A particular solution and a fundamental solution set are given for the nonhomogeneous equation below and its corresponding homogeneous equation. (a) Find a general solution to the nonhomogeneous equation. (b) Find the solution that satisfies the specified initial conditions. 2xy'' – 8y" = - 16; x>0 y(1) = 4, y'(1) = - 5, y'"(1) = - 28; Yp = x?; (1, x, xº}
A particular solution and a fundamental solution set are given for the nonhomogeneous equation below and its corresponding homogeneous equation. (a) Find a general solution to the nonhomogeneous equation. (b) Find the solution that satisfies the specified initial conditions. 2xy'' – 8y" = - 16; x>0 y(1) = 4, y'(1) = - 5, y'"(1) = - 28; Yp = x?; (1, x, xº}
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: Solving Nonhomogeneous Equations
A particular solution and a fundamental solution set are given for the nonhomogeneous equation below, along with its corresponding homogeneous equation. In this exercise, you are to:
(a) Find a general solution to the nonhomogeneous equation.
(b) Find the solution that satisfies the specified initial conditions.
**Given:**
\[ 2xy''' - 8y'' = -16; \quad x > 0 \]
**Initial Conditions:**
\[ y(1) = 4, \quad y'(1) = -5, \quad y''(1) = -28; \]
**Particular Solution:**
\[ y_p = x^2 \]
**Fundamental Solution Set:**
\[ \{1, x, x^6\} \]
### Steps:
**(a) Find a general solution to the nonhomogeneous equation.**
The general solution \( y(x) \) is given by:
\[ y(x) = \boxed{} \]
(Note: Fill in the blank with the correct solution in practice.)
---
**Explanation of the Fundamental Solution Set and Particular Solution:**
- The particular solution \( y_p = x^2 \) is a solution to the nonhomogeneous differential equation.
- The fundamental solution set \( \{1, x, x^6\} \) corresponds to the solutions of the associated homogeneous equation \( 2xy''' - 8y'' = 0 \).
To find the general solution, we combine the particular solution with the general solution of the homogeneous equation. The general solution to the nonhomogeneous differential equation is therefore a combination of the particular solution and the linear combination of the solutions from the fundamental solution set.
\[ y(x) = y_h(x) + y_p(x) \]
Where \( y_h(x) \) is the general solution of the homogeneous equation and \( y_p(x) \) is a particular solution.
The homogeneous solution can be written as:
\[ y_h(x) = C_1 \cdot 1 + C_2 \cdot x + C_3 \cdot x^6 \]
So, by combining the particular solution, the general solution is:
\[ y(x) = C_1 + C_2 \cdot x + C_3 \cdot x^6 + x^2 \]
Where \( C_1, C_2, \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffae14c32-d1e2-4020-8be9-d90c8f175447%2Fe14bc7d4-f4fe-48f6-8d0e-d8dc01adb379%2Fr93vftj.png&w=3840&q=75)
Transcribed Image Text:### Differential Equations: Solving Nonhomogeneous Equations
A particular solution and a fundamental solution set are given for the nonhomogeneous equation below, along with its corresponding homogeneous equation. In this exercise, you are to:
(a) Find a general solution to the nonhomogeneous equation.
(b) Find the solution that satisfies the specified initial conditions.
**Given:**
\[ 2xy''' - 8y'' = -16; \quad x > 0 \]
**Initial Conditions:**
\[ y(1) = 4, \quad y'(1) = -5, \quad y''(1) = -28; \]
**Particular Solution:**
\[ y_p = x^2 \]
**Fundamental Solution Set:**
\[ \{1, x, x^6\} \]
### Steps:
**(a) Find a general solution to the nonhomogeneous equation.**
The general solution \( y(x) \) is given by:
\[ y(x) = \boxed{} \]
(Note: Fill in the blank with the correct solution in practice.)
---
**Explanation of the Fundamental Solution Set and Particular Solution:**
- The particular solution \( y_p = x^2 \) is a solution to the nonhomogeneous differential equation.
- The fundamental solution set \( \{1, x, x^6\} \) corresponds to the solutions of the associated homogeneous equation \( 2xy''' - 8y'' = 0 \).
To find the general solution, we combine the particular solution with the general solution of the homogeneous equation. The general solution to the nonhomogeneous differential equation is therefore a combination of the particular solution and the linear combination of the solutions from the fundamental solution set.
\[ y(x) = y_h(x) + y_p(x) \]
Where \( y_h(x) \) is the general solution of the homogeneous equation and \( y_p(x) \) is a particular solution.
The homogeneous solution can be written as:
\[ y_h(x) = C_1 \cdot 1 + C_2 \cdot x + C_3 \cdot x^6 \]
So, by combining the particular solution, the general solution is:
\[ y(x) = C_1 + C_2 \cdot x + C_3 \cdot x^6 + x^2 \]
Where \( C_1, C_2, \)
![### Solving Differential Equations with Initial Conditions
#### (b) Find the solution that satisfies the initial conditions \( y(1) = 4 \), \( y'(1) = -5 \), \( y''(1) = -28 \).
To find the solution of the differential equation that satisfies the given initial conditions, follow these steps:
1. **Identify the form of the solution:**
If the differential equation is homogeneous, we will look for solutions in the form of exponentials or polynomials depending on the nature of the differential equation.
2. **Apply the initial conditions:**
Use the given values at \( x = 1 \) to determine the constants of integration.
3. **Verify the solution:**
Substitute the found solution back into the original differential equation to ensure that it satisfies the equation as well as the initial conditions.
Given the specific values \( y(1) = 4 \), \( y'(1) = -5 \), and \( y''(1) = -28 \), we need to find the function \( y(x) \) that can satisfy these requirements.
Please consult the specific form of the differential equation to proceed with these steps:
\[ y'' + p(x)y' + q(x)y = g(x) \]
The above steps will guide you through integrating and applying these conditions accurately.
#### Example
Say the differential equation is of the form \( y'' + 4y' + 4y = 0 \). The characteristic equation will be \( r^2 + 4r + 4 = 0 \), which simplifies to \( (r+2)^2 = 0 \). Hence, \( r = -2 \), which gives the solution \( y(x) = (C_1 + C_2 x)e^{-2x} \).
- Using \( y(1) = 4 \):
\[ (C_1 + C_2 \cdot 1)e^{-2 \cdot 1} = 4 \implies C_1 + C_2 = 4e^2 \]
- Using \( y'(1) = -5 \):
\[ (-2C_1 + C_2 \cdot -2 + C_2 (-2)e^{-2 \cdot 1}) = -5 \implies -2C_1 - C_2 = -5e^2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffae14c32-d1e2-4020-8be9-d90c8f175447%2Fe14bc7d4-f4fe-48f6-8d0e-d8dc01adb379%2Fuvhhpi.png&w=3840&q=75)
Transcribed Image Text:### Solving Differential Equations with Initial Conditions
#### (b) Find the solution that satisfies the initial conditions \( y(1) = 4 \), \( y'(1) = -5 \), \( y''(1) = -28 \).
To find the solution of the differential equation that satisfies the given initial conditions, follow these steps:
1. **Identify the form of the solution:**
If the differential equation is homogeneous, we will look for solutions in the form of exponentials or polynomials depending on the nature of the differential equation.
2. **Apply the initial conditions:**
Use the given values at \( x = 1 \) to determine the constants of integration.
3. **Verify the solution:**
Substitute the found solution back into the original differential equation to ensure that it satisfies the equation as well as the initial conditions.
Given the specific values \( y(1) = 4 \), \( y'(1) = -5 \), and \( y''(1) = -28 \), we need to find the function \( y(x) \) that can satisfy these requirements.
Please consult the specific form of the differential equation to proceed with these steps:
\[ y'' + p(x)y' + q(x)y = g(x) \]
The above steps will guide you through integrating and applying these conditions accurately.
#### Example
Say the differential equation is of the form \( y'' + 4y' + 4y = 0 \). The characteristic equation will be \( r^2 + 4r + 4 = 0 \), which simplifies to \( (r+2)^2 = 0 \). Hence, \( r = -2 \), which gives the solution \( y(x) = (C_1 + C_2 x)e^{-2x} \).
- Using \( y(1) = 4 \):
\[ (C_1 + C_2 \cdot 1)e^{-2 \cdot 1} = 4 \implies C_1 + C_2 = 4e^2 \]
- Using \( y'(1) = -5 \):
\[ (-2C_1 + C_2 \cdot -2 + C_2 (-2)e^{-2 \cdot 1}) = -5 \implies -2C_1 - C_2 = -5e^2
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