Problem 12 - Differential Equations and Linear Algebra

Question

Solve the given non homogeneous system. $$x_{1}^{\prime}=x_{1}+2 x_{2}+5 e^{4 t}, \quad x_{2}^{\prime}=2 x_{1}+x_{2}$$

Step-by-Step Solution

Answer

The general solution for the given non-homogeneous system is: \[ \vec{x}(t) = c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} e^{3t} + c_2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^{-t} + \begin{pmatrix} \frac{5}{8} \\ \frac{5}{16} \end{pmatrix} e^{4t}. \]

Step 1: Write down the homogeneous system

From the given system, the homogeneous system will have the same coefficients but with no forcing term: \[ \begin{cases} x_{1}^{\prime} = x_{1} + 2x_{2} \\ x_{2}^{\prime} = 2x_{1} + x_{2} \end{cases} \]

Step 2: Compute the eigenvalues and eigenvectors of the matrix

We write down the matrix of coefficients and compute its eigenvalues and eigenvectors: \(A= \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \) To find the eigenvalues, we'll solve the characteristic equation: \(|A - \lambda I| = 0\) where \(I\) is the identity matrix. \(|A - \lambda I| = \begin{vmatrix} 1 - \lambda & 2 \\ 2 & 1 - \lambda \end{vmatrix} = (1-\lambda)^{2} - (2)(2) = 0\) \(\lambda^2 - 2\lambda - 3 = (\lambda-3)(\lambda+1)=0\) So the eigenvalues are \(\lambda_1 = 3\) and \(\lambda_2 = -1\). Next, we'll find the eigenvectors associated with each eigenvalue: Eigenvectors for \(\lambda_1 = 3\): \( (A-3 I) \vec{v}_1 = 0 \Rightarrow \begin{pmatrix} -2 & 2 \\ 2 & -2 \end{pmatrix} \vec{v}_1 = 0 \) From the equation, we can deduce that the eigenvector is \(\vec{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) Eigenvectors for \(\lambda_2 = -1\): \( (A+1 I) \vec{v}_2 = 0 \Rightarrow \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix} \vec{v}_2 = 0 \) From the equation, we can deduce that the eigenvector is \(\vec{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\)

Step 3: Solve the homogeneous system

Now that we have the eigenvalues and eigenvectors, we can write down the general solution for the homogeneous system: \[ \vec{x}_h(t) = c_1 \vec{v}_1 e^{\lambda_1 t} + c_2 \vec{v}_2 e^{\lambda_2 t} = c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} e^{3t} + c_2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^{-t}, \] where \(c_1\) and \(c_2\) are constants to be determined by the initial conditions.

Step 4: Find a particular solution for the non-homogeneous system

The non-homogeneous term in the given system is \(5 e^{4t}\), so we will try to find a solution of the form \(\vec{x}_p(t) = \begin{pmatrix} a(t) \\ b(t) \end{pmatrix} e^{4t}\). Plugging this into the non-homogeneous system, we get: \[ \begin{pmatrix} a'(t) \\ b'(t) \end{pmatrix} e^{4t} = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} a(t) \\ b(t) \end{pmatrix} e^{4t} + \begin{pmatrix} 5 \\ 0 \end{pmatrix} e^{4t} \] Divide by \(e^{4t}\) and differentiate with respect to \(t\): \[ \begin{pmatrix} a''(t) - 4a'(t) \\ b''(t) - 4b'(t) \end{pmatrix} = \begin{pmatrix} -8a(t) + 16b(t) \\ 16a(t) - 8b(t) \end{pmatrix} + \begin{pmatrix} 5 \\ 0 \end{pmatrix} \] Comparing the components of the matrices, we obtain the following system of equations: \[ \begin{cases} a''(t) - 4a'(t) = -8a(t) + 16b(t) + 5 \\ b''(t) - 4b'(t) = 16a(t) - 8b(t) \end{cases} \] To make the right-hand side zero, we will set \(a(t) = \frac{5}{8}\) and \(b(t) = \frac{5}{16}\). Therefore, a particular solution for the non-homogeneous system is \(\vec{x}_p(t) = \begin{pmatrix} \frac{5}{8} \\ \frac{5}{16} \end{pmatrix} e^{4t}\).

Step 5: Write down the general solution of the non-homogeneous system

Finally, combining the general solution of the homogeneous system and the particular solution we found, the general solution for given non-homogeneous system will be: \[ \vec{x}(t) = \vec{x}_h(t) + \vec{x}_p(t) = c_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} e^{3t} + c_2 \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^{-t} + \begin{pmatrix} \frac{5}{8} \\ \frac{5}{16} \end{pmatrix} e^{4t}. \]

Key Concepts

eigenvalues and eigenvectors

Eigenvalues and eigenvectors are key concepts when solving differential equations involving matrices. They help us understand the dynamics of a system.
To find the eigenvalues of a matrix, we need to solve the characteristic equation, which is formulated as \(|A - \lambda I| = 0\). Here, \(A\) is the original matrix, \(\lambda\) represents eigenvalues, and \(I\) is the identity matrix.
In our exercise, the matrix is \(A = \begin{pmatrix} 1 & 2 \ 2 & 1 \end{pmatrix}\). Solving \((\lambda^2 - 2\lambda - 3 = 0)\), we find its eigenvalues are \(3\) and \(-1\).

• For each eigenvalue, we solve \((A - \lambda I) v = 0\) to find corresponding eigenvectors.
• For \(\lambda_1 = 3\), we find \(v_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix}\).
• For \(\lambda_2 = -1\), \(v_2 = \begin{pmatrix} 1 \ -1 \end{pmatrix}\).

Eigenvalues indicate growth or decay rates in the system, while eigenvectors show the direction in which these changes occur.

particular solution

A particular solution targets the non-homogeneous part of a differential equation. It solves the equation without the constants of the general solution.
In our case, the non-homogeneous part is \(5e^{4t}\). We propose a solution of the form \(\vec{x}_p(t) = \begin{pmatrix} a(t) \ b(t) \end{pmatrix}e^{4t}\).
Plugging this form into the original equations allows us to solve for \(a(t)\) and \(b(t)\). Through substitutions and some calculations, we find:

• \(a(t) = \frac{5}{8}\),
• \(b(t) = \frac{5}{16}\)

Hence, the particular solution is \(\vec{x}_p(t) = \begin{pmatrix} \frac{5}{8} \ \frac{5}{16} \end{pmatrix} e^{4t}\). This is the part of the solution that counterbalances the external forcing term \(5e^{4t}\), allowing us to agree with the non-homogeneous nature of the system.

general solution

The general solution of a differential equation combines the solutions of both its homogeneous and particular forms. This provides a complete picture of the system's behavior.
For our non-homogeneous system, the solution starts with solving its homogeneous version, which we derived with eigenvalues and eigenvectors.
The general solution is written as:
\(\vec{x}(t) = \vec{x}_h(t) + \vec{x}_p(t) \), where:

• \(\vec{x}_h(t) = c_1 \begin{pmatrix} 1 \ 1 \end{pmatrix} e^{3t} + c_2 \begin{pmatrix} 1 \ -1 \end{pmatrix} e^{-t}\), representing solutions to the homogeneous part
• \(\vec{x}_p(t) = \begin{pmatrix} \frac{5}{8} \ \frac{5}{16} \end{pmatrix} e^{4t}\), the particular solution

Constants \(c_1\) and \(c_2\) are determined through initial conditions not provided here.
The general solution captures all possible behaviors of the system, including natural dynamics and external influences.

homogeneous system

A homogeneous system of differential equations has no external forcing terms, represented by zeros on the right side of the equation. The homogeneous system derived from the given problem is:
\(\begin{cases} x_{1}^{\prime} = x_{1} + 2x_{2} \ x_{2}^{\prime} = 2x_{1} + x_{2} \end{cases}\).
Solving this allows us to discover the system's inherent characteristics using eigenvalues and eigenvectors. These eigenvalues and eigenvectors depict how the system would evolve over time without external disturbances.

• Eigenvalues indicate rates of growth or decay.
• Eigenvectors reveal the direction of these rates in the state space.

Essentially, the homogeneous solution provides the foundation onto which particular solutions build, showing the system's natural behavior, absent of any external forcing influence.