Use power series to find the general solution of the given differential equation. y" - 6xy' + 9y = 0
Use power series to find the general solution of the given differential equation. y" - 6xy' + 9y = 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Use power series to solve
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### Using Power Series to Solve a Differential Equation
To solve the differential equation
\[y'' - 6xy' + 9y = 0\]
using power series, follow these steps. Below are potential solutions. Select the correct one. Here, \(c_0\) and \(c_1\) are arbitrary constants.
#### Choose the correct solution below:
- **A.**
\[
y = c_0 \left( 1 - \frac{9}{2}x^2 - \frac{9}{8}x^4 + \ldots \right) + c_1 \left( x - \frac{1}{2}x^3 - \frac{9}{40}x^5 + \ldots \right)
\]
- **B.**
\[
y = c_0 \left( x - \frac{9}{2}x^3 - \frac{9}{8}x^5 + \ldots \right) + c_1 \left( 1 - \frac{1}{2}x^2 - \frac{9}{40}x^4 + \ldots \right)
\]
- **C.**
\[
y = c_0 \left( x + \frac{9}{2}x^3 + \frac{1}{4}x^5 + \ldots \right) + c_1 \left( 1 - \frac{1}{2}x^2 + \frac{9}{20}x^4 + \ldots \right)
\]
- **D.**
\[
y = c_0 \left( 1 - \frac{9}{2}x^2 + \frac{1}{4}x^4 + \ldots \right) + c_1 \left( x - \frac{1}{2}x^3 + \frac{9}{20}x^5 + \ldots \right)
\]
---
Consider each form and recognize that they are constructed from the power series approach to solving differential equations, ensuring that they satisfy the original equation given.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F32ac3e12-ae1c-4f9c-8855-d6c71d330ed6%2F93b03858-e651-40a2-b600-5265d3555953%2Fqql90pf_processed.png&w=3840&q=75)
Transcribed Image Text:---
### Using Power Series to Solve a Differential Equation
To solve the differential equation
\[y'' - 6xy' + 9y = 0\]
using power series, follow these steps. Below are potential solutions. Select the correct one. Here, \(c_0\) and \(c_1\) are arbitrary constants.
#### Choose the correct solution below:
- **A.**
\[
y = c_0 \left( 1 - \frac{9}{2}x^2 - \frac{9}{8}x^4 + \ldots \right) + c_1 \left( x - \frac{1}{2}x^3 - \frac{9}{40}x^5 + \ldots \right)
\]
- **B.**
\[
y = c_0 \left( x - \frac{9}{2}x^3 - \frac{9}{8}x^5 + \ldots \right) + c_1 \left( 1 - \frac{1}{2}x^2 - \frac{9}{40}x^4 + \ldots \right)
\]
- **C.**
\[
y = c_0 \left( x + \frac{9}{2}x^3 + \frac{1}{4}x^5 + \ldots \right) + c_1 \left( 1 - \frac{1}{2}x^2 + \frac{9}{20}x^4 + \ldots \right)
\]
- **D.**
\[
y = c_0 \left( 1 - \frac{9}{2}x^2 + \frac{1}{4}x^4 + \ldots \right) + c_1 \left( x - \frac{1}{2}x^3 + \frac{9}{20}x^5 + \ldots \right)
\]
---
Consider each form and recognize that they are constructed from the power series approach to solving differential equations, ensuring that they satisfy the original equation given.
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