Question

Find the particular solution of the linear system that satisfies the stated initial conditions. $$ \begin{aligned} &x^{\prime}=3 x+5 y \\ &y^{\prime}=-2 x+5 y \\ &x(0)=5, y(0)=-1 \end{aligned} $$

Short answer

The particular solution of the linear system that satisfies the given initial conditions is: \[ \begin{pmatrix} x(t) \\ y(t) \\ \end{pmatrix} = (2 - i)e^{(4 + 3i)t} \begin{pmatrix} 1 \\ \frac{1 + 3i}{2} \end{pmatrix} + (3 + i)e^{(4 - 3i)t} \begin{pmatrix} 1 \\ \frac{1 - 3i}{2} \end{pmatrix}. \]

Step-by-step solution

Rewrite the system in matrix form

We can rewrite the given system of first-order differential equations in matrix form as \[ \begin{pmatrix} x'(t) \\ y'(t) \\ \end{pmatrix} = \begin{pmatrix} 3 & 5 \\ -2 & 5 \\ \end{pmatrix} \begin{pmatrix} x(t) \\ y(t) \\ \end{pmatrix}. \]

Solve the resulting single-variable equation

The given problem is a linear, constant coefficients equation of first order. To solve it, find the eigenvalues and eigenvectors of the matrix. The characteristic equation is given by: \[ \text{det}(A - \lambda I) = 0, \] where A is the coefficient matrix, \(\lambda\) denotes the eigenvalues, and I represents the identity matrix. The determinant can be calculated as follows: \[ \text{det} \begin{pmatrix} 3-\lambda & 5 \\ -2 & 5-\lambda \end{pmatrix} = (3-\lambda)(5-\lambda) - (-2)(5) = \lambda^{2} - 8\lambda + 25. \]

Find the eigenvalues and eigenvectors

By solving the characteristic equation, we get the eigenvalues: \[ \lambda^{2} - 8\lambda + 25 = 0 \] \[ \lambda_1 = 4 + 3i, \quad \lambda_2 = 4 - 3i \] Now, we will find the eigenvectors corresponding to the eigenvalues. For \(\lambda_1 = 4 + 3i\), we solve the equation \((A - \lambda_1 I) v_1 = 0\). We get \[ \begin{pmatrix} -1-3i & 5 \\ -2 & 1-3i \end{pmatrix} \begin{pmatrix} v_{11} \\ v_{12} \end{pmatrix} = 0, \] and choose v_1 to be \[ v_1 = \begin{pmatrix} 1 \\ \frac{1 + 3i}{2} \end{pmatrix}. \] Similarly, for \(\lambda_2 = 4 - 3i\), we solve the equation \((A - \lambda_2 I) v_2 = 0\) and choose the eigenvectors to be \[ v_2 = \begin{pmatrix} 1 \\ \frac{1 - 3i}{2} \end{pmatrix}. \]

Construct the general solution

The general solution of the homogeneous system is given by \[ \begin{pmatrix} x(t) \\ y(t) \\ \end{pmatrix} = c_1 e^{(4 + 3i)t} \begin{pmatrix} 1 \\ \frac{1 + 3i}{2} \end{pmatrix} + c_2 e^{(4 - 3i)t} \begin{pmatrix} 1 \\ \frac{1 - 3i}{2} \end{pmatrix}, \] where \(c_1\) and \(c_2\) are constants.

Apply the initial conditions and find the particular solution

We are given the initial conditions \(x(0) = 5\) and \(y(0) = -1\). Substituting these into the general solution, we get: \[ \begin{pmatrix} 5 \\ -1 \\ \end{pmatrix} = c_1 e^{0} \begin{pmatrix} 1 \\ \frac{1 + 3i}{2} \end{pmatrix} + c_2 e^{0} \begin{pmatrix} 1 \\ \frac{1 - 3i}{2} \end{pmatrix}. \] Solving for the constants \(c_1\) and \(c_2\), we find that \(c_1 = 2 - i\) and \(c_2 = 3 + i\).

Write the particular solution

Now that we have the values of the constants, we can write the particular solution of the linear system that satisfies the given initial conditions as: \[ \begin{pmatrix} x(t) \\ y(t) \\ \end{pmatrix} = (2 - i)e^{(4 + 3i)t} \begin{pmatrix} 1 \\ \frac{1 + 3i}{2} \end{pmatrix} + (3 + i)e^{(4 - 3i)t} \begin{pmatrix} 1 \\ \frac{1 - 3i}{2} \end{pmatrix}. \]

Key concepts

Linear System of Differential Equations

A linear system of differential equations is a set of equations where each equation is a linear combination of the derivatives of one or more functions. In its simplest form, it can be represented as a matrix equation, where the matrix contains the coefficients linking the dependent variables and their derivatives. Understanding how to convert these equations into matrix form paves the way for employing more advanced techniques, like finding eigenvalues and eigenvectors, which are essential for solving the system.

For example, the given system in the exercise:
\[\begin{aligned}&x'(t)=3 x(t)+5 y(t) \&y'(t)=-2 x(t)+5 y(t) \end{aligned}\]
can be expressed in matrix form as a linear combination of the variables and their derivatives, allowing us to use tools from linear algebra to find a solution.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra. They prove especially useful when solving systems of differential equations, like the one presented in the exercise. The eigenvalues are special scalars associated with a matrix that, when multiplied with its corresponding eigenvector, yield a vector that is a scalar multiple of the original vector.

In our problem, solving the characteristic equation for the matrix yields the eigenvalues. These eigenvalues are crucial in forming the general solution to the differential equations. For each eigenvalue, there’s a corresponding eigenvector, which when combined, form the parts of the solution that explain how the system evolves over time. In a sense, finding the eigenvalues is like finding the heartbeat of the system, and the eigenvectors show in which direction the system states move as they pulse with the beat of the eigenvalues.

Homogeneous System

A homogeneous linear system of differential equations is one in which the constant terms are zero – meaning, there is no external force or influence acting on the system. Mathematically, this boils down to the simplification that the system can be solved in terms of the variables and their derivatives alone, without the need to account for outside factors.

When we deal with a homogeneous system, we search for solutions that describe the inherent behavior of the system. In the case of the exercise, once the eigenvalues and eigenvectors are known, the next step is building the general solution that depicts how the system behaves over time purely based on its internal dynamics. In real-world scenarios, this ‘pure’ behavior rarely exists in isolation, but understanding it lays the foundation for introducing complexity and external influences.

Particular Solution Initial Conditions

Once the general solution to a homogeneous system of differential equations has been established, the particular solution is identified by applying the initial conditions. Initial conditions serve as a snapshot of the system at a specific instant, providing a starting point for the solution to evolve from. They allow us to calculate specific values for the constants in our general solution, thus tailoring the solution to an individual scenario.

The given initial conditions locate us precisely in the 'state-space' of our problem. By plugging them into the general solution, we can solve for the constants, transforming a broad set of potential solutions into the single, unique trajectory that the system will follow. This interplay between the general solution and the initial conditions highlights the importance of both the system's structure and the system's initial state in determining its future behavior.