Question

Find the general solution of the given system of equations. $$ \mathbf{x}^{\prime}=\left(\begin{array}{rr}{-\frac{5}{4}} & {\frac{3}{4}} \\\ {\frac{3}{4}} & {-\frac{5}{4}}\end{array}\right) \mathbf{x}+\left(\begin{array}{c}{2 t} \\ {e^{t}}\end{array}\right) $$

Short answer

Answer: The general solution to the given system of linear first-order differential equations is $$ \mathbf{x}(t) = c_1 \left(\begin{array}{c}{1} \\\ {1}\end{array}\right) + c_2 \left(\begin{array}{c}{1} \\\ {-1}\end{array}\right) e^{-\frac{5}{2}t} + \left(\begin{array}{c}{-4t^2 + 4t - 8} \\\ {4e^t}\end{array}\right) $$

Step-by-step solution

Finding the Homogeneous Solution

To find the homogeneous solution, consider the system $$ \mathbf{x}^{\prime} = \mathbf{A}\mathbf{x} $$ Here matrix \(\mathbf{A} = \left(\begin{array}{rr}{-\frac{5}{4}} & {\frac{3}{4}} \\\ {\frac{3}{4}} & {-\frac{5}{4}}\end{array}\right)\). To get the homogeneous solution, we must find eigenvalues and eigenvectors of \(\mathbf{A}\). Start by finding the characteristic equation of \(\mathbf{A}\), given by $$ \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 $$ Here \(\mathbf{I}\) is the identity matrix. Substituting the given matrix \(\mathbf{A}\), we get: $$ \text{det}\left(\begin{array}{cc}{-\frac{5}{4}-\lambda} & {\frac{3}{4}} \\\ {\frac{3}{4}} & {-\frac{5}{4}-\lambda}\end{array}\right)=\left(-\frac{5}{4}-\lambda\right)^2-\left(\frac{3}{4}\right)^2 = \lambda^2-(-\frac{5}{2})\lambda$$ Now find the eigenvalues by solving the characteristic equation: $$ \lambda^2 -(-\frac{5}{2})\lambda = \lambda (\lambda + \frac{5}{2}) = 0 $$ The eigenvalues are \(\lambda_1 = 0\) and \(\lambda_2 = -\frac{5}{2}\). Next, find the corresponding eigenvectors by solving the equation \((\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = 0\) for each eigenvalue: For \(\lambda_1 = 0\): $$\left(\begin{array}{cc}{-\frac{5}{4}} & {\frac{3}{4}} \\\ {\frac{3}{4}} & {-\frac{5}{4}}\end{array}\right)\left(\begin{array}{c}{v_{11}} \\\ {v_{12}}\end{array}\right) = \left(\begin{array}{c}{0} \\\ {0}\end{array}\right)$$ We get a corresponding eigenvector \(\mathbf{v}_1 = \left(\begin{array}{c}{1} \\\ {1}\end{array}\right)\). For \(\lambda_2 = -\frac{5}{2}\): $$\left(\begin{array}{cc}{\frac{3}{4}} & {\frac{3}{4}} \\\ {\frac{3}{4}} & {\frac{3}{4}}\end{array}\right)\left(\begin{array}{c}{v_{21}} \\\ {v_{22}}\end{array}\right) = \left(\begin{array}{c}{0} \\\ {0}\end{array}\right)$$ We get a corresponding eigenvector \(\mathbf{v}_2 = \left(\begin{array}{c}{1} \\\ {-1}\end{array}\right)\). Hence, the homogeneous solution is given by $$ \mathbf{x}_h(t) = c_1 \mathbf{v}_1 e^{\lambda_1 t} + c_2 \mathbf{v}_2 e^{\lambda_2 t} = c_1 \left(\begin{array}{c}{1} \\\ {1}\end{array}\right) + c_2 \left(\begin{array}{c}{1} \\\ {-1}\end{array}\right) e^{-\frac{5}{2}t} $$

Finding the Particular Solution

To find a particular solution \(\mathbf{x}_p(t)\), we must find the form of \(\mathbf{x}_p(t)\) that solves the non-homogeneous equation: $$\mathbf{x}^{\prime} = \mathbf{A}\mathbf{x} + \mathbf{b}(t)$$ with \(\mathbf{b}(t) = \left(\begin{array}{c}{2t} \\\ {e^{t}}\end{array}\right)\). Since \(\mathbf{b}(t)\) is a polynomial and an exponential function, we can assume the form of \(\mathbf{x}_p(t)\) as: $$\mathbf{x}_p(t) = \left(\begin{array}{c}{at^2 + bt + c} \\\ {de^t}\end{array}\right)$$ Differentiate \(\mathbf{x}_p(t)\) with respect to \(t\) to obtain the first derivative: $$\mathbf{x}_p^{\prime}(t) = \left(\begin{array}{c}{2at + b} \\\ {de^t}\end{array}\right)$$ Now, substituting \(\mathbf{x}_p(t)\) and \(\mathbf{x}_p^{\prime}(t)\) into the non-homogeneous equation, we get: $$\left(\begin{array}{c}{2at + b} \\\ {de^t}\end{array}\right) = \left(\begin{array}{rr}{-\frac{5}{4}} & {\frac{3}{4}} \\\ {\frac{3}{4}} & {-\frac{5}{4}}\end{array}\right)\left(\begin{array}{c}{at^2 + bt + c} \\\ {de^t}\end{array}\right) + \left(\begin{array}{c}{2t} \\\ {e^{t}}\end{array}\right)$$ Solving the system of equations for \(a\), \(b\), \(c\), and \(d\), we get the particular solution: $$\mathbf{x}_p(t) = \left(\begin{array}{c}{-4t^2 + 4t - 8} \\\ {4e^t}\end{array}\right)$$

Finding the General Solution

To find the general solution, add the homogeneous solution and the particular solution: $$ \mathbf{x}(t) = \mathbf{x}_h(t) + \mathbf{x}_p(t) $$ $$ \mathbf{x}(t) = c_1 \left(\begin{array}{c}{1} \\\ {1}\end{array}\right) + c_2 \left(\begin{array}{c}{1} \\\ {-1}\end{array}\right) e^{-\frac{5}{2}t} + \left(\begin{array}{c}{-4t^2 + 4t - 8} \\\ {4e^t}\end{array}\right) $$ This is the general solution to the given system of equations.

Key concepts

Homogeneous Solution

When dealing with linear systems of differential equations, a critical step is finding the homogeneous solution. The process begins by examining the system without any external forces or inputs, thus focusing on the equation \( \mathbf{x}' = \mathbf{A}\mathbf{x} \). Here, \( \mathbf{A} \) represents a matrix that contains the coefficients of the system.

To solve this, we seek solutions of the form \( \mathbf{x}_h(t) = \mathbf{v} e^{\lambda t} \), which led us to the concept of eigenvalues and eigenvectors. To find these, we calculate the characteristic equation \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where \( \mathbf{I} \) is the identity matrix.

By substituting the given matrix \( \mathbf{A} \), we solve the quadratic characteristic equation for \( \lambda \). The solutions \( \lambda_1 = 0 \) and \( \lambda_2 = -\frac{5}{2} \) are the eigenvalues. Each eigenvalue corresponds to an eigenvector, which is found by solving \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = 0 \).

• For \( \lambda_1 = 0 \), the eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
• For \( \lambda_2 = -\frac{5}{2} \), the eigenvector is \( \mathbf{v}_2 = \begin{pmatrix} 1 \ -1 \end{pmatrix} \).

The homogeneous solution consequently becomes a linear combination of these eigenvectors, characterized by arbitrary constants \( c_1 \) and \( c_2 \). Thus, we have
\( \mathbf{x}_h(t) = c_1 \begin{pmatrix} 1 \ 1 \end{pmatrix} + c_2 \begin{pmatrix} 1 \ -1 \end{pmatrix} e^{-\frac{5}{2}t} \).

Eigenvalues and Eigenvectors

Discovering eigenvalues and eigenvectors is a fundamental step in solving systems of linear differential equations. They help us understand the behavior of these systems over time.

To begin, we find the characteristic equation by computing \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \). This equation is derived from the matrix subtraction of \( \mathbf{A} \) and \( \lambda \mathbf{I} \), where \( \mathbf{I} \) is the identity matrix. The result is a polynomial whose roots are the eigenvalues \( \lambda \).

• For example, with \( \mathbf{A} = \begin{pmatrix} -\frac{5}{4} & \frac{3}{4} \ \frac{3}{4} & -\frac{5}{4} \end{pmatrix} \), we calculated that the eigenvalues are \( \lambda_1 = 0 \) and \( \lambda_2 = -\frac{5}{2} \).

Once the eigenvalues are known, we find the eigenvectors by solving \( (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = 0 \). This process involves substituting each eigenvalue back into the matrix equation and solving for the vector \( \mathbf{v} \), which satisfies the equation.

• For \( \lambda_1 = 0 \), the eigenvector is \( \mathbf{v}_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
• For \( \lambda_2 = -\frac{5}{2} \), the eigenvector is \( \mathbf{v}_2 = \begin{pmatrix} 1 \ -1 \end{pmatrix} \).

Eigenvectors indicate the direction in which the system evolves, while the corresponding eigenvalues determine the rate at which the system moves along these directions.

Particular Solution

Finding the particular solution of a non-homogeneous system of differential equations is key to obtaining the complete system behavior. The system \( \mathbf{x}^{\prime} = \mathbf{A}\mathbf{x} + \mathbf{b}(t) \) includes an external input or forcing function \( \mathbf{b}(t) \), such as \( \begin{pmatrix} 2t \ e^t \end{pmatrix} \).

To tackle this, we assume a form for the particular solution \( \mathbf{x}_p(t) \) that mirrors the structure of \( \mathbf{b}(t) \). Since \( \mathbf{b}(t) \) combines both polynomial and exponential terms, we use a trial solution of similar types:
\( \mathbf{x}_p(t) = \begin{pmatrix} at^2 + bt + c \ de^t \end{pmatrix} \).

We differentiate this assumed solution to find \( \mathbf{x}_p^{\prime}(t) \), which will then be substituted back into the original differential equation. This substitution allows us to equate and solve for the unknown coefficients \( a, b, c, \text{ and } d \).

• Through this process, we derive that
  \( \mathbf{x}_p(t) = \begin{pmatrix} -4t^2 + 4t - 8 \ 4e^t \end{pmatrix} \) satisfies the non-homogeneous system.

Thus, the particular solution captures the response of the system directly driven by \( \mathbf{b}(t) \), linking into the broader general solution \( \mathbf{x}(t) \) when combined with the homogeneous solution.