Question

In each exercise, find the general solution of the homogeneous linear system and then solve the given initial value problem. $$ \begin{array}{lll} y_{1}^{\prime} & =2 y_{1}+y_{2}+2 y_{3}, & y_{1}(0) & =4 \\ y_{2}^{\prime} & = & 3 y_{2}+2 y_{3}, & y_{2}(0) & =3 \\ y_{3}^{\prime} & = & y_{3}, & y_{3}(0) & =-1 \end{array} $$

Short answer

Question: Find the particular solution of the initial value problem for the given homogeneous linear system of differential equations, with the initial conditions \(y_1(0) = 4\), \(y_2(0) = 3\), and \(y_3(0) = -1\). System of differential equations: $$y_1' = 2y_1 + y_2 + 2y_3$$ $$y_2' = 3y_2 + 2y_3$$ $$y_3' = y_1$$ Answer: The particular solution to the initial value problem is: $$\begin{bmatrix} y_1(t) \\ y_2(t) \\ y_3(t) \end{bmatrix} = -\frac{3}{2}e^{2t}\begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix} + \frac{11}{2}e^{3t}\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} - e^{t}\begin{bmatrix} -2 \\ -1 \\ 1 \end{bmatrix}.$$

Step-by-step solution

Rewrite the system of differential equations in matrix form

To start, we can rewrite the given system of differential equations in matrix form: $$\begin{bmatrix} y_1' \\ y_2' \\ y_3' \end{bmatrix} = \begin{bmatrix} 2 & 1 & 2 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix}.$$ This matrix form allows us to find the general solution more easily.

Find the eigenvalues and eigenvectors of the matrix

Now we need to find the eigenvalues (λ) and their corresponding eigenvectors of the coefficient matrix: $$\begin{bmatrix} 2 & 1 & 2 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix} = \begin{bmatrix} 2-\lambda & 1 & 2 \\ 0 & 3-\lambda & 2 \\ 0 & 0 & 1-\lambda \end{bmatrix}.$$ To find the eigenvalues, we need to calculate the determinant of this matrix: $$(2-\lambda)((3-\lambda)(1-\lambda)) = 0.$$ The eigenvalues are λ1 = 2, λ2 = 3, and λ3 = 1. Note that the matrix is upper triangular, so the eigenvalues appear directly on the diagonal. Now, for each eigenvalue, we need to find its corresponding eigenvector. We substitute the eigenvalue into the matrix equation: For λ1 = 2: $$\begin{bmatrix} 0 & 1 & 2 \\ 0 & 1 & 2 \\ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$$ The eigenvector for λ1 is obtained by solving this system, which yields: $v_1 = \begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix}.$$ For λ2 = 3: $$\begin{bmatrix} -1 & 1 & 2 \\ 0 & 0 & 2 \\ 0 & 0 & -2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$$ The eigenvector for λ2 is: $v_2 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}.$$ For λ3 = 1: $$\begin{bmatrix} 1 & 1 & 2 \\ 0 & 2 & 2 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}.$$ The eigenvector for λ3 is: $v_3 = \begin{bmatrix} -2 \\ -1 \\ 1 \end{bmatrix}.$$

Find the general solution of the homogeneous linear system

Using the obtained eigenvalues and eigenvectors, we can find the general solution to the system of differential equations: $$\begin{bmatrix} y_1(t) \\ y_2(t) \\ y_3(t) \end{bmatrix} = c_1e^{2t}\begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix} + c_2e^{3t}\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + c_3e^{t}\begin{bmatrix} -2 \\ -1 \\ 1 \end{bmatrix}.$$

Apply the initial conditions and solve for the coefficients

We are given the following initial conditions: \(y_1(0) = 4\), \(y_2(0) = 3\), and \(y_3(0) = -1\). We plug these values into the general solution to solve for the coefficients \(c_1\), \(c_2\), and \(c_3\): $$\begin{bmatrix} 4 \\ 3 \\ -1 \end{bmatrix} = c_1\begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix} + c_2\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + c_3\begin{bmatrix} -2 \\ -1 \\ 1 \end{bmatrix}.$$ Solving this system, we get \(c_1 = -\frac{3}{2}\), \(c_2 = \frac{11}{2}\), and \(c_3 = -1\).

Write down the particular solution of the initial value problem

Now substitute the coefficients back into the general solution to get the particular solution for the initial value problem: $$\begin{bmatrix} y_1(t) \\ y_2(t) \\ y_3(t) \end{bmatrix} = -\frac{3}{2}e^{2t}\begin{bmatrix} 1 \\ -2 \\ 0 \end{bmatrix} + \frac{11}{2}e^{3t}\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} - e^{t}\begin{bmatrix} -2 \\ -1 \\ 1 \end{bmatrix}.$$

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are central concepts in the analysis of linear systems. They provide insight into the behavior of the system over time. When we deal with a matrix in a system of linear differential equations, determining its eigenvalues can show us the nature of the solutions.

Eigenvalues, often denoted by \( \lambda \), are scalars that satisfy the equation \( A\mathbf{v} = \lambda\mathbf{v} \). \( \mathbf{v} \) represents the eigenvector corresponding to the eigenvalue \( \lambda \). This equation essentially means that when a matrix \( A \) acts on \( \mathbf{v} \), the action stretches \( \mathbf{v} \) by the scale factor of \( \lambda \) without changing its direction.

Finding the eigenvalues requires solving the characteristic equation, which emerges from setting the determinant of \( \mathbf{A} - \lambda \mathbf{I} \) to zero. The eigenvectors are found by substituting each eigenvalue back into the equation \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0} \) and solving for \( \mathbf{v} \).

Understanding eigenvalues and eigenvectors helps us determine how systems evolve, determining stability and oscillatory behaviors, which are crucial for responding to initial conditions in differential equations.

Matrix Form Differential Equations

Rewriting a system of differential equations in matrix form provides a structured way to handle and solve them. This form is especially useful in the context of linear systems and involves arranging the equations and their coefficients into matrix multiplication.

Consider the system:

• \( y' = A\mathbf{y} \), where \( y' \) is a vector of the derivatives of \( y_1, y_2, \ldots, y_n \)
• \( \mathbf{y} \) is a vector of the functions \( y_1, y_2, \ldots, y_n \)
• \( A \) is the coefficient matrix containing the constants from each linear equation.

By framing a differential system in this way, one can utilize linear algebra techniques like diagonalization or direct computation of solutions.

This matrix form makes it feasible to leverage advanced mathematical tools to find solutions, analyze system behavior, and accommodate initial conditions or other constraints effectively.

Initial Value Problem Solution

Solving an initial value problem (IVP) involves finding a solution to a differential equation that meets specified conditions at a starting point, typically \( t = 0 \). In the given problem, the initial conditions are \( y_1(0) = 4 \), \( y_2(0) = 3 \), and \( y_3(0) = -1 \).

The process of solving the initial value problem after finding the general solution of the differential equation is crucial. This specific solution respects the initial conditions by substituting them into the general solution and solving for the constants.

Usually, the general solution appears as a linear combination of eigenvectors multiplied by exponential functions of time (\( e^{\lambda t} \)). Applying initial conditions involves setting up equations to solve for constants (such as \( c_1, c_2, \) and \( c_3 \)) in the general solution. This results in a particular solution that specifies exact behavior at the start and guides how the system changes over time.