Find two power series solutions of the given differential equation about the ordinary point x = 0. y" + x²y = 0 1 4 Oy=1- -X + 12 1 Oy=1-x + 6 14 Oy = 1- x² + 6 O y = 1 + 1 672 14 -x + 1 252 Oy=1++++++ 12 672 1 252 8 1 252 8 8 - and y = x - ... and y = x - + + ... and y = x + and y = x - and y = x + 1 20 1 12 5 -x³ + 1x5 20 1 -X + +5 12 1 12 + + + 1 1440 1 5 ·X 672 1 1440 1 672 1 672 9
Find two power series solutions of the given differential equation about the ordinary point x = 0. y" + x²y = 0 1 4 Oy=1- -X + 12 1 Oy=1-x + 6 14 Oy = 1- x² + 6 O y = 1 + 1 672 14 -x + 1 252 Oy=1++++++ 12 672 1 252 8 1 252 8 8 - and y = x - ... and y = x - + + ... and y = x + and y = x - and y = x + 1 20 1 12 5 -x³ + 1x5 20 1 -X + +5 12 1 12 + + + 1 1440 1 5 ·X 672 1 1440 1 672 1 672 9
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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![### Power Series Solutions of Differential Equations
The problem is to find two power series solutions of the differential equation:
\[ y'' + x^2y = 0 \]
This equation is to be solved about the ordinary point \( x = 0 \).
Here are the possible solutions:
1. **Option A:**
- \( y = 1 - \frac{1}{12}x^4 + \frac{1}{672}x^8 - \ldots \)
- \( y = x - \frac{1}{20}x^5 + \frac{1}{1440}x^9 - \ldots \)
2. **Option B:**
- \( y = 1 - \frac{1}{6}x^2 + \frac{1}{252}x^4 - \ldots \)
- \( y = x - \frac{1}{12}x^3 + \frac{1}{672}x^5 - \ldots \)
3. **Option C:**
- \( y = 1 + \frac{1}{12}x^4 + \frac{1}{672}x^8 + \ldots \)
- \( y = x + \frac{1}{20}x^5 + \frac{1}{1440}x^9 + \ldots \)
4. **Option D:**
- \( y = 1 - \frac{1}{6}x^4 + \frac{1}{252}x^8 - \ldots \)
- \( y = x - \frac{1}{12}x^5 + \frac{1}{672}x^9 - \ldots \)
5. **Option E:**
- \( y = 1 + \frac{1}{6}x^4 + \frac{1}{252}x^8 + \ldots \)
- \( y = x + \frac{1}{12}x^5 + \frac{1}{672}x^9 + \ldots \)
### Explanation
The above solutions are expressed as power series expansions, which are infinite sums involving powers of \( x \). Each option provides a pair of such series solutions, and the pattern of coefficients follows particular calculations based on solving the differential equation using power series methods. The correct pair of series will](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa103160f-6870-471c-8ca1-5357643d2f88%2Fd6078cbb-a074-4f54-bd5f-ee43164a3b3a%2Fyuzly4c_processed.png&w=3840&q=75)
Transcribed Image Text:### Power Series Solutions of Differential Equations
The problem is to find two power series solutions of the differential equation:
\[ y'' + x^2y = 0 \]
This equation is to be solved about the ordinary point \( x = 0 \).
Here are the possible solutions:
1. **Option A:**
- \( y = 1 - \frac{1}{12}x^4 + \frac{1}{672}x^8 - \ldots \)
- \( y = x - \frac{1}{20}x^5 + \frac{1}{1440}x^9 - \ldots \)
2. **Option B:**
- \( y = 1 - \frac{1}{6}x^2 + \frac{1}{252}x^4 - \ldots \)
- \( y = x - \frac{1}{12}x^3 + \frac{1}{672}x^5 - \ldots \)
3. **Option C:**
- \( y = 1 + \frac{1}{12}x^4 + \frac{1}{672}x^8 + \ldots \)
- \( y = x + \frac{1}{20}x^5 + \frac{1}{1440}x^9 + \ldots \)
4. **Option D:**
- \( y = 1 - \frac{1}{6}x^4 + \frac{1}{252}x^8 - \ldots \)
- \( y = x - \frac{1}{12}x^5 + \frac{1}{672}x^9 - \ldots \)
5. **Option E:**
- \( y = 1 + \frac{1}{6}x^4 + \frac{1}{252}x^8 + \ldots \)
- \( y = x + \frac{1}{12}x^5 + \frac{1}{672}x^9 + \ldots \)
### Explanation
The above solutions are expressed as power series expansions, which are infinite sums involving powers of \( x \). Each option provides a pair of such series solutions, and the pattern of coefficients follows particular calculations based on solving the differential equation using power series methods. The correct pair of series will
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