Problem 25 - Differential Equations and Linear Algebra

Question

Use the variation-of-parameters method to determine a particular solution to the nonhomogeneous linear system \(\mathbf{x}^{\prime}=A \mathbf{x}+\mathbf{b} .\) Also find the general solution to the system. $$A=\left[\begin{array}{rr} -6 & 1 \\ 6 & -5 \end{array}\right], \mathbf{b}=\left[\begin{array}{c} 1 \\ e^{-t} \end{array}\right]$$

Step-by-Step Solution

Answer

The general solution to the given nonhomogeneous linear system is: $$\mathbf{x}(t) = \left[\begin{array}{cc} 3e^{-3t}& -e^{-8t} \\\ e^{-3t} & 2e^{-8t} \end{array}\right]\left[\begin{array}{c} c_1 \\\ c_2 \end{array}\right] + \left[\begin{array}{c} \frac{1}{11}e^{-3t} + \frac{1}{12}e^{-8t} \\\ -\frac{1}{11} + \frac{1}{6}e^{-8t} \end{array}\right]$$ where \(c_1\) and \(c_2\) are constants.

Step 1: Finding the Fundamental Matrix Φ

First, we need to find the fundamental matrix \(\Phi\) of the homogeneous system \(\mathbf{x}^{\prime}=A \mathbf{x}\), where $$A=\left[\begin{array}{rr} -6 & 1 \\\ 6 & -5 \end{array}\right]$$ To do this, we'll need to find the eigenvalues and eigenvectors of matrix A. The characteristic equation is given by: $$\det(A - \lambda I) = 0$$ Which, for matrix A, becomes: $$\det \left(\begin{array}{cc} -6-\lambda & 1 \\\ 6 & -5-\lambda \end{array}\right) = (\lambda + 6) (\lambda + 5) - 6 = \lambda^2 + 11\lambda + 24$$ We can find the eigenvalues by solving this equation: $$\lambda^2 + 11\lambda + 24 = 0$$ Factorization of the equation gives us: $$(\lambda + 3)(\lambda + 8) = 0$$ Thus, we have two eigenvalues: \(\lambda_1 = -3\) and \(\lambda_2 = -8\). Next, we find the corresponding eigenvectors for each eigenvalue: For \(\lambda_1 = -3\): $$(A + 3I)\mathbf{v_1} = \left[\begin{array}{rr} -3 & 1 \\\ 6 & -2 \end{array}\right]\mathbf{v_1}=0$$ From the first row, we get \(v_{11}-3v_{12}=0\), and we can write the eigenvector \(\mathbf{v_1}\) as: $$\mathbf{v_1} = \left[\begin{array}{c} 3v_{12} \\\ v_{12} \end{array}\right]$$ For \(\lambda_2 = -8\): $$(A + 8I)\mathbf{v_2} = \left[\begin{array}{rr} 2 & 1 \\\ 6 & 3 \end{array}\right]\mathbf{v_2}=0$$ Using the first row, we get \(2v_{21}+v_{22}=0\), so the eigenvector \(\mathbf{v_2}\) can be written as: $$\mathbf{v_2} = \left[\begin{array}{c} -v_{22} \\\ 2v_{22} \end{array}\right]$$ Now we form the fundamental matrix \(\Phi\), using \(\mathbf{v_1}\) and \(\mathbf{v_2}\) and their respective eigenvalues: $$\Phi(t) = \left[\begin{array}{cc} 3e^{-3t} & -e^{-8t} \\\ e^{-3t} & 2e^{-8t} \end{array}\right]$$

Step 2: Determining the Particular Solution

Now we'll determine the particular solution \(\mathbf{x_p}\) to the nonhomogeneous system \(\mathbf{x}^{\prime}=A \mathbf{x}+\mathbf{b}\), using the variation of parameters method: $$\mathbf{x_p} = \Phi(t)\mathbf{v_p}(t)$$ Where \(\mathbf{x_p}\) is the particular solution and \(\mathbf{v_p}(t)\) is an unknown vector function. To find \(\mathbf{v_p}(t)\), we'll differentiate \(\mathbf{x_p}\) and plug it back into our original equation: $$\mathbf{x_p}^{\prime}=\Phi(t)\mathbf{v_p}^{\prime}(t) + \Phi^{\prime}(t)\mathbf{v_p}(t)$$ and so, $$\mathbf{v_p}^{\prime}(t)=\Phi^{-1}(t)(\mathbf{b}(t)-\Phi^{\prime}(t)\mathbf{v_p}(t))$$ We compute \(\Phi^{-1}(t)\), the inverse of the fundamental matrix: $$\Phi^{-1}(t) = \frac{1}{5e^{-11t}}\left[\begin{array}{cc} 2e^{-8t} & e^{-8t} \\\ -e^{-3t} & 3e^{-3t} \end{array}\right]$$ We also need the derivative of \(\Phi(t)\): $$\Phi^{\prime}(t) = \left[\begin{array}{cc} -9e^{-3t} & 8e^{-8t} \\\ -3e^{-3t} & -16e^{-8t} \end{array}\right]$$ Next, plug in \(\bf{b}(t)\) and \(\Phi'(t)\mathbf{v_p}(t)\) into the equation for \(\mathbf{v_p}^{\prime}(t)\) and integrate: $$\mathbf{v_p}^{\prime}(t)=\frac{1}{5e^{-11t}}\left[\begin{array}{c} (1 - 8e^{-8t}v_{p2}) \\\ (e^{-t} - 16e^{-8t}v_{p2}) \end{array}\right]$$ Integrating both sides, we get: $$\mathbf{v_p}(t)=\left[\begin{array}{c} -\frac{1}{11}(e^{3t} - 1) \\\ -\frac{1}{e^{11t}} \int e^{11t}e^{-t} dt \end{array}\right]$$ For the integral, we use integration by parts: \[ \int e^{11t}e^{-t} dt = -\frac{1}{12}e^{10t} \\ \] We then have: $$\mathbf{v_p}(t)=\left[\begin{array}{c} -\frac{1}{11}(e^{3t} - 1) \\\ -\frac{1}{12} (e^{-t} - 1) \end{array}\right]$$ Thus, the particular solution \(\mathbf{x_p}(t)=\Phi(t)\mathbf{v_p}(t)\) is given by: $$\mathbf{x_p}(t) = \left[\begin{array}{cc} 3e^{-3t} & -e^{-8t} \\\ e^{-3t} & 2e^{-8t} \end{array}\right]\left[\begin{array}{c} -\frac{1}{11}(e^{3t} - 1) \\\ -\frac{1}{12} (e^{-t} - 1) \end{array}\right] = \left[\begin{array}{c} \frac{1}{11}e^{-3t} + \frac{1}{12}e^{-8t} \\\ -\frac{1}{11} + \frac{1}{6}e^{-8t} \end{array}\right]$$

Step 3: Writing the General Solution

The general solution is given by the sum of the solution of the homogeneous system and the particular solution: $$\mathbf{x}(t) = \Phi(t)\mathbf{v} + \mathbf{x_p}(t)$$ Where \(\mathbf{v}\) is an arbitrary constant vector. Thus, the general solution is $$\mathbf{x}(t) = \left[\begin{array}{cc} 3e^{-3t} & -e^{-8t} \\\ e^{-3t} & 2e^{-8t} \end{array}\right]\left[\begin{array}{c} c_1 \\\ c_2 \end{array}\right] + \left[\begin{array}{c} \frac{1}{11}e^{-3t} + \frac{1}{12}e^{-8t} \\\ -\frac{1}{11} + \frac{1}{6}e^{-8t} \end{array}\right]$$ Where \(c_1\) and \(c_2\) are constants.

Key Concepts

Fundamental Matrix

The fundamental matrix is crucial to understanding systems of differential equations. In the context of solving a homogeneous system \(\mathbf{x}^{\prime}=A\mathbf{x}\), with a matrix \(A\) given, the fundamental matrix is a key player. Essentially, it is a matrix composed of solutions that can solve any initial value problem for this system.To build the fundamental matrix \(\Phi(t)\), we leverage eigenvalues and eigenvectors of matrix \(A\). Once we find these, they form the columns that together make \(\Phi(t)\). Each solution corresponding to an eigenvalue and eigenvector is represented by columns in this matrix. This not only allows us to express solutions for homogeneous systems but also forms the backbone for finding solutions in non-homogeneous cases.In this exercise, the fundamental matrix is calculated using eigenvalues found from the given matrix \(A\) and their corresponding eigenvectors. This matrix establishes the basis for all solutions to the homogeneous equation.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are foundational concepts when dealing with linear transformations encapsulated by matrices. They are especially important in solving linear differential equations, as they allow us to determine the behavior of systems.For a given matrix \(A\), finding eigenvalues involves solving the characteristic equation: \(\text{det}(A - \lambda I) = 0\). This exercise reveals \(\lambda_1 = -3\) and \(\lambda_2 = -8\) as the eigenvalues for matrix \(A\). Each eigenvalue corresponds to an eigenvector, providing us insight into how each dimension in space is stretched or shrunk by the transformation.Eigenvectors \(\mathbf{v_1}\) and \(\mathbf{v_2}\) were computed by substituting each eigenvalue back into the modified matrix equation to solve for vectors that satisfy the equation \(A\mathbf{v} = \lambda\mathbf{v}\). These eigenvectors form part of the fundamental matrix, defining specific trajectories of the system's state over time.

Nonhomogeneous Linear System

Nonhomogeneous linear systems introduce an external force or driving function into the system in the form of a vector \(\mathbf{b}(t)\). These systems are expressed as \(\mathbf{x}^{\prime} = A\mathbf{x} + \mathbf{b}\). The challenge is to find a particular solution \(\mathbf{x_p}\) that satisfies this equation.In the exercise provided, the method of variation of parameters is used to find \(\mathbf{x_p}\). This involves initially finding the fundamental matrix, then using it to determine another matrix function \(\mathbf{v_p}(t)\). The particular solution is constructed as \(\mathbf{x_p} = \Phi(t)\mathbf{v_p}(t)\), incorporating the effects of the non-homogeneous component. Variation of parameters is advantageous because it leverages the solutions of the homogeneous system, modifying them to account for the system's nonhomogeneity - that is, the matrix \(A\) and the function \(\mathbf{b}(t)\) interacting together to deviate from homogeneity.

General Solution of Differential Equations

The general solution of a differential equation involves combining the solutions of both the homogeneous and non-homogeneous systems. In this system of equations, the solution \(\mathbf{x}(t)\) becomes a superposition of the homogeneous solution and the particular solution.The homogeneous solution is denoted as \(\Phi(t)\mathbf{v}\), where \(\Phi(t)\) is the fundamental matrix and \(\mathbf{v}\) is a vector of arbitrary constants. This component describes all possible behaviors of the system without external forces.By adding the particular solution \(\mathbf{x_p}(t)\) to the homogeneous solution, we account for both inherent system dynamics and external inputs. The general solution then provides a complete description of the system's behavior. In summary, \(\mathbf{x}(t) = \Phi(t)\mathbf{v} + \mathbf{x_p}(t)\), ensuring that all potential system scenarios are included, capturing both natural and driven states.