Chapter 17: Problem 5
Solve the equations in Exercises by the method of undetermined coefficients. $$y^{\prime \prime}+y=\cos 3 x$$
Short Answer
Expert verified
The general solution is \( y = C_1 \cos x + C_2 \sin x - \frac{1}{8} \cos 3x \).
Step by step solution
01
Identify the Homogeneous Equation
The given differential equation is \( y'' + y = \cos 3x \). Start by considering the corresponding homogeneous equation: \( y'' + y = 0 \).
02
Solve the Homogeneous Equation
Solve the homogeneous equation \( y'' + y = 0 \). The characteristic equation is \( r^2 + 1 = 0 \), which gives roots \( r = \pm i \). This means the solution to the homogeneous equation is \( y_h = C_1 \cos x + C_2 \sin x \).
03
Assume a Particular Solution Form
Since the non-homogeneous term is \( \cos 3x \), assume a particular solution of the form \( y_p = A \cos 3x + B \sin 3x \), where \( A \) and \( B \) are constants to be determined.
04
Compute Derivatives of Particular Solution
Calculate the derivatives: \( y_p' = -3A \sin 3x + 3B \cos 3x \) and \( y_p'' = -9A \cos 3x - 9B \sin 3x \).
05
Substitute and Solve for Coefficients
Substitute \( y_p \) and its derivatives into the differential equation: \( (-9A \cos 3x - 9B \sin 3x) + (A \cos 3x + B \sin 3x) = \cos 3x \). Simplify to get: \( (-8A) \cos 3x + (-8B) \sin 3x = \cos 3x \). Thus, \( -8A = 1 \) and \( -8B = 0 \). Solve these to get \( A = -\frac{1}{8} \) and \( B = 0 \).
06
Write the General Solution
Combine the homogeneous and particular solutions to write the general solution: \( y = y_h + y_p = C_1 \cos x + C_2 \sin x - \frac{1}{8} \cos 3x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Equation
A homogeneous equation is a type of linear differential equation that equals zero when all terms are on one side. In our given exercise, the differential equation is \( y'' + y = \cos 3x \). To find the related homogeneous equation, we simply set the right side to zero, resulting in \( y'' + y = 0 \).
Solving a homogeneous equation involves finding the characteristic equation; in this case, it is \( r^2 + 1 = 0 \), with roots \( r = \pm i \). These roots indicate that the solution will be a combination of trigonometric functions.
Solving a homogeneous equation involves finding the characteristic equation; in this case, it is \( r^2 + 1 = 0 \), with roots \( r = \pm i \). These roots indicate that the solution will be a combination of trigonometric functions.
- The general solution to our homogeneous equation is \( y_h = C_1 \cos x + C_2 \sin x \).
- \( C_1 \) and \( C_2 \) are arbitrary constants determined by any given initial or boundary conditions.
Particular Solution
A particular solution is a specific solution to a non-homogeneous differential equation that includes additional terms. These terms account for any non-zero parts of the equation, in contrast to a homogeneous equation that equals zero.
In our exercise, we have a non-homogeneous differential equation: \( y'' + y = \cos 3x \). Given that the non-homogeneous term is \( \cos 3x \), we hypothesize a particular solution of the form \( y_p = A \cos 3x + B \sin 3x \). Here
In our exercise, we have a non-homogeneous differential equation: \( y'' + y = \cos 3x \). Given that the non-homogeneous term is \( \cos 3x \), we hypothesize a particular solution of the form \( y_p = A \cos 3x + B \sin 3x \). Here
- \( A \) and \( B \) are coefficients that need to be determined.
- \(-8A = 1\), solving gives \( A = -\frac{1}{8} \)
- \(-8B = 0\), solving gives \( B = 0 \)
Differential Equations
Differential equations are mathematical equations that involve derivatives, revealing how a particular variable changes with respect to another. In simple terms, they explain how things evolve over time or other dimensions.
These equations can be classified broadly into two types: homogeneous and non-homogeneous. A homogeneous differential equation, like \( y'' + y = 0 \), has all terms on one side equalling zero, while a non-homogeneous equation includes an additional term on the other side, such as \( \cos 3x \) in our example \( y'' + y = \cos 3x \).
Solving differential equations is essential across many fields:
These equations can be classified broadly into two types: homogeneous and non-homogeneous. A homogeneous differential equation, like \( y'' + y = 0 \), has all terms on one side equalling zero, while a non-homogeneous equation includes an additional term on the other side, such as \( \cos 3x \) in our example \( y'' + y = \cos 3x \).
Solving differential equations is essential across many fields:
- In physics, they model dynamic systems such as motion or thermodynamics.
- In engineering, they describe electrical circuits or structural analysis.
- In biology, they can represent population growth or the spread of diseases.