Solve the given differential equation by undetermined coefficients. y" - 2y' + 17y = ex cos(4x) y(x) =
Solve the given differential equation by undetermined coefficients. y" - 2y' + 17y = ex cos(4x) y(x) =
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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![**Title: Solving Differential Equations Using the Method of Undetermined Coefficients**
**Introduction:**
In this section, we will solve the given differential equation by applying the method of undetermined coefficients. This technique is often utilized for linear differential equations with constant coefficients where the non-homogeneous term is a particular type of function (such as an exponential, polynomial, sine, cosine, or a combination of these).
**Problem Statement:**
Solve the following differential equation:
\[ y'' - 2y' + 17y = e^x \cos(4x) \]
**Solution:**
First, let's break down the differential equation:
1. **Homogeneous Equation:** The homogeneous part of the equation is:
\[ y'' - 2y' + 17y = 0 \]
2. **Particular Solution:** The particular solution \(y_p(x)\) needs to be determined based on the non-homogeneous term \( e^x \cos(4x) \).
**Homogeneous Solution:**
To find the complimentary solution, \(y_c(x)\), we solve the characteristic equation associated with the homogeneous part:
\[ r^2 - 2r + 17 = 0 \]
Using the quadratic formula:
\[ r = \frac{2 \pm \sqrt{4 - 68}}{2} = 1 \pm 4i \]
Hence, the complementary solution is:
\[ y_c(x) = e^x (C_1 \cos(4x) + C_2 \sin(4x)) \]
**Particular Solution:**
For the particular solution \(y_p(x)\), we assume a form that resembles the non-homogeneous term. Given \( e^x \cos(4x) \), we try:
\[ y_p(x) = e^x (A \cos(4x) + B \sin(4x)) \]
Differentiating this assumed solution twice, we can substitute it back into the original differential equation to determine the coefficients \(A\) and \(B\).
**Final Solution:**
The final solution to the differential equation is:
\[ y(x) = y_c(x) + y_p(x) \]
\[ y(x) = e^x (C_1 \cos(4x) + C_2 \sin(4x)) + e^x (A \cos(4x) + B](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F21e8b9dc-8a46-44e8-b3f1-f32e58de6fd1%2F98819701-991f-46ef-8370-d3618538690a%2F0st7yad_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Solving Differential Equations Using the Method of Undetermined Coefficients**
**Introduction:**
In this section, we will solve the given differential equation by applying the method of undetermined coefficients. This technique is often utilized for linear differential equations with constant coefficients where the non-homogeneous term is a particular type of function (such as an exponential, polynomial, sine, cosine, or a combination of these).
**Problem Statement:**
Solve the following differential equation:
\[ y'' - 2y' + 17y = e^x \cos(4x) \]
**Solution:**
First, let's break down the differential equation:
1. **Homogeneous Equation:** The homogeneous part of the equation is:
\[ y'' - 2y' + 17y = 0 \]
2. **Particular Solution:** The particular solution \(y_p(x)\) needs to be determined based on the non-homogeneous term \( e^x \cos(4x) \).
**Homogeneous Solution:**
To find the complimentary solution, \(y_c(x)\), we solve the characteristic equation associated with the homogeneous part:
\[ r^2 - 2r + 17 = 0 \]
Using the quadratic formula:
\[ r = \frac{2 \pm \sqrt{4 - 68}}{2} = 1 \pm 4i \]
Hence, the complementary solution is:
\[ y_c(x) = e^x (C_1 \cos(4x) + C_2 \sin(4x)) \]
**Particular Solution:**
For the particular solution \(y_p(x)\), we assume a form that resembles the non-homogeneous term. Given \( e^x \cos(4x) \), we try:
\[ y_p(x) = e^x (A \cos(4x) + B \sin(4x)) \]
Differentiating this assumed solution twice, we can substitute it back into the original differential equation to determine the coefficients \(A\) and \(B\).
**Final Solution:**
The final solution to the differential equation is:
\[ y(x) = y_c(x) + y_p(x) \]
\[ y(x) = e^x (C_1 \cos(4x) + C_2 \sin(4x)) + e^x (A \cos(4x) + B
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