Problem 14 - Differential Equations and Linear Algebra

Question

Solve the given non homogeneous system. $$x_{1}^{\prime}=x_{1}+x_{2}+e^{2 t}, \quad x_{2}^{\prime}=3 x_{1}-x_{2}+5 e^{2 t}$$

Step-by-Step Solution

Answer

The general solution of the given non-homogeneous system of first-order linear differential equations is given by: \[ \boldsymbol{x}(t) = C_1 e^{(1 + i\sqrt{3})t} \begin{pmatrix} 1 \\ -1 + i\sqrt{3} \end{pmatrix} + C_2 e^{(1 - i\sqrt{3})t} \begin{pmatrix} 1 \\ -1 - i\sqrt{3} \end{pmatrix} + \begin{pmatrix} 1 \\ -2 \end{pmatrix} e^{2t} \] where \(C_1\) and \(C_2\) are constants.

Step 1: Find the general solution of the corresponding homogeneous system

To find the general solution of the corresponding homogeneous system, we need to consider the following system of equations without the non-homogeneous terms: \[ x_1' - x_1 - x_2 = 0 \] \[ x_2' - 3x_1 + x_2 = 0 \] We can rewrite it in matrix form: \[ \begin{pmatrix} x_1' \\ x_2' \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ -3 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \] Next, we find the eigenvalues and eigenvectors of the matrix.

Step 2: Find the eigenvalues and eigenvectors

To find the eigenvalues, we solve the characteristic equation: \[ \text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0 \] where \(\mathbf{A}\) is the matrix of the homogeneous system, \(\lambda\) is the eigenvalue, and \(\mathbf{I}\) is the identity matrix. After calculating the determinant, we have: \[ (1 - \lambda)(1 - \lambda) - (1)(-3) = 0\] \[ \lambda^2 - 2\lambda + 4 = 0\] The roots of this equation are \(\lambda_1 = 1 + i\sqrt{3}\) and \(\lambda_2 = 1 - i\sqrt{3}\). Now we will find the eigenvectors corresponding to each eigenvalue. For each eigenvalue, we will solve the linear equation: \[ (\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = 0\] For \(\lambda_1 = 1 + i\sqrt{3}\), the linear equation becomes: \[ \begin{pmatrix} -i\sqrt{3} & 1 \\ -3 & -i\sqrt{3} \end{pmatrix} \begin{pmatrix} v_{11} \\ v_{12} \end{pmatrix} = 0\] And for \(\lambda_2 = 1 - i\sqrt{3}\), the linear equation becomes: \[ \begin{pmatrix} i\sqrt{3} & 1 \\ -3 & i\sqrt{3} \end{pmatrix} \begin{pmatrix} v_{21} \\ v_{22} \end{pmatrix} = 0\] After solving these equations, we obtain the eigenvectors \(v_1 = \begin{pmatrix} 1 \\ -1 + i\sqrt{3} \end{pmatrix}\) and \(v_2 = \begin{pmatrix} 1 \\ -1 - i\sqrt{3} \end{pmatrix}\).

Step 3: Write down the general solution for the homogeneous system

Using the eigenvalues and eigenvectors, we can now write down the general solution for the homogeneous system: \[ \boldsymbol{x}_h(t) = C_1 e^{(1 + i\sqrt{3})t} \begin{pmatrix} 1 \\ -1 + i\sqrt{3} \end{pmatrix} + C_2 e^{(1 - i\sqrt{3})t} \begin{pmatrix} 1 \\ -1 - i\sqrt{3} \end{pmatrix} \]

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

To find a particular solution for the non-homogeneous system, we assume a particular solution of the form: \[ \boldsymbol{x}_p(t) = \begin{pmatrix} A \\ B \end{pmatrix} e^{2t}\] After plugging \(\boldsymbol{x}_p(t)\) into the non-homogeneous system and solving for \(A\) and \(B\), we get: \[ A = 1 \quad B = -2 \] So, the particular solution is: \[ \boldsymbol{x}_p(t) = \begin{pmatrix} 1 \\ -2 \end{pmatrix} e^{2t}\]

Step 5: Combine both solutions and write down the general solution of the non-homogeneous system

Finally, we can combine the general solution of the homogeneous system, \(\boldsymbol{x}_h(t)\), and the particular solution, \(\boldsymbol{x}_p(t)\), to get the general solution of the non-homogeneous system: \[ \boldsymbol{x}(t) = \boldsymbol{x}_h(t) + \boldsymbol{x}_p(t) = C_1 e^{(1 + i\sqrt{3})t} \begin{pmatrix} 1 \\ -1 + i\sqrt{3} \end{pmatrix} + C_2 e^{(1 - i\sqrt{3})t} \begin{pmatrix} 1 \\ -1 - i\sqrt{3} \end{pmatrix} + \begin{pmatrix} 1 \\ -2 \end{pmatrix} e^{2t} \] where \(C_1\) and \(C_2\) are constants.

Key Concepts

Non-homogeneous systems

A non-homogeneous system of differential equations includes terms that are not solely dependent on the functions themselves, making them a tad more complex than homogeneous systems. In homogeneous systems, each equation is set to zero, while in non-homogeneous systems, extra functions like exponential or sine waves, also known as forcing terms, are present.
For example, in our given system:

• \(x_1' = x_1 + x_2 + e^{2t}\)
• \(x_2' = 3x_1 - x_2 + 5e^{2t}\)

The terms \(e^{2t}\) and \(5e^{2t}\) make this system non-homogeneous. These terms represent external forces or inputs that modify the behavior of the system. Understanding these terms is crucial since they influence the solution, diversifying it beyond what would be seen in the absence of such terms.

Eigenvalues and eigenvectors

Finding eigenvalues and eigenvectors is a vital step in solving linear systems of equations. They help transform and simplify the system from complex differential equations into solvable expressions. Given a matrix \(\mathbf{A}\), we first calculate the eigenvalues by solving the characteristic equation:\[\text{det}(\mathbf{A} - \lambda \mathbf{I}) = 0\]Here, \(\lambda\) represents our eigenvalues, and \(\mathbf{I}\) is the identity matrix.
In our example, the eigenvalues were found to be complex: \(\lambda_1 = 1 + i\sqrt{3}\) and \(\lambda_2 = 1 - i\sqrt{3}\). Using these, we compute eigenvectors, which are non-zero vectors invariant to scaling by \(\mathbf{A}\). Solving \((\mathbf{A} - \lambda \mathbf{I})\mathbf{v} = 0\) for each \(\lambda\) determines the direction in which the system evolves over time.

• Eigenvector for \(\lambda_1\): \(\begin{pmatrix} 1 \ -1 + i\sqrt{3} \end{pmatrix}\)
• Eigenvector for \(\lambda_2\): \(\begin{pmatrix} 1 \ -1 - i\sqrt{3} \end{pmatrix}\)

General solution

The general solution to a differential equation includes all possible solutions derived from the associated homogeneous system. It provides a comprehensive view of how the state of the system evolves over time. By combining the structure of the eigenvectors and the exponential of the eigenvalues, we can build the general solution as follows:\[\boldsymbol{x}_h(t) = C_1 e^{(1 + i\sqrt{3})t} \begin{pmatrix} 1 \ -1 + i\sqrt{3} \end{pmatrix} + C_2 e^{(1 - i\sqrt{3})t} \begin{pmatrix} 1 \ -1 - i\sqrt{3} \end{pmatrix}\]Here, \(C_1\) and \(C_2\) are arbitrary constants, indicating the multitude of solutions that cover all initial condition scenarios. Understanding the oscillatory nature induced by the imaginary components of the eigenvalues \(i\sqrt{3}\) is crucial here, as it signifies a tendency to rotate or spiral through the phase space trajectory.

Particular solution

For a non-homogeneous system, the particular solution addresses the specific changes induced by non-homogeneous terms like \(e^{2t}\). Unlike the general solution, which accounts for a system's inherent properties, the particular solution targets the distinct, constant forcing functions. In our problem, we hypothesized a particular solution of\[\boldsymbol{x}_p(t) = \begin{pmatrix} A \ B \end{pmatrix} e^{2t}\]After substituting into the system, we found:

• \(A = 1\)
• \(B = -2\)

This gives us the particular solution \(\boldsymbol{x}_p(t) = \begin{pmatrix} 1 \ -2 \end{pmatrix} e^{2t}\). This solution captures the effect of the external forces, ensuring our full system solution reflects both inherent dynamics and the impact of non-homogeneous terms.