Question

In each exercise, find the general solution of the homogeneous linear system and then solve the given initial value problem. $$ \begin{array}{ll} y_{1}^{\prime}=-y_{1}-y_{2}, & y_{1}(0)=-2 \\ y_{2}^{\prime}=6 y_{1}+4 y_{2}, & y_{2}(0)=6 \end{array} $$

Short answer

In conclusion, the solution for the given initial value problem (IVP) of the system of linear homogeneous differential equations is: $$ \begin{array}{l} y_1(t) = -4e^{2t} + 2e^{-5t}, \\ y_2(t) = 12e^{2t} + 8e^{-5t}. \end{array} $$

Step-by-step solution

1. Create the matrix equation

To represent the system of linear differential equations in a matrix equation, we can use the following equation: $$y'(t) = Ay(t),$$ where \(y(t) = \begin{bmatrix} y_1(t) \\ y_2(t) \end{bmatrix}\), and $$A = \begin{bmatrix} -1 & -1 \\ 6 & 4 \end{bmatrix}.$$ So, the matrix equation for the given system is $$y'(t) = \begin{bmatrix} -1 & -1 \\ 6 & 4 \end{bmatrix} y(t).$$

2. Find eigenvalues and eigenvectors

Next, we need to find the eigenvalues and eigenvectors of the coefficient matrix \(A\). To find the eigenvalues, we will solve the characteristic equation, given by: $$\det(A - \lambda I) = 0.$$ The characteristic equation for matrix \(A\) is \(\det(\begin{bmatrix} -1-\lambda & -1 \\ 6 & 4-\lambda \end{bmatrix}) = 0\). This results in \((\lambda+1)(\lambda-4) + 6 = \lambda^2 - 3\lambda - 10 = 0\). Solving for \(\lambda\), we have \(\lambda_1 = 2\) and \(\lambda_2 = -5\). Now, we must calculate eigenvectors for each eigenvalue: 1. For \(\lambda_1 = 2\), solve the equation \((A - 2I)v_1 = 0\). We have $$\begin{bmatrix} -3 & -1 \\ 6 & 2 \end{bmatrix} \begin{bmatrix} v_{11} \\ v_{12} \end{bmatrix} = 0.$$ The system is redundant and has one free variable. By choosing \(v_{11} = 1\), we get \(v_{12} = -3\). So, the eigenvector corresponding to \(\lambda_1\) is \(v_1 = \begin{bmatrix} 1 \\ -3 \end{bmatrix}\). 2. For \(\lambda_2 = -5\), solve the equation \((A + 5I)v_2 = 0\). We have $$\begin{bmatrix} 4 & -1 \\ 6 & 9 \end{bmatrix} \begin{bmatrix} v_{21} \\ v_{22} \end{bmatrix} = 0.$$ Again, the system is redundant and has one free variable. By choosing \(v_{21} = 1\), we get \(v_{22} = 4\). So, the eigenvector corresponding to \(\lambda_2\) is \(v_2 = \begin{bmatrix} 1 \\ 4 \end{bmatrix}\).

3. Write the general solution

Now that we have the eigenvalues and eigenvectors, we can write the general solution to the homogeneous linear system. Since the eigenvalues are real and distinct, the general solution is given by: $$y(t) = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t} v_2 = c_1 e^{2t}\begin{bmatrix} 1 \\ -3 \end{bmatrix} + c_2 e^{-5t}\begin{bmatrix} 1 \\ 4 \end{bmatrix}.$$

4. Use initial conditions to find specific constants

With the general solution, we now apply the initial conditions, \(y_1(0)=-2\) and \(y_2(0)=6\), to find the constants \(c_1\) and \(c_2\). $$y(0) = c_1 \begin{bmatrix} 1 \\ -3 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 4 \end{bmatrix} = \begin{bmatrix} -2 \\ 6 \end{bmatrix}.$$ Setting up and solving the system of equations, we have \begin{align*} c_1 + c_2 &= -2, \\ -3c_1 + 4c_2 &= 6. \end{align*} Solving the system, we find \(c_1 = -4\) and \(c_2 = 2\). So, the solution to the initial value problem is $$y(t) = -4e^{2t}\begin{bmatrix} 1 \\ -3 \end{bmatrix} + 2e^{-5t}\begin{bmatrix} 1 \\ 4 \end{bmatrix}.$$ Thus, we have found the solution for the given IVP: $$ \begin{array}{l} y_1(t) = -4e^{2t} + 2e^{-5t}, \\ y_2(t) = 12e^{2t} + 8e^{-5t}. \end{array} $$

Key concepts

General Solution of Differential Equations

Understanding the general solution to differential equations is crucial for students studying advanced mathematics or physics. A differential equation relates a function with its derivatives, explaining how a quantity changes over time or space. The general solution of a homogeneous linear differential equation is a formula that contains all possible solutions to the equation. It usually involves arbitrary constants, which can vary to produce particular solutions based on specific initial conditions or boundary values.

In the matrix form of a system of linear differential equations like the one in our exercise, the general solution is an expression that combines exponentials of the eigenvalues and their corresponding eigenvectors of the system's matrix. These exponential terms are multiplied by arbitrary constants, which will later be determined using initial conditions. This approach, using eigenvalues and eigenvectors, simplifies finding the general solution, especially when dealing with multiple coupled differential equations.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation along with a specified value, called the initial condition, which the solution must satisfy at a given point. The problem consists of finding a function that not only satisfies the differential equation but also meets the prescribed initial conditions. This additional information allows us to find the specific or particular solution out of the general solution.

In our textbook exercise, after determining the general solution involving the arbitrary constants, we use the given initial values, namely, the values of both functions when the independent variable (in this case, time t) is zero, to solve for these constants. The result is a unique solution that precisely fits the initial state of the system being analyzed.

Eigenvalues and Eigenvectors

The concepts of eigenvalues and eigenvectors are fundamental in various fields of science and engineering, particularly in the study of linear transformations and systems of differential equations. An eigenvalue is a scalar that signifies the factor by which an eigenvector is scaled during a linear transformation represented by a matrix.

In the context of solving a system of differential equations, eigenvalues help us determine the nature of the system's behaviour and the corresponding eigenvector provides the direction of this behaviour. For a given matrix A, eigenvalues are found by solving the characteristic equation \(\det(A - \lambda I) = 0\), where \(\lambda\) is an eigenvalue and I is the identity matrix. This equation often results in a polynomial which roots are the eigenvalues. Each eigenvalue is then used to find its corresponding eigenvector by solving the equation \( (A - \lambda I)v = 0 \), where \(v\) is the eigenvector associated with eigenvalue \(\lambda\).

Matrix Representation of Differential Equations

When we encounter systems of differential equations, representing them in matrix form can tremendously simplify the process of finding solutions. The matrix representation encapsulates the system into a compact and manageable form, using matrix multiplications and vector spaces.

A linear system of differential equations can be written as \(y'(t) = Ay(t)\), where \(y(t)\) is a vector of functions, A is a matrix of coefficients, and \(y'(t)\) is the vector of their derivatives. This form is particularly useful since it allows us to apply linear algebra techniques like finding eigenvalues and eigenvectors to solve the system. Once we have the eigenvalues and eigenvectors, they are used to build the system's general solution, which can predict the system's behaviour over time. This method is elegant and powerful, significantly reducing the complexity inherent in working with multiple equations.