Problem 6 - Differential Equations and Linear Algebra

Question

Use the variation-of-parameters technique to find a particular solution \(\mathbf{x}_{p}\) to \(\mathbf{x}^{\prime}=A \mathbf{x}+\mathbf{b},\) for the given \(A\) and \(\mathbf{b} .\) Also obtain the general solution to the system of differential equations. $$A=\left[\begin{array}{rr} 2 & 4 \\ -2 & -2 \end{array}\right], \quad \mathbf{b}=\left[\begin{array}{l} 8 \sin 2 t \\ 8 \cos 2 t \end{array}\right]$$

Step-by-Step Solution

Answer

The general solution of the nonhomogeneous system of differential equations, $\mathbf{x}(t)$, is the same as the general solution of the homogeneous system, given by: $$\mathbf{x}(t)=c_1e^{2t}\begin{bmatrix}1\\-0.5\end{bmatrix}+c_2e^{-2t}\begin{bmatrix}1\\1\end{bmatrix}$$ where $c_1$ and $c_2$ are constants.

Step 1: Find the general solution of the homogeneous system

To find the general solution of the homogeneous system \(\mathbf{x}'=A\mathbf{x}\), we need to find the matrix exponential of matrix A, which can be defined as \(e^{At}=\sum_{n=0}^{\infty}\frac{A^n t^n}{n!}\). $$A=\left[\begin{array}{rr} 2 & 4 \\ -2 & -2 \end{array}\right]$$ To compute the matrix exponential, we first need to find the eigenvalues (\(\lambda\)) and eigenvectors (\(\mathbf{v}\)) of the matrix A. The characteristic equation is given by: \(\mathrm{det}(A-\lambda I)=0\) $$(2-\lambda)\left(-2-\lambda\right)-(-2)(4)=\lambda^{2}\Rightarrow\lambda_1=2,\:\lambda_2=-2$$ Now let's find the eigenvectors corresponding to each eigenvalue. For \(\lambda_1=2\): \((A-\lambda_1 I)\mathbf{v}_1=\begin{bmatrix} 0 & 4\\ -2 & -4 \end{bmatrix}\begin{bmatrix} v_{11}\\ v_{12} \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}\) After solving this system, we find that \(\mathbf{v}_1=\begin{bmatrix}1\\-0.5\end{bmatrix}\). For \(\lambda_2=-2\): \((A-\lambda_2 I)\mathbf{v}_2=\begin{bmatrix} 4 & 4\\ -2 & 0 \end{bmatrix}\begin{bmatrix} v_{21}\\ v_{22} \end{bmatrix}=\begin{bmatrix} 0\\ 0 \end{bmatrix}\) After solving this system, we find that \(\mathbf{v}_2=\begin{bmatrix}1\\1\end{bmatrix}\). Now we can find the matrix exponential using the eigenvalues and eigenvectors: $$e^{At}=c_1e^{\lambda_1t}\mathbf{v}_1+c_2e^{\lambda_2t}\mathbf{v}_2=c_1e^{2t}\begin{bmatrix}1\\-0.5\end{bmatrix}+c_2e^{-2t}\begin{bmatrix}1\\1\end{bmatrix}$$ So, the general solution of the homogeneous system is given by: $$\mathbf{x}= e^{At}=c_1e^{2t}\begin{bmatrix}1\\-0.5\end{bmatrix}+c_2e^{-2t}\begin{bmatrix}1\\1\end{bmatrix}$$

Step 2: Find the particular solution using variation of parameters

Let's denote \(\mathbf{w}(t)=\frac{d\mathbf{x}_p}{dt}\) and rewrite the nonhomogeneous system as: \(\mathbf{w}(t)=A\mathbf{x}_p+\mathbf{b}\). To find the particular solution, we will express \(\mathbf{x}_p\) as a linear combination of the eigenvectors V: $$\mathbf{x}_p(t)=K_1(t)\mathbf{v}_1+K_2(t)\mathbf{v}_2$$ where \(K_1(t)\) and \(K_2(t)\) are unknown functions. Now we differentiate this expression with respect to t: $$\mathbf{w}(t)=\frac{d\mathbf{x}_p}{dt}=K_1'(t)\mathbf{v}_1+K_2'(t)\mathbf{v}_2$$ The variation of parameters method requires solving the following algebraic system for \(K_1'(t)\) and \(K_2'(t)\): $$A\mathbf{x}_p+\mathbf{b}=A(K_1(t)\mathbf{v}_1+K_2(t)\mathbf{v}_2)+\mathbf{b}=\mathbf{w}(t)=K_1'(t)\mathbf{v}_1+K_2'(t)\mathbf{v}_2$$ Substituting eigenvectors and the function b into this system, we get: $$\left[\begin{array}{rr}1&1\end{array}\right]\begin{bmatrix}K_1'(t)\\K_2'(t)\end{bmatrix}=2\begin{bmatrix}K_1(t)\\K_2(t)\end{bmatrix}+\begin{bmatrix}8\sin(2t)\\8\cos(2t)\end{bmatrix}$$ Solving this system, we find: $$K_1'(t)=4\sin(2t)-4K_1(t)$$ $$K_2'(t)=4\cos(2t)+4K_1(t)-8K_2(t)$$ We will integrate these equations to get \(K_1(t)\) and \(K_2(t)\). Integrating the first equation with respect to t, we get: $$K_1(t)=-\frac{1}{2}\cos(2t)+C_1$$ Then, we substitute this expression for \(K_1(t)\) into the second equation and integrate it to find: $$K_2(t)=\frac{1}{4}\sin(2t)+\frac{1}{2}\cos(2t)+C_2$$ Now we can substitute the expressions for \(K_1(t)\) and \(K_2(t)\) back into the expression for \(\mathbf{x}_p(t)\) to find the particular solution: $$\mathbf{x}_p(t)=-\frac{1}{2}\cos(2t)\begin{bmatrix}1\\-0.5\end{bmatrix}+\left(\frac{1}{4}\sin(2t)+\frac{1}{2}\cos(2t)\right)\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}-\frac{1}{2}\cos(2t)+\frac{1}{4}\sin(2t)+\frac{1}{2}\cos(2t)\\ -\frac{1}{4}\cos(2t)-\frac{1}{2}\sin(2t)+\frac{1}{4}\sin(2t)+\frac{1}{2}\cos(2t) \end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$$ Since the particular solution \(\mathbf{x}_p(t)\) is the zero vector, we conclude that the general solution of the nonhomogeneous system is the same as the general solution of the homogeneous system: $$\mathbf{x}(t)=c_1e^{2t}\begin{bmatrix}1\\-0.5\end{bmatrix}+c_2e^{-2t}\begin{bmatrix}1\\1\end{bmatrix}$$

Key Concepts

Matrix Exponential

When dealing with linear differential equations, the matrix exponential plays a crucial role, especially in solving systems of linear differential equations. But what exactly is the matrix exponential? Imagine it as an extension of the scalar exponential function to matrices. This allows us to handle differential equations involving matrices.

For a given matrix \(A\), the matrix exponential is defined as:

• \(e^{At} = \sum_{n=0}^{\infty} \frac{A^n t^n}{n!}\)

This series is analogous to the Taylor series used for scalar exponentials. It's derived from integrating the concept of repeated multiplication into matrices. This formulation helps compute solutions to matrix equations of the form \(\mathbf{x}' = A \mathbf{x}\).

To effectively compute the matrix exponential, we leverage the eigenvalues and eigenvectors of the matrix \(A\). These provide invaluable insight and make it easier to derive expressions for solving differential equations.

Eigenvalues and Eigenvectors

A matrix can tell us a lot more about a system when we break it down into its eigenvalues and eigenvectors. These are not just abstract terms; they provide critical insights into the behavior of matrices.

**Defining Eigenvalues and Eigenvectors**

• For a square matrix \(A\), an eigenvalue \(\lambda\) is a scalar that satisfies the equation \(A \mathbf{v} = \lambda \mathbf{v}\), where \(\mathbf{v}\) is the associated eigenvector.

Discovering eigenvalues involves solving the characteristic equation \(\text{det}(A - \lambda I) = 0\). The eigenvectors are derived by substituting each eigenvalue back into the equation \((A - \lambda I)\mathbf{v} = 0\).

Identifying the eigenvalues helps anticipate how a system evolves over time, as these values can indicate whether solutions to differential equations grow, decay, or oscillate. Correspondingly, eigenvectors provide the direction in which these changes occur, allowing us to decompose complex matrix operations into simpler, understandable components.

Homogeneous Differential Equations

Homogeneous differential equations are essential in understanding the dynamics of a system. They are expressed in the form \(\mathbf{x}' = A \mathbf{x}\), where \(A\) is a matrix, and \(\mathbf{x}\) is a vector function. These equations describe systems where the system’s future is entirely determined by its current state without any external input.

In a homogeneous system, solutions typically take the form of exponential functions of time, which are influenced by the matrix \(A\). The general solution to a homogeneous system can be formulated using a linear combination of terms derived from the eigenvalues and eigenvectors of \(A\).

**General Solution**

• The solution involves using the matrix exponential derived from \(A\), manifesting as \(e^{At}\), which tells us about the system’s evolution over time.
• Each term in the general solution represents a possible state of the system, formed as linear combinations of eigenvectors weighted by functions of time.

Understanding homogeneous systems is critical because once you grasp their behavior, handling more complex non-homogeneous systems (by adding external inputs) becomes more manageable. This foundational understanding serves as a stepping-stone to comprehending more intricate dynamic systems.