, y'(0) = 2.
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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Hi, I tried to solve this differential equation involving series methods, but I think the a2, a3, a4 factors might be wrong because I ended up with an "x" in the answer and there should just be number digits. Could I get some help please?
(x+1)y" + xy = 0
where initial conditions are y(0) = 1, y'(0) = 2.
Thank you.
![### Solving Differential Equations Using Power Series Method
Consider the differential equation:
\[ (x+1)y'' + xy = 0 \]
with the initial conditions:
\[ y(0) = 1 \quad \text{and} \quad y'(0) = 2 \]
These initial conditions translate to the power series coefficients as:
\[ a_0 = 1 \]
\[ a_1 = 2 \]
Assume a solution of the form:
\[ y = \sum_{n=0}^{\infty} a_n x^n \]
### Deriving Representations for \( y' \) and \( y'' \)
Taking the derivatives, we get:
\[ y' = \sum_{n=1}^{\infty} n a_n x^{n-1} \Rightarrow y' = \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n \]
\[ y'' = \sum_{n=2}^{\infty} n (n-1) a_n x^{n-2} \Rightarrow y'' = \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n \]
### Substituting into the Original Differential Equation
By substituting \( y \) and \( y'' \) into the original equation, we have:
\[ (x+1) \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n + x \sum_{n=0}^{\infty} a_n x^n = 0 \]
Combining terms and simplifying:
\[ \sum_{n=0}^{\infty} [(n+1)(n) a_{n+1} x^n + (n+2)(n+1) a_{n+2} x^n + x a_n] x^n = 0 \]
### Simplification Leads to Recurrence Relation
The above simplifies to:
\[ \sum_{n=0}^{\infty} \left[ (n+1)(n) a_{n+1} x^n + (n+2)(n+1) a_{n+2} x^n + x a_n \right] = 0 \]
Leading to the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9d5b5049-d8dd-402a-aa3b-0cfd97dc82be%2F17c70960-ef65-4ee7-b98a-de417231350b%2Fyviihw_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving Differential Equations Using Power Series Method
Consider the differential equation:
\[ (x+1)y'' + xy = 0 \]
with the initial conditions:
\[ y(0) = 1 \quad \text{and} \quad y'(0) = 2 \]
These initial conditions translate to the power series coefficients as:
\[ a_0 = 1 \]
\[ a_1 = 2 \]
Assume a solution of the form:
\[ y = \sum_{n=0}^{\infty} a_n x^n \]
### Deriving Representations for \( y' \) and \( y'' \)
Taking the derivatives, we get:
\[ y' = \sum_{n=1}^{\infty} n a_n x^{n-1} \Rightarrow y' = \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n \]
\[ y'' = \sum_{n=2}^{\infty} n (n-1) a_n x^{n-2} \Rightarrow y'' = \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n \]
### Substituting into the Original Differential Equation
By substituting \( y \) and \( y'' \) into the original equation, we have:
\[ (x+1) \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n + x \sum_{n=0}^{\infty} a_n x^n = 0 \]
Combining terms and simplifying:
\[ \sum_{n=0}^{\infty} [(n+1)(n) a_{n+1} x^n + (n+2)(n+1) a_{n+2} x^n + x a_n] x^n = 0 \]
### Simplification Leads to Recurrence Relation
The above simplifies to:
\[ \sum_{n=0}^{\infty} \left[ (n+1)(n) a_{n+1} x^n + (n+2)(n+1) a_{n+2} x^n + x a_n \right] = 0 \]
Leading to the
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