Problem 10 - Advanced Engineering Mathematics

Question

In Problems 9 and 10, solve the given initial-value problem. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rr} 1 & -1 \\ 1 & 3 \end{array}\right) \mathbf{X}+\left(\begin{array}{c} t \\ t+1 \end{array}\right), \quad \mathbf{X}(0)=\left(\begin{array}{l} 0 \\ 2 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \( \mathbf{X}(t) = e^{2t} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + e^{2t}t \begin{pmatrix} 1 \\ 1 \end{pmatrix} + e^{2t} \begin{pmatrix} 0 \\ 1 \end{pmatrix} + \begin{pmatrix} -1 \\ t \end{pmatrix} \).

Step 1: Identify the System Components

The differential equation given is \( \mathbf{X}' = A\mathbf{X} + \mathbf{g}(t) \), with initial condition \( \mathbf{X}(0) = \begin{pmatrix} 0 \ 2 \end{pmatrix} \), where \( A = \begin{pmatrix} 1 & -1 \ 1 & 3 \end{pmatrix} \) and \( \mathbf{g}(t) = \begin{pmatrix} t \ t+1 \end{pmatrix} \). We want to find \( \mathbf{X}(t) = \begin{pmatrix} x_1(t) \ x_2(t) \end{pmatrix} \).

Step 2: Solve the Homogeneous Equation

To solve \( \mathbf{X}' = A\mathbf{X} \), consider the homogeneous equation \( \mathbf{Y}' = A\mathbf{Y} \). Find the eigenvalues of matrix \( A \). These are the roots of the characteristic equation \( \lambda^2 - 4\lambda + 4 = 0 \). The roots are both \( \lambda = 2 \), which means the solution involves a repeated root.

Step 3: Form the General Solution for the Homogeneous Part

Since we have repeated eigenvalues, the general solution for \( \mathbf{Y}(t) \) is \( \mathbf{Y}(t) = c_1 e^{2t} \begin{pmatrix} 1 \ 1 \end{pmatrix} + c_2 e^{2t} \left( \begin{pmatrix} 1 \ 1 \end{pmatrix}t + \begin{pmatrix} 0 \ 1 \end{pmatrix} \right) \), where \( \begin{pmatrix} 1 \ 1 \end{pmatrix} \) is the eigenvector and \( \begin{pmatrix} 0 \ 1 \end{pmatrix} \) is a generalized eigenvector.

Step 4: Guess the Particular Solution

For \( \mathbf{X}' = A\mathbf{X} + \mathbf{g}(t) \), a particular solution of the form \( \mathbf{X}_p(t) = At + B \) can be proposed. Substitute it into the non-homogeneous equation to solve for constants \( A \) and \( B \).

Step 5: Solve for Particular Solution Constants

By substituting \( \mathbf{X}_p(t) = \begin{pmatrix} at + b \ ct + d \end{pmatrix} \) into the non-homogeneous part of the differential equation, equate coefficients to find \( a = 0, b = -1, c = 1, d = 0 \). This gives the particular solution \( \mathbf{X}_p(t) = \begin{pmatrix} -1 \ t \end{pmatrix} \).

Step 6: Write the General Solution

Combine the solutions: \( \mathbf{X}(t) = \mathbf{Y}(t) + \mathbf{X}_p(t) = c_1 e^{2t} \begin{pmatrix} 1 \ 1 \end{pmatrix} + c_2 e^{2t} \left( \begin{pmatrix} 1 \ 1 \end{pmatrix}t + \begin{pmatrix} 0 \ 1 \end{pmatrix} \right) + \begin{pmatrix} -1 \ t \end{pmatrix} \).

Step 7: Apply Initial Conditions

To find constants \( c_1 \) and \( c_2 \), use \( \mathbf{X}(0) = \begin{pmatrix} 0 \ 2 \end{pmatrix} \). Plug \( t = 0 \) into the general solution and solve, leading to \( c_1 = 1 \) and \( c_2 = 1 \).

Step 8: Final Solution

The solution, satisfying the initial-value problem, is \( \mathbf{X}(t) = e^{2t} \begin{pmatrix} 1 \ 1 \end{pmatrix} + e^{2t}t \begin{pmatrix} 1 \ 1 \end{pmatrix} + e^{2t} \begin{pmatrix} 0 \ 1 \end{pmatrix} + \begin{pmatrix} -1 \ t \end{pmatrix} \).

Key Concepts

Systems of Differential Equations

Making sense of systems of differential equations can be quite tricky at first glance, but let's break it down. These systems consist of more than one differential equation that works with multiple functions. In our exercise, we are dealing with a system where changes in variables are coupled together. The notation \( \mathbf{X}' = A\mathbf{X} + \mathbf{g}(t) \) is a standard way of expressing this.

In such equations, \( \mathbf{X} \) represents a vector that holds all the functions we are interested in. These functions change over time based on their current states and any present outside influences, which in our case are encapsulated in \( \mathbf{g}(t) \). The matrix \( A \) is known as the coefficient matrix, containing numbers that tell how much each function influences the others.

• Homogeneous Systems: If \( \mathbf{g}(t) \) were zero, all changes in the system would rely solely on the states and interactions dictated by \( A \).
• Non-homogeneous Systems: Non-zero \( \mathbf{g}(t) \) implies an external influence, such as time-dependent inputs, changing the nature of these interactions.

This exercise touches upon such an interaction by adding the time-dependent vector \( \mathbf{g}(t) \), making our system non-homogeneous. The initial conditions give specific values that help guide us to the precise solution required.

Eigenvalues and Eigenvectors

When dealing with systems of differential equations, eigenvalues and eigenvectors pop up often due to their importance in solving these complex systems. In simple terms, an eigenvalue is a special number associated with a matrix that gives insight into the system's behavior.

For our matrix \( A = \begin{pmatrix} 1 & -1 \ 1 & 3 \end{pmatrix} \), we find eigenvalues by solving the characteristic equation: \( \det(A - \lambda I) = 0 \), resulting in \( \lambda^2 - 4\lambda + 4 = 0 \). In this case, both roots are \( \lambda = 2 \), indicating a repeated root. This shows how our system behaves in a very specific manner.

• Eigenvalues (\( \lambda \)): Numbers that reveal important natural frequencies or rates of the system.
• Eigenvectors: Vectors associated with each eigenvalue that represent invariant directions under the matrix transformation.
• Repeated Eigenvalues: Lead to solutions involving generalized eigenvectors, which add complexity to our solution.

The presence of repeated eigenvalues means our solutions require not just basic eigenvectors, but also generalized ones to cover all possible behaviors of the system, which is seen in our exercise solution.

Particular and General Solutions

Solving differential equations often involves finding both a particular and general solution. In our exercise, we dealt with these layers of solutions to tackle the non-homogeneous system.

The general solution addresses the homogeneous part of the equation \( \mathbf{X}' = A\mathbf{X} \), capturing the essence of the system without external influences. It's often expressed using eigenvalues and eigenvectors derived from \( A \), as seen in our solution with the vector \( \begin{pmatrix} 1 \ 1 \end{pmatrix} \) and a generalized vector \( \begin{pmatrix} 0 \ 1 \end{pmatrix} \).

• Homogeneous Solution: Combines components from eigenvalues to reflect how internal system dynamics evolve.
• Particular Solution: Solves the equation including the external forces or inputs, \( \mathbf{g}(t) \).

For the non-homogeneous part, a particular solution is determined by testing forms that fit \( \mathbf{g}(t) \). Here, a simple linear form \( At + B \) was initially proposed, testing constants to achieve an adequate fit. By adding this to the homogeneous solution, the complete path of each variable is established.
The final step involves adapting this combined result to the specific conditions provided, known as initial values, producing a final solution tailored specifically to the problem at hand.