Question

In Exercises \(11-20\) find a particular solution, given that \(Y\) is a fundamental matrix for the complementary system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{cc} \frac{1}{t-1} & -\frac{e^{-t}}{t-1} \\ \frac{e^{t}}{t+1} & \frac{1}{t+1} \end{array}\right] \mathbf{y}+\left[\begin{array}{c} t^{2}-1 \\ t^{2}-1 \end{array}\right] ; \quad Y=\left[\begin{array}{cc} t & e^{-t} \\ e^{t} & t \end{array}\right] $$

Short answer

Question: Find the particular solution of the given system of linear differential equations with the given fundamental matrix Y and forcing vector f(t): \(Y = \left[\begin{array}{cc} t & e^{-t} \\ e^{t} & t \end{array}\right]\) \(\mathbf{f}(t) = \left[\begin{array}{c} t^{2}-1 \\ t^{2}-1 \end{array}\right]\) Answer: The particular solution of the given system is: \(\mathbf{y_p} = \left[\begin{array}{c} t^2-t^2e^{-t}-e^{-2t}-e^{-t}t \\ t^2e^t-e^{2t}+t^2-e^{-t}t^2 \end{array}\right]\)

Step-by-step solution

Compute the inverse of the fundamental matrix

To find the inverse of the given matrix \(Y\), we need to calculate the determinant \(\Delta_{Y}\) and the adjoint matrix \(adj(Y)\). Then, the inverse can be computed as: $$Y^{-1} = \frac{1}{\Delta_{Y}} adj(Y)$$ We have the matrix \(Y\) as: \(Y=\left[\begin{array}{cc} t & e^{-t} \\ e^{t} & t \end{array}\right]\) Let's find the determinant \(\Delta_{Y}\): $$\Delta_{Y} = \det(Y) = t \cdot t - e^{-t} \cdot e^{t}= t^2 - e^{-t}e^t = t^2 -1$$ Now, let's find the adjoint matrix \(adj(Y)\): \(adj(Y) = \left[ \begin{array}{cc} t & -e^{t} \\ -e^{-t} & t \end{array} \right]\) Finally, let's compute the inverse \(Y^{-1}\): $$Y^{-1}=\frac{1}{t^2 -1} \left[\begin{array}{cc} t & -e^{t} \\ -e^{-t} & t \end{array}\right]$$

Compute the product \(Y^{-1} \mathbf{f}(t)\)

The given forcing vector \(\mathbf{f}(t)\) is: \(\mathbf{f}(t) = \left[\begin{array}{c} t^{2}-1 \\ t^{2}-1 \end{array}\right]\) Now compute the product \(Y^{-1} \mathbf{f}(t)\): $$Y^{-1} \mathbf{f}(t) = \frac{1}{t^2 -1} \left[\begin{array}{cc} t & -e^{t} \\ -e^{-t} & t \end{array}\right] \left[\begin{array}{c} t^{2}-1 \\ t^{2}-1 \end{array}\right] = \left[\begin{array}{c} 1-e^{t} \\ e^{-t}+1 \end{array}\right]$$

Integrate the product from step 2

Now, we need to integrate the product computed in step 2: $$\int Y^{-1} \mathbf{f}(t) dt = \int \left[\begin{array}{c} 1-e^{t} \\ e^{-t}+1 \end{array}\right] dt = \left[\begin{array}{c} t-e^{t} \\ -e^{-t}+t \end{array}\right] + \mathbf{C}$$

Compute the product \(Y\) times the integral from step 3

Finally, compute the product of the fundamental matrix \(Y\) and the integral obtained in step 3: $$\mathbf{y_p} = Y \int Y^{-1} \mathbf{f}(t) dt = \left[\begin{array}{cc} t & e^{-t} \\ e^{t} & t \end{array}\right] \left[\begin{array}{c} t-e^{t} \\ -e^{-t}+t \end{array}\right] = \left[\begin{array}{c} t^2-te^{-t}-(e^{-t}+t)e^{-t} \\ e^{t}(t-e^t)+t(-e^{-t}+t) \end{array}\right] = \left[\begin{array}{c} t^2-t^2e^{-t}-e^{-2t}-e^{-t}t \\ t^2e^t-e^{2t}+t^2-e^{-t}t^2 \end{array}\right]$$ So, the particular solution of the given system is: $$\mathbf{y_p} = \left[\begin{array}{c} t^2-t^2e^{-t}-e^{-2t}-e^{-t}t \\ t^2e^t-e^{2t}+t^2-e^{-t}t^2 \end{array}\right]$$

Key concepts

Matrix Inversion

Matrix inversion is a fundamental concept in linear algebra. It allows us to find matrices that can reverse the effects of square matrices. If a matrix is invertible, we can express it in a way where multiplying by its inverse results in the identity matrix. It's like finding the reverse operation to undo a previous one. To compute the inverse of a matrix, we commonly use the formula:

• Find the determinant of the matrix.
• Calculate the adjoint (or adjugate) matrix.
• Use these values to find the inverse as: \( Y^{-1} = \frac{adj(Y)}{\Delta_{Y}} \)

This process is crucial for solving systems of linear equations, as it helps to manipulate matrices to achieve desired results. In our example, the original matrix is \( Y = \begin{bmatrix} t & e^{-t} \ e^{t} & t \end{bmatrix} \), and its inverse is \( Y^{-1} = \frac{1}{t^2 -1} \begin{bmatrix} t & -e^{t} \ -e^{-t} & t \end{bmatrix} \). Practitioners should ensure the determinant, \( \Delta_{Y} = t^2 -1 \), is non-zero to confirm the matrix is invertible.

Particular Solution

A particular solution in differential equations is a specific solution that satisfies both the non-homogeneous differential equation and its initial conditions. While a general solution includes arbitrary constants and represents a family of solutions, the particular solution pinpoints exactly one scenario among these possibilities. This is often obtained by adding a specific function to the homogeneous solution. In the context of our exercise:

• The forcing term or input \( \mathbf{f}(t) = \begin{bmatrix} t^{2}-1 \ t^{2}-1 \end{bmatrix} \)
• The particular solution \( \mathbf{y_p} = \begin{bmatrix} t^2-t^2e^{-t}-e^{-2t}-e^{-t}t \ t^2e^t-e^{2t}+t^2-e^{-t}t^2 \end{bmatrix} \)

In essence, the goal is to find one solution that fits the entire system's conditions, ensuring it matches up with a given input. The process often involves using the inverse of a fundamental matrix and integration to solve the equation as required.

Fundamental Matrix

The concept of a fundamental matrix is essential in solving systems of linear differential equations. It essentially represents a set of solutions to the homogeneous differential equation that can be used to construct the general solution. Here, the fundamental matrix is given by \( Y = \begin{bmatrix} t & e^{-t} \ e^{t} & t \end{bmatrix} \), which helps provide the structure needed for the solution.

• A fundamental matrix is used to express solutions to differential equations compactly.
• It can be inverted (if possible) to find specific solutions or manipulate the equations further.

The fundamental matrix is particularly valuable when dealing with non-homogeneous systems, as it allows us to transition to the complementary or general solution before addressing other terms. It's like having a baseline from which modifications or specific solutions are derived.

Integration Techniques

Integration techniques are vital when working with differential equations, especially when finding particular solutions. Integration helps determine functions whose derivatives match given criteria. In this exercise, after computing the product \( Y^{-1} \mathbf{f}(t) \), we integrate component-wise:

• Integrate terms like \( 1-e^{t} \) and \( e^{-t}+1 \) over the variable \(t\)
• Compute indefinite integrals to get expressions like \( t-e^{t} \) and \( -e^{-t}+t \)

These integration steps are straightforward, but each step must be done carefully to ensure accuracy in finding the particular solution. Integration not only modifies terms to the specific form needed but also provides the constants needed to adjust the equation's fit to specific conditions or initial values. Integration, paired with other techniques, helps to distill the solution uniquely suited to the differential system's requirements.