Question

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{ll}{2} & {-5} \\ {1} & {-2}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{0} \\ {\cos t}\end{array}\right), \quad 0

Short answer

Question: Determine the general solution of the non-homogeneous system of linear differential equations given by: $$\begin{cases}x'(t) = 2x(t) - 5y(t)\\ y'(t) = x(t) - 2y(t) + \cos(t) \end{cases}$$ Answer: The general solution of the given system of equations is: $$\mathbf{x}(t) = C_1 e^{t}\left(\begin{array}{c}{5}\ \\ 1\end{array}\right) + C_2 e^{-t}\left(\begin{array}{c}{5}\ \\ 3\end{array}\right) + \left(\begin{array}{c}{-\frac{5}{21}\cos{t}}\ \\ \frac{1}{21}\cos{t} - \frac{5}{21}\sin{t}\end{array}\right)$$

Step-by-step solution

Find the eigenvalues and eigenvectors of the matrix

We need to find the eigenvalues by solving the following characteristic equation: $$\text{det} \begin{vmatrix} 2-\lambda & -5 \\ 1 & -2-\lambda\end{vmatrix} = 0$$ Calculating the determinant, we get: $$(2-\lambda)(-2-\lambda)-(-5\cdot 1)=\lambda^2 - 1$$ Solve for the eigenvalues: $$\lambda^2 - 1=0 \Rightarrow \lambda_1=1, \lambda_2=-1$$ Now, find the eigenvectors for each eigenvalue: For \(\lambda_1 = 1\): $$\begin{cases}(2-1)v_{11} - 5v_{12} = 0\\ v_{21} - (2-1)v_{22}=0 \end{cases} \Rightarrow \begin{cases}v_{11} - 5v_{12} = 0\\ v_{21} - v_{22}=0 \end{cases}$$ With arbitrary constants \(c_1\) and \(c_2\) as constants of integration, the eigenvector is: $$\mathbf{v}_1\left(\begin{array}{c}{5c_1}\ \\ c_1\end{array}\right) = c_1\left(\begin{array}{c}{5}\ \\ 1\end{array}\right)$$ For \(\lambda_2 = -1\): $$\begin{cases}(2+1)v_{11} - 5v_{12} = 0\\ v_{21} - (2+1)v_{22}=0 \end{cases} \Rightarrow \begin{cases}3v_{11} - 5v_{12} = 0\\ v_{21} - 3v_{22}=0 \end{cases}$$ The eigenvector is: $$\mathbf{v}_2\left(\begin{array}{c}{5c_2}\ \\ 3c_2\end{array}\right) = c_2\left(\begin{array}{c}{5}\ \\ 3\end{array}\right)$$

Find the general solution of the homogeneous system

The general solution of the homogeneous system is given by a linear combination of eigenvectors and their corresponding exponential terms: $$\mathbf{x_h}(t)=C_1 e^{\lambda_1 t}\mathbf{v}_1 + C_2 e^{\lambda_2 t}\mathbf{v}_2 = C_1 e^{t}\left(\begin{array}{c}{5}\ \\ 1\end{array}\right) + C_2 e^{-t}\left(\begin{array}{c}{5}\ \\ 3\end{array}\right)$$

Find a particular solution of the non-homogeneous system

We can use the method of undetermined coefficients to find a particular solution of the non-homogeneous system. Assume a particular solution of the form: $$\mathbf{x_p}(t) = \left(\begin{array}{c}{A \cos{t}}\ \\ B \cos{t} + C \sin{t}\end{array}\right)$$ Differentiate \(\mathbf{x_p}(t)\): $$\mathbf{x_p'}(t) = \left(\begin{array}{c}{-A \sin{t}}\ \\ -B \sin{t} + C \cos{t}\end{array}\right)$$ Plug \(\mathbf{x_p}(t)\) and \(\mathbf{x_p'}(t)\) into the non-homogeneous system and equate coefficients of like terms: $$\begin{cases}(-A \sin{t})=2A \cos{t} - 5B \cos{t}\\ (-B \sin{t} + C \cos{t})=A \cos{t} - 2B \cos{t} + \cos{t}\end{cases}$$ Equate the coefficients to get: $$\begin{cases}A=-5B\\ 2A - 5B=-1\\ C=A-2B\\ -B=C \end{cases}$$ Solve for A, B, and C: $$A=-\frac{5}{21}, B=\frac{1}{21}, C=-\frac{5}{21}$$ The particular solution is: $$\mathbf{x_p}(t) = \left(\begin{array}{c}{-5/21 \cos{t}}\ \\ \frac{1}{21} \cos{t} - \frac{5}{21} \sin{t}\end{array}\right)$$

Find the general solution

Add the general solution of the homogeneous system (Step 2) with the particular solution (Step 3): $$\mathbf{x}(t) = \mathbf{x_h}(t) + \mathbf{x_p}(t) = C_1 e^{t}\left(\begin{array}{c}{5}\ \\ 1\end{array}\right) + C_2 e^{-t}\left(\begin{array}{c}{5}\ \\ 3\end{array}\right) + \left(\begin{array}{c}{-5/21 \cos{t}}\ \\ 1/21 \cos{t} - 5/21 \sin{t}\end{array}\right)$$ This is the general solution of the given system of equations.

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra, playing a crucial role in the analysis of linear transformations. An eigenvalue is a special scalar associated with a linear system of equations; it's a value \( \lambda \) for which there exists a non-zero vector \( \mathbf{v} \)—called an eigenvector—such that when the linear transformation represented by matrix \( A \) acts on \( \mathbf{v} \) the output is a scalar multiple of \( \mathbf{v} \), or \( A\mathbf{v} = \lambda\mathbf{v} \).

Eigenvectors and eigenvalues provide insight into the properties of a matrix, such as whether a system of equations will have non-trivial solutions. They are particularly useful in solving differential equations, as seen in the provided exercise, where the eigenvectors serve as a basis for the solution space of the homogeneous equation.

Homogeneous System

A homogeneous system of linear equations is one where all constant terms are zero, formally written as \( A\mathbf{x} = \mathbf{0} \), where \( A \) is a matrix, \( \mathbf{x} \) is the vector of unknowns, and \( \mathbf{0} \) is the zero vector. The significance of a homogeneous system hinges on the fact that it always has at least one solution: the trivial solution where all variables are zero.

The solution space of a homogeneous system may contain infinite solutions, which form a vector space spanned by the system's eigenvectors. In the context of the exercise, solving the homogeneous system involves finding a combination of eigenvectors multiplied by exponential functions of eigenvalues, which lays the foundation for constructing the complete solution to the entire system.

Non-Homogeneous System

In contrast to a homogeneous system, a non-homogeneous system includes non-zero constants, given by \( A\mathbf{x} = \mathbf{b} \), where \( \mathbf{b} \) is non-zero. Non-homogeneous systems often arise when dealing with external forces or inputs, as seen in the exercise with the adding of \( \cos t \) on the right side of the equation.

One way to tackle such systems is to find a particular solution \( \mathbf{x_p} \) that satisfies the non-homogeneous equation. Methods like undetermined coefficients allow us to guess a form for \( \mathbf{x_p} \), as shown in the solution process for the exercise. This particular solution, when added to the general solution of the related homogeneous system, gives the complete solution to the non-homogeneous system.

Characteristic Equation

The characteristic equation of a matrix is a critical tool for finding its eigenvalues. It is derived from the equation \( \det(A - \lambda I) = 0 \) where \( A \) is a square matrix, \( \lambda \) is a scalar, \( I \) is the identity matrix, and \( \det \) refers to the determinant of a matrix. The solutions to this equation provide us with the eigenvalues of \( A \).

The process of solving the characteristic equation can often be simplified by recognizing patterns in the matrix, such as a diagonal or triangular form. In the given exercise, the characteristic equation yields a quadratic equation that is straightforward to solve, revealing the eigenvalues that are essential to build the general solution of the system.