Consider the linear differential equation ay" - xy' + y = 0. (a) Verify that y = x is a solution to this DE. (i.e., show that y1 = I satisfies the DE.) (b) Use the method of reduction of order to find a second lin. indep. solution y2. Hint: For reduction of order, suppose that y2 = uY1.
Consider the linear differential equation ay" - xy' + y = 0. (a) Verify that y = x is a solution to this DE. (i.e., show that y1 = I satisfies the DE.) (b) Use the method of reduction of order to find a second lin. indep. solution y2. Hint: For reduction of order, suppose that y2 = uY1.
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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## Example Problem: Solving a Linear Differential Equation
Consider the linear differential equation:
\[ x^3 y'' - xy' + y = 0 \]
### Steps to Solve:
**(a)** Verify that \( y_1 = x \) is a solution to this differential equation (i.e., show that \( y_1 = x \) satisfies the DE).
**(b)** Use the method of reduction of order to find a second linearly independent solution \( y_2 \).
**Hint:** For reduction of order, suppose that \( y_2 = uy_1 \).
---
### Solution Outline:
1. **Verification of \( y_1 = x \) as a Solution:**
- To verify that \( y_1 = x \) is a solution, substitute \( y_1 = x \) into the differential equation and confirm that the left-hand side equals zero.
2. **Reduction of Order:**
- Given the first solution \( y_1 = x \), find the second solution by assuming \( y_2 = uy_1 \), where \( u \) is a function of \( x \).
- Derive the necessary expressions and substitute back into the original differential equation to solve for \( u \).
---
### Detailed Solution Process:
**Verification:**
1. Let \( y_1 = x \)
2. Compute the required derivatives:
- \( y_1' = 1 \)
- \( y_1'' = 0 \)
3. Substitute these into the differential equation \( x^3 y'' - xy' + y \):
\[
x^3 (0) - x(1) + x = 0
\]
4. Simplify to show:
\[
-x + x = 0
\]
This confirms that \( y_1 = x \) satisfies the differential equation.
**Reduction of Order:**
1. Assume \( y_2 = uy_1 = ux \)
2. Compute the derivatives needed for substitution:
- \( y_2' = u'x + u \)
- \( y_2'' = u''x + 2u' \)
3. Substitute \( y_2, y_2', y_2'' \) into the differential equation:
\[ x^3 (u''x + 2u') - x(u'x +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0b9151a0-b51c-4e98-8bd5-faad9be5df3c%2F0f37d549-2e37-4fa5-987c-2871df9b0dc2%2Fe2pwbjc.jpeg&w=3840&q=75)
Transcribed Image Text:---
## Example Problem: Solving a Linear Differential Equation
Consider the linear differential equation:
\[ x^3 y'' - xy' + y = 0 \]
### Steps to Solve:
**(a)** Verify that \( y_1 = x \) is a solution to this differential equation (i.e., show that \( y_1 = x \) satisfies the DE).
**(b)** Use the method of reduction of order to find a second linearly independent solution \( y_2 \).
**Hint:** For reduction of order, suppose that \( y_2 = uy_1 \).
---
### Solution Outline:
1. **Verification of \( y_1 = x \) as a Solution:**
- To verify that \( y_1 = x \) is a solution, substitute \( y_1 = x \) into the differential equation and confirm that the left-hand side equals zero.
2. **Reduction of Order:**
- Given the first solution \( y_1 = x \), find the second solution by assuming \( y_2 = uy_1 \), where \( u \) is a function of \( x \).
- Derive the necessary expressions and substitute back into the original differential equation to solve for \( u \).
---
### Detailed Solution Process:
**Verification:**
1. Let \( y_1 = x \)
2. Compute the required derivatives:
- \( y_1' = 1 \)
- \( y_1'' = 0 \)
3. Substitute these into the differential equation \( x^3 y'' - xy' + y \):
\[
x^3 (0) - x(1) + x = 0
\]
4. Simplify to show:
\[
-x + x = 0
\]
This confirms that \( y_1 = x \) satisfies the differential equation.
**Reduction of Order:**
1. Assume \( y_2 = uy_1 = ux \)
2. Compute the derivatives needed for substitution:
- \( y_2' = u'x + u \)
- \( y_2'' = u''x + 2u' \)
3. Substitute \( y_2, y_2', y_2'' \) into the differential equation:
\[ x^3 (u''x + 2u') - x(u'x +
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