Question

Let \(A\) be an invertible diagonalizable \(n \times n\) matrix and let \(\mathbf{b}\) be an \(n\) -column of constant functions. We can solve the system \(\mathbf{f}^{\prime}=A \mathbf{f}+\mathbf{b}\) as follows: a. If \(\mathbf{g}\) satisfies \(\mathbf{g}^{\prime}=A \mathbf{g}\) (using Theorem 3.5 .2 ), show that \(\mathbf{f}=\mathbf{g}-A^{-1} \mathbf{b}\) is a solution to \(\mathbf{f}^{\prime}=A \mathbf{f}+\mathbf{b}\). b. Show that every solution to \(\mathbf{f}^{\prime}=A \mathbf{f}+\mathbf{b}\) arises as in (a) for some solution \(\mathbf{g}\) to \(\mathbf{g}^{\prime}=A \mathbf{g}\).

Short answer

Every solution \( \mathbf{f}(t) \) is of the form \( \mathbf{g}(t) - A^{-1}\mathbf{b} \), derived from any solution \( \mathbf{g}(t) \) to \( \mathbf{g}^{\prime} = A\mathbf{g} \).

Step-by-step solution

Differentiating the Proposed Solution

We are given that \( \mathbf{f} = \mathbf{g} - A^{-1} \mathbf{b} \) and need to show it satisfies \( \mathbf{f}^{\prime} = A \mathbf{f} + \mathbf{b} \). Start by taking the derivative of \( \mathbf{f} \). Since \( \mathbf{g} \) satisfies \( \mathbf{g}^{\prime} = A \mathbf{g} \), we have:\[\mathbf{f}^{\prime} = \mathbf{g}^{\prime} - \frac{d}{dt}(A^{-1} \mathbf{b}).\] Since \( \mathbf{b} \) is a constant column vector, its derivative is zero: \[ \frac{d}{dt}(A^{-1} \mathbf{b}) = \mathbf{0}. \] Thus, \( \mathbf{f}^{\prime} = \mathbf{g}^{\prime}. \)

Substitute \( \mathbf{g}^{\prime} \) into \( \mathbf{f}^{\prime} \)

Now substitute \( \mathbf{g}^{\prime} = A \mathbf{g} \) into the expression for \( \mathbf{f}^{\prime} \):\[ \mathbf{f}^{\prime} = A \mathbf{g}. \] Next, express \( \mathbf{f} \) in terms of \( \mathbf{g} \):\[ \mathbf{f} = \mathbf{g} - A^{-1} \mathbf{b}. \] Thus, \[ A \mathbf{f} = A(\mathbf{g} - A^{-1} \mathbf{b}) = A\mathbf{g} - \mathbf{b}. \]

Show Equality of Expressions

Now, substitute \( A \mathbf{f} + \mathbf{b} \) into the expression:\[ A\mathbf{f} + \mathbf{b} = (A\mathbf{g} - \mathbf{b}) + \mathbf{b} = A\mathbf{g}. \] Thus, we see that \( \mathbf{f}^{\prime} = A\mathbf{f} + \mathbf{b} \), which confirms that \( \mathbf{f} = \mathbf{g} - A^{-1} \mathbf{b} \) is indeed a solution to the system.

Showing All Solutions Are of This Form

Consider any solution \( \mathbf{f} \) to \( \mathbf{f}^{\prime} = A\mathbf{f} + \mathbf{b} \). Rewriting, this gives us:\[ \mathbf{f}^{\prime} - A\mathbf{f} = \mathbf{b}. \] If \( \mathbf{g} \) is a particular solution to \( \mathbf{g}^{\prime} = A \mathbf{g} \), then for \( \mathbf{f} \) to satisfy the equation, it must be of the form\[ \mathbf{f} = \mathbf{g} - A^{-1} \mathbf{b} + \mathbf{h}, \] where \( \mathbf{h} \) is any homogeneous solution such that \( \mathbf{h}^{\prime} = A \mathbf{h} \). All such homogeneous solutions involve a linear combination of solutions proportional to \( \mathbf{g} \). Therefore, every solution arises from \( \mathbf{g} \) plus a constant shift \( A^{-1} \mathbf{b} \), as derived initially.

Key concepts

System of Differential Equations

A system of differential equations involves multiple equations that relate different functions and their derivatives. These systems are essential in modeling various natural and engineering phenomena. They help describe how different variables change with respect to each other over time.

The general form of a system of linear differential equations can be represented as:

• \[ \frac{d\mathbf{y}}{dt} = A\mathbf{y} + \mathbf{b} \]

This equation suggests that the change in a vector \(\mathbf{y}\) over time depends linearly on its current state, as described by matrix \(A\), and an external input represented by the vector \(\mathbf{b}\). Understanding how to solve these systems is crucial as they appear in various fields such as physics, economics, and biology.

Solve these systems by finding functions that satisfy each equation simultaneously. This often requires knowledge of linear algebra and the properties of matrices. Specifically, when the matrix \(A\) is diagonalizable, the process of finding solutions becomes more straightforward as it can be transformed into a simpler form, where each differential equation can be solved independently.

Invertible Matrix

An invertible matrix, also known as a non-singular or non-degenerate matrix, is a square matrix that has an inverse. If \(A\) is an invertible matrix, there exists another matrix \(A^{-1}\) such that \(AA^{-1} = A^{-1}A = I\), where \(I\) is the identity matrix.

Understanding invertible matrices is crucial because they play a key role in solving systems of equations. Specifically, in differential equations, if a matrix \(A\) is invertible and appears in a system like \(\mathbf{f}^{\prime}=A \mathbf{f}+\mathbf{b}\), it allows us to manipulate and solve the system effectively by removing dependencies or isolating different terms.

For instance, in the solution provided, the invertibility of \(A\) enables us to express solutions like \(\mathbf{f} = \mathbf{g} - A^{-1} \mathbf{b} \) during the solving process. This operation ensures that the corrective term \(A^{-1}\mathbf{b}\) can be subtracted to refine \(\mathbf{f}\), making invertible matrices a powerful tool in linear algebra and differential equations.

Homogeneous Solution

A homogeneous solution refers to a solution of a system of differential equations where the non-homogeneous part (or external input) is zero. For example, in the equation \(\mathbf{g}^{\prime} = A \mathbf{g}\), the vector \(\mathbf{g}\) is a homogeneous solution if it satisfies this equality without any additional terms.

Homogeneous solutions are pivotal, as they form the core solutions that follow from the inherent properties of the system, ignoring external influences. They provide a basis for constructing more complex solutions that incorporate external functions. In the context of the original exercise, identifying homogeneous solutions \(\mathbf{h}\) helps in solving the broader system \(\mathbf{f}^{\prime} = A \mathbf{f} + \mathbf{b}\).

We can express any solution \(\mathbf{f}\) to the non-homogeneous system as a combination of a particular solution and all homogeneous solutions. Specifically, \(\mathbf{f} = \mathbf{g} - A^{-1} \mathbf{b} + \mathbf{h}\), where \(\mathbf{h}\) is a homogeneous solution. By understanding homogeneous solutions, learners unlock the potential to interpret and solve complex differential systems by constructing comprehensive solutions from fundamental components.