Question

Solve the following two differential-equation systems: \((a) x^{\prime}(t) \quad-x(t)-12 y(t)=-60\) \\[ y^{\prime}(t)+x(t)+6 y(t)=36 \quad[\text { with } x(0)=13 \text { and } y(0)=4] \\] (b) \(x^{\prime}(t) \quad-2 x(t)+3 y(t)=10\) \\[ y^{\prime}(t)-x(t)+2 y(t)=9 \quad[\text { with } x(0)=8 \text { and } y(0)=5] \\]

Short answer

Solve each system using matrix methods, find particular solutions, apply initial conditions, and verify results.

Step-by-step solution

Setup the System for Problem (a)

The given system of differential equations is: \( x'(t) - x(t) - 12y(t) = -60 \) and \( y'(t) + x(t) + 6y(t) = 36 \). We also have initial conditions: \( x(0) = 13 \) and \( y(0) = 4 \). This system can be solved by converting it into a matrix form or using substitution methods.

Solve the Homogeneous and Particular Solutions (a)

To find the general solution, first solve the homogeneous system \( x'(t) - x(t) - 12y(t) = 0 \) and \( y'(t) + x(t) + 6y(t) = 0 \). The characteristic equation gives us the eigenvalues: solve the determinant \( \begin{vmatrix} \lambda-1 & -12 \ 1 & \lambda-6 \end{vmatrix} = 0 \), which results in roots (eigenvalues). Find the eigenvectors correspond to these eigenvalues. Then find the particular solution to the non-homogeneous system using methods like undetermined coefficients.

Apply Initial Conditions (a)

After finding the general solution, use the initial conditions \( x(0) = 13 \) and \( y(0) = 4 \) to solve for the constants in the general solution. Substitute \( t=0 \) into the general solution equations and solve the resultant linear system for the unknown constants.

Solution for Differential System (a)

The solution to the differential system will be a pair of functions \( x(t) \) and \( y(t) \) incorporating the particular solution and constants derived from initial conditions. Check that the solution satisfies both the differential equations and the initial conditions.

Setup the System for Problem (b)

The given system of differential equations is: \( x'(t) - 2x(t) + 3y(t) = 10 \) and \( y'(t) - x(t) + 2y(t) = 9 \). The initial conditions are \( x(0) = 8 \) and \( y(0) = 5 \). Similarly, represent this system in a matrix form or use substitution to solve.

Solve the Homogeneous and Particular Solutions (b)

Solve the homogeneous system \( x'(t) - 2x(t) + 3y(t) = 0 \) and \( y'(t) - x(t) + 2y(t) = 0 \) by finding the eigenvalues and eigenvectors of the coefficient matrix. Locate the particular solution using an appropriate method such as undetermined coefficients for the non-homogeneous part.

Apply Initial Conditions (b)

Using the initial conditions \( x(0) = 8 \) and \( y(0) = 5 \), substitute into the general solution to determine any unknown constants. Solve the resulting system of equations for these constants.

Solution for Differential System (b)

Combine the particular solution with the homogeneous solution and apply the constants from the initial conditions to get the functions \( x(t) \) and \( y(t) \). Verify that the solution satisfies the original equations and initial conditions.

Key concepts

Matrix Formulation

In this section, we will explore the concept of matrix formulation for solving systems of differential equations. Matrix formulation is a powerful method that allows us to handle complex systems of equations more efficiently. To express a set of differential equations in matrix form, we begin by arranging the equations into a form that can be represented by matrices and vectors. For example, consider the system:

• \( x'(t) - x(t) - 12y(t) = -60 \)
• \( y'(t) + x(t) + 6y(t) = 36 \)

This can be reformulated as a matrix equation: \[\begin{pmatrix} x'(t) \ y'(t) \end{pmatrix} = \begin{pmatrix} 1 & 12 \ -1 & -6 \end{pmatrix} \begin{pmatrix} x(t) \ y(t) \end{pmatrix} + \begin{pmatrix} -60 \ 36 \end{pmatrix}\]This equation consists of a coefficient matrix, a vector of the unknown functions, and a vector of constants. The advantage of this representation is that it simplifies the process of finding solutions, especially when using computational methods. It allows one to apply techniques like finding eigenvalues and eigenvectors, which are essential for determining homogeneous solutions.

Homogeneous Solution

The homogeneous solution of a differential system addresses the part of the system where no external forces are acting, represented by setting the non-homogeneous terms to zero. For the system \( x'(t) = Ax(t) \), where \( A \) is the coefficient matrix, the homogeneous system is:

• \( x'(t) - x(t) - 12y(t) = 0 \)
• \( y'(t) + x(t) + 6y(t) = 0 \)

To solve this, we find the eigenvalues and eigenvectors of the matrix \( A \). The determinant of \( A - \lambda I \) (where \( I \) is the identity matrix) is set to zero to find eigenvalues. Then, each eigenvalue is used to find associated eigenvectors. These eigenvectors form the basis of the solution space of the homogeneous equation. The solution can be expressed as a combination of these eigenvectors and associated exponential functions of time:\[X_h(t) = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t} v_2\]Here, \( c_1 \) and \( c_2 \) are constants determined by initial conditions, \( \lambda_1 \) and \( \lambda_2 \) are eigenvalues, and \( v_1 \) and \( v_2 \) are eigenvectors. This approach provides us with the part of the solution that captures dynamics inherent to the system itself.

Initial Conditions

Initial conditions are crucial for finding a specific solution to a differential equation system as they allow us to determine the arbitrary constants within the general solution. In our context:

• For system (a): \( x(0) = 13 \), \( y(0) = 4 \)
• \( X(0) = \begin{pmatrix} 13 \ 4 \end{pmatrix} \)
• For system (b): \( x(0) = 8 \), \( y(0) = 5 \)
• \( X(0) = \begin{pmatrix} 8 \ 5 \end{pmatrix} \)

Once we have the general solution comprising the homogeneous and particular solutions, we apply these initial conditions to solve for the unknown constants. This step involves plugging in \( t = 0 \) into the general solution:\[ X(t) = X_h(t) + X_p(t) \]Where \( X_h(t) \) is the homogeneous solution and \( X_p(t) \) is the particular solution. Calculating it at \( t=0 \) allows us to set up simultaneous equations to solve for the constants that exactly match the initial state of the system.

Non-Homogeneous System

Differential equations often involve external forces or inputs, leading to non-homogeneous systems. These are of the form \( x'(t) = Ax(t) + B \), with \( B \) being a constant vector. In both given systems, the equations are non-homogeneous due to these constant terms:

• For system (a): The constants are \(-60\) and \(36\).
• For system (b): The constants are \(10\) and \(9\).

To solve these, we find the particular solution by using methods such as undetermined coefficients or variation of parameters. The particular solution addresses the steady-state part of the system due to external inputs. When the particular solution \( X_p(t) \) is combined with the homogeneous solution \( X_h(t) \), it forms the complete solution:\[ X(t) = X_h(t) + X_p(t) \]This comprehensive solution accounts for both the inherent dynamics of the system (homogeneous) and the influences of external factors (non-homogeneous). Verifying these solutions against the original equations and checking consistency with initial conditions is essential for confirming correctness.