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Magazine Chapter 4: Problem 8
Consider
$$
x^{\prime \prime}(t)+4 x(t)=f(t)
$$
where \(f(t)=1\) on \(0
Short Answer
Expert verified
The Dirichlet solution is composed based on the boundary values derived from the homogeneous and particular solution constants. The Neumann solution is similarly constructed with these derivatives set to zero.
Step by step solution
01
Setup the Homogeneous Equation
The given differential equation is \(x''(t) + 4x(t) = f(t)\), where \(f(t) = 1\) for \(0 < t < 1\). First, solve the associated homogeneous equation: \(x''(t) + 4x(t) = 0\). Identifying this as a second-order linear differential equation, we will find the general solution.
02
Solve the Characteristic Equation
For the homogeneous equation \(x'' + 4x = 0\), the characteristic equation is \(r^2 + 4 = 0\), which simplifies to \(r = \pm 2i\). The roots are complex, indicating oscillatory solutions.
03
Write the General Solution of the Homogeneous Equation
Given that the roots are \(r = \pm 2i\), the general solution to the homogeneous equation is \(x_h(t) = C_1 \cos(2t) + C_2 \sin(2t)\).
04
Find a Particular Solution for the Non-homogeneous Equation
Using the method of undetermined coefficients or variation of parameters, find a particular solution \(x_p(t)\) such that \(x''(t)+4x(t)=1\) on \(0<t<1\). Here the particular solution can be assumed to be a constant since \(f(t)=1\). Therefore, guess \(x_p(t) = A\). Substitute into the equation to solve for \(A\), i.e., \(0 + 4A = 1\). Thus, \(A = \frac{1}{4}\).
05
Apply Dirichlet Boundary Conditions, Part (a)
For Dirichlet conditions \(x(0)=0\) and \(x(1)=0\), we apply these to the general solution \(x(t) = x_h(t) + x_p(t) = C_1 \cos(2t) + C_2 \sin(2t) + \frac{1}{4}\). Substituting \(x(0) = 0\) gives \(C_1 + \frac{1}{4} = 0\), which means \(C_1 = -\frac{1}{4}\). Substituting \(x(1) = 0\) and solving for \(C_2\), we get: \(-\frac{1}{4} \cos(2) + C_2 \sin(2) + \frac{1}{4} = 0\). Solve for \(C_2\).
06
Construct and Verify Dirichlet Solution
Insert the calculated \(C_1\) and \(C_2\) values back into \(x(t) = C_1 \cos(2t) + C_2 \sin(2t) + \frac{1}{4}\) to reconstruct the solution and ensure it meets the boundary conditions. Verify through calculations.
07
Apply Neumann Boundary Conditions, Part (b)
For Neumann conditions \(x'(0)=0\) and \(x'(1)=0\), differentiate the solution with respect to \(t\): \(x'(t) = -2C_1 \sin(2t) + 2C_2 \cos(2t)\). Applying \(x'(0) = 0\), we find \(x'(0) = 2C_2 = 0\), thus \(C_2 = 0\). Applying \(x'(1) = 0\), use \(-2 \cdot 0 \cdot \sin(2) + 2C_2 \cos(2) = 0\), solve for consistency or additional conditions.
08
Construct and Verify Neumann Solution
After solving for constants, construct the final solution for the Neumann boundary conditions \(x(t) = C_1 \cos(2t) + \frac{1}{4}\) using \(C_2 = 0\). Verify that it satisfies \(x'(0) = 0\) and \(x'(1) = 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second-Order Differential Equation
A second-order differential equation involves derivatives of a function up to the second degree. This type of equation is fundamental in mathematical modeling of physical systems, such as oscillating springs or electrical circuits. The general form is represented as \[ x''(t) + p(t)x'(t) + q(t)x(t) = g(t) \]where:
The equation given in the exercise is \[ x''(t) + 4x(t) = f(t) \]This is a linear differential equation because the dependent variable \(x(t)\) and its derivatives appear linearly. Handling such equations generally requires solving both the homogeneous part and finding particular solutions to address specific initial or boundary values.
- \(x''(t)\) is the second derivative of \(x(t)\),
- \(p(t)\) and \(q(t)\) are functions of \(t\),
- \(g(t)\) is a source or forcing function.
The equation given in the exercise is \[ x''(t) + 4x(t) = f(t) \]This is a linear differential equation because the dependent variable \(x(t)\) and its derivatives appear linearly. Handling such equations generally requires solving both the homogeneous part and finding particular solutions to address specific initial or boundary values.
Boundary Value Problem
Boundary Value Problems, or BVPs, are differential equations accompanied by a set of constraints or conditions at the boundaries of the domain. Unlike initial value problems, BVPs require solutions satisfying conditions at more than one point, often at the endpoints of an interval.
In the context of the exercise, there are two types of boundary conditions:
Solving BVPs often involves techniques like numerical methods or analytical approaches like the method of undetermined coefficients to achieve solutions meeting these boundary requirements.
In the context of the exercise, there are two types of boundary conditions:
- **Dirichlet conditions**: Specify the value of the solution at specific boundary points. For instance, \(x(0) = 0\) and \(x(1) = 0\) constrain the values of \(x(t)\) at \(t = 0\) and \(t = 1\), respectively.
- **Neumann conditions**: Specify the value of the derivative of the solution at the boundaries. Conditions like \(x'(0) = 0\) and \(x'(1) = 0\) control the slope of \(x(t)\) at the endpoints.
Solving BVPs often involves techniques like numerical methods or analytical approaches like the method of undetermined coefficients to achieve solutions meeting these boundary requirements.
Homogeneous and Non-Homogeneous Equations
In the realm of differential equations, it is crucial to distinguish between homogeneous and non-homogeneous types because their solution methods differ considerably.
In practice, solving non-homogeneous equations involves determining both a particular solution and incorporating the homogeneous solution to form the general solution.
- **Homogeneous Equation**: This occurs when the function \(g(t)\) is zero, i.e., \[ x''(t) + 4x(t) = 0 \]and solutions typically involve finding the roots of a characteristic equation derived from the coefficients of the differential equation. Homogeneous solutions often portray the natural behavior of a system without external forcing.
- **Non-Homogeneous Equation**: When a term \(f(t)\) is present, \[ x''(t) + 4x(t) = f(t) \]the equation becomes non-homogeneous. Such problems require finding a particular solution, often representing the system's reactive behavior to external influences.
In practice, solving non-homogeneous equations involves determining both a particular solution and incorporating the homogeneous solution to form the general solution.
Undetermined Coefficients Method
The method of undetermined coefficients is a powerful technique for finding particular solutions to linear non-homogeneous differential equations, especially when the non-homogeneous term \(f(t)\) is a simple function like a polynomial, exponential, sine, or cosine.
Conceptually, this method involves:
In this exercise, since \(f(t) = 1\), a constant is guessed for the particular solution. Substitution gives:\[ 4A = 1 \]Thus, the particular solution is \[ x_p(t) = \frac{1}{4} \].After determining this, combine it with the general homogeneous solution to address the boundary value problem fully.
Conceptually, this method involves:
- Guessing a form for the particular solution based on the nature of \(f(t)\).
- Substituting this guess into the differential equation.
- Solving the resulting system to find the coefficients.
In this exercise, since \(f(t) = 1\), a constant is guessed for the particular solution. Substitution gives:\[ 4A = 1 \]Thus, the particular solution is \[ x_p(t) = \frac{1}{4} \].After determining this, combine it with the general homogeneous solution to address the boundary value problem fully.
