Chapter 7: Problem 62
Solve the following equations using the method of undetermined coefficients. $$y^{\prime \prime}+9 y=e^{x} \cos x$$
Short Answer
Expert verified
General solution is \( y = C_1 \cos(3x) + C_2 \sin(3x) + e^x \left( \frac{7}{53} \cos x - \frac{2}{53} \sin x \right) \)."
Step by step solution
01
Find the Complementary Solution
Begin by finding the solution to the homogeneous equation associated with the given differential equation. For the homogeneous equation \( y'' + 9y = 0 \), assume a trial solution of \( y = e^{rx} \). Substituting into the equation gives \( r^2 + 9 = 0 \), which gives roots \( r = \pm 3i \). This corresponds to the complementary solution \( y_c = C_1 \cos(3x) + C_2 \sin(3x) \).
02
Propose a Particular Solution Form
For the non-homogeneous equation \( y'' + 9y = e^x \cos x \), propose a particular solution of the form \( y_p = e^x(A \cos x + B \sin x) \) based on the form of the right-hand side and the method of undetermined coefficients.
03
Determine the Derivatives
Calculate the first derivative \( y_p' \) and the second derivative \( y_p'' \) using the product rule:\( y_p' = e^x(-A \sin x + B \cos x) + e^x(A \cos x + B \sin x) \)\( y_p'' = e^x(-A \cos x - B \sin x) + 2e^x(-A \sin x + B \cos x) + e^x(-A \cos x - B \sin x) \).
04
Plug the Particular Solution into the Differential Equation
Substitute \( y_p \) and its derivatives into the original differential equation:\( y_p'' + 9y_p = e^x(-2A \cos x - 2B \sin x) + 9e^x(A \cos x + B \sin x) = e^x \cos x \).Combine like terms to obtain:\( (7A - 2B)e^x \cos x + (7B + 2A)e^x \sin x = e^x \cos x \).
05
Equate Coefficients
Match coefficients of \( e^x \cos x \) and \( e^x \sin x \) from both sides of the equation:1. \( 7A - 2B = 1 \)2. \( 7B + 2A = 0 \)Solve this system of equations to find \( A \) and \( B \).
06
Solve for A and B
From the equations:1. \( 7A - 2B = 1 \)2. \( 7B + 2A = 0 \)From the second equation, \( B = -\frac{2}{7}A \).Substitute \( B \) into the first equation:\( 7A - 2(-\frac{2}{7}A) = 1 \)\( 7A + \frac{4}{7}A = 1 \)\( \frac{53}{7}A = 1 \)\( A = \frac{7}{53} \)Then, \( B = -\frac{2}{7}\cdot\frac{7}{53} = -\frac{2}{53} \).
07
Write the General Solution
Combine the complementary and particular solutions. The general solution is:\( y = C_1 \cos(3x) + C_2 \sin(3x) + e^x \left( \frac{7}{53} \cos x - \frac{2}{53} \sin x \right) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
complementary solution
When tackling a differential equation, the first step is often finding the complementary solution. This is the solution to the associated homogeneous equation, where the non-homogeneous term is set to zero.
For example, given the differential equation \( y'' + 9y = e^x \cos x \), the complementary equation would be \( y'' + 9y = 0 \).
To solve, we suggest a solution of the form \( y = e^{rx} \). Substituting into the homogeneous equation, we find the characteristic equation: \( r^2 + 9 = 0 \). The roots are \( r = \pm 3i \), which indicate a complex scenario, leading to a complementary solution:
For example, given the differential equation \( y'' + 9y = e^x \cos x \), the complementary equation would be \( y'' + 9y = 0 \).
To solve, we suggest a solution of the form \( y = e^{rx} \). Substituting into the homogeneous equation, we find the characteristic equation: \( r^2 + 9 = 0 \). The roots are \( r = \pm 3i \), which indicate a complex scenario, leading to a complementary solution:
- \( y_c = C_1 \cos(3x) + C_2 \sin(3x) \)
particular solution
Once the complementary solution is found, the focus shifts to finding a particular solution for the non-homogeneous differential equation. The goal is to find a specific solution that satisfies the entire differential equation.
Based on the method of undetermined coefficients, the form of the particular solution depends on the non-homogeneous term \( e^x \cos x \) itself.
Through substitution into the original differential equation, derivatives of this trial solution are computed. The product rule plays a critical role here in evaluating these derivatives:
Based on the method of undetermined coefficients, the form of the particular solution depends on the non-homogeneous term \( e^x \cos x \) itself.
- For \( e^x \cos x \), we propose a particular solution of the form: \( y_p = e^x(A \cos x + B \sin x) \)
Through substitution into the original differential equation, derivatives of this trial solution are computed. The product rule plays a critical role here in evaluating these derivatives:
- \( y_p' = e^x(-A \sin x + B \cos x) + e^x(A \cos x + B \sin x) \)
- \( y_p'' = e^x(-A \cos x - B \sin x) + 2e^x(-A \sin x + B \cos x) + e^x(-A \cos x - B \sin x) \)
non-homogeneous differential equation
A non-homogeneous differential equation is one that includes an external source term, often making the equation more complex to solve.
For an equation like \( y'' + 9y = e^x \cos x \), the right-hand side, \( e^x \cos x \), serves as the non-homogeneous term that affects the form of our solution. In contrast to homogeneous equations where the right-hand side is zero, these non-zero terms require a particular solution.
The general solution to a non-homogeneous differential equation is a combination of its complementary and particular solutions:
For an equation like \( y'' + 9y = e^x \cos x \), the right-hand side, \( e^x \cos x \), serves as the non-homogeneous term that affects the form of our solution. In contrast to homogeneous equations where the right-hand side is zero, these non-zero terms require a particular solution.
The general solution to a non-homogeneous differential equation is a combination of its complementary and particular solutions:
- Complementary Solution (\(y_c\)): Resolves the homogeneous part and captures the natural response of the system, not influenced by external forces.
- Particular Solution (\(y_p\)): Account for the non-homogeneous part, the specific influence of external forces on the system.
- \( y = y_c + y_p \)