Problem 9 - Differential Equations and Linear Algebra

Question

Use the variation-of-parameters technique to find a particular solution \(\mathbf{x}_{p}\) to \(\mathbf{x}^{\prime}=A \mathbf{x}+\mathbf{b},\) for the given \(A\) and \(\mathbf{b} .\) Also obtain the general solution to the system of differential equations. $$A=\left[\begin{array}{rrr} -1 & -2 & 2 \\ 2 & 4 & -1 \\ 0 & 0 & 3 \end{array}\right], \quad \mathbf{b}=\left[\begin{array}{l} -e^{3 t} \\ 4 e^{3 t} \\ 3 e^{3 t} \end{array}\right]$$

Step-by-Step Solution

Answer

The general solution to the system of differential equations is given by: $$\mathbf{x}(t) = X(t) \mathbf{c} + \mathbf{x}_p(t) = \left[\begin{array}{ccc}\displaystyle e^{-t}(t^2 - 2t + 2)& e^{-t}(2t - 4)& -2e^{-t}+te^{3t}+2e^{3t} \\ \displaystyle -2e^{-t}(t-1) & e^{-t}(1 - 2t)& 2e^{-t}+2te^{3t}-4e^{3t} \\ \displaystyle 0 & 0 & e^{3t} \end{array}\right]\left[\begin{array}{l}C_1\\C_2\\C_3\end{array}\right]+\left[\begin{array}{l}\displaystyle (e^{-t}(t^2 - 2t + 2))\left(-\frac{1}{3}t^3 + t^2 + C_1\right) + (e^{-t}(2t - 4))\left(-t^3 + 2t^2 + C_2\right) + (-2e^{-t}+te^{3t}+2e^{3t}) C_3\\ \displaystyle (-2e^{-t}(t-1))\left(-\frac{1}{3}t^3 + t^2 + C_1\right) + (e^{-t}(1 - 2t))\left(-t^3 + 2t^2 + C_2\right) + (2e^{-t}+2te^{3t}-4e^{3t}) C_3\\ \displaystyle e^{3t}C_3 \end{array}\right]$$ Where \(\mathbf{c}=\left[\begin{array}{l}C_1\\C_2\\C_3\end{array}\right]\) is a constant vector containing arbitrary constants.

Step 1: Find a fundamental matrix for the homogeneous system

First, we are going to find a fundamental matrix \(X(t)\) for the homogeneous system by finding the matrix exponential of \(A\). \(X(t) = e^{At}\) The given matrix \(A\) is: $$ A=\left[\begin{array}{rrr} -1 & -2 & 2 \\ 2 & 4 & -1 \\ 0 & 0 & 3 \end{array}\right] $$ The matrix exponential is obtained as follows: $$X(t) = e^{At} = I + At + \frac{(At)^2}{2!} + \frac{(At)^3}{3!} + \cdots$$ Computing, we have $$X(t) = \left[\begin{array}{ccc}\displaystyle e^{-t}(t^2 - 2t + 2)& e^{-t}(2t - 4)& -2e^{-t}+te^{3t}+2e^{3t} \\ \displaystyle -2e^{-t}(t-1) & e^{-t}(1 - 2t)& 2e^{-t}+2te^{3t}-4e^{3t} \\ \displaystyle 0 & 0 & e^{3t} \end{array}\right]$$

Step 2: Find the particular solution using the variation of parameters

Now, it is time to use the variation of parameters formula to find the particular solution \(\mathbf{x}_p\). The formula states: $$\mathbf{x}_p(t) = X(t)\int X^{-1}(t)\mathbf{b} dt$$ First, calculate the inverse of \(X(t)\) and then multiply it by \(b\): $$X^{-1}(t)\mathbf{b} = \left[\begin{array}{ccc}\displaystyle e^{t}(t^2 - 2t + 2)& -2e^{t}(t-1)& 0 \\ \displaystyle e^{t}(2t - 4)& e^{t}(1 - 2t) & 0 \\ \displaystyle -2e^{t}+te^{-3t}+2e^{-3t}& 2e^{t}+2te^{-3t}-4e^{-3t} & e^{-3t} \end{array}\right]\left[\begin{array}{l} -e^{3 t} \\ 4 e^{3 t} \\ 3 e^{3 t} \end{array}\right] = \left[\begin{array}{l} -t^2 + 2t \\ -3t^2 + 6t \\ 0 \end{array}\right]$$ Now, we integrate each element of this vector and obtain: $$\int X^{-1}(t)\mathbf{b} dt = \left[\begin{array}{l} -\frac{1}{3}t^3 + t^2 + C_1\\ -t^3 + 2t^2 + C_2 \\ C_3 \end{array}\right]$$ Now we multiply the result by \(X(t)\) to obtain \(\mathbf{x}_p(t)\): $$\mathbf{x}_p(t) = X(t) \left[\begin{array}{l} -\frac{1}{3}t^3 + t^2 + C_1\\ -t^3 + 2t^2 + C_2 \\ C_3 \end{array}\right] = \left[\begin{array}{l}\displaystyle (e^{-t}(t^2 - 2t + 2))\left(-\frac{1}{3}t^3 + t^2 + C_1\right) + (e^{-t}(2t - 4))\left(-t^3 + 2t^2 + C_2\right) + (-2e^{-t}+te^{3t}+2e^{3t}) C_3\\ \displaystyle (-2e^{-t}(t-1))\left(-\frac{1}{3}t^3 + t^2 + C_1\right) + (e^{-t}(1 - 2t))\left(-t^3 + 2t^2 + C_2\right) + (2e^{-t}+2te^{3t}-4e^{3t}) C_3\\ \displaystyle e^{3t}C_3 \end{array}\right]$$

Step 3: Obtain the general solution

Finally, to obtain the general solution to the system of differential equations, we combine the solutions of the homogeneous system and the particular solution. $$\mathbf{x}(t) = X(t) \mathbf{c} + \mathbf{x}_p(t)$$ Where \(\mathbf{c}=\left[\begin{array}{l}C_1\\C_2\\C_3\end{array}\right]\) is a constant vector containing arbitrary constants. This forms the general solution to the system of differential equations.

Key Concepts

Fundamental Matrix

A fundamental matrix is a solution matrix used to solve systems of linear homogeneous differential equations. It consists of multiple solutions that are linearly independent. This solution matrix plays a crucial role in finding other solutions to the differential system.
For our given matrix \(A\), we need to find the fundamental matrix \(X(t)\). To do this, we calculate the matrix exponential \(e^{At}\). This involves using the power series expansion:

• \(I\) is the identity matrix, which acts as 1 in matrix terms.
• Terms like \(At, \frac{(At)^2}{2!}, \frac{(At)^3}{3!}\), etc., are calculated iteratively.

This series provides us with a matrix where each entry is a time-dependent function, representing the fundamental matrix \(X(t)\).

Matrix Exponential

Matrix exponentials are a fascinating way to express the solution of linear homogeneous differential equations. It provides a path from linear algebra to differential equations, and is critical for understanding the dynamics of systems.
Simply put, the matrix exponential \(e^{At}\) transforms the vector \(\mathbf{x}(t)\) through time, based on the effects of matrix \(A\). It helps determine how solutions evolve.
Calculating the matrix exponential involves taking powers of \(A\) and scaling them. This could be seen through:

• \(e^{At} = I + At + \frac{(At)^2}{2!} + \frac{(At)^3}{3!} + \cdots\)
• This allows for evaluating the essence of \(A\) over time, encapsulating complex dynamics as a series expansion.

The beauty of the matrix exponential is it rolls linear algebra into differential equations seamlessly.

Particular Solution

The particular solution is focused on solving non-homogeneous differential equations, enriching the general solution by introducing an additional component.
Using the technique known as variation of parameters, we aim to find a specific solution that satisfies \(\mathbf{x}' = A \mathbf{x} + \mathbf{b}\). This involves:

• Computing the inverse of the fundamental matrix \(X^{-1}(t)\).
• Multiplying \(X^{-1}(t)\) by \(\mathbf{b}\), the non-homogeneous part.
• Integrating the resulting vector - this gives the solution to the particular problem.

This meticulous process helps pinpoint exactly how the particular source \(\mathbf{b}\) affects the system over time.

Differential Equations

Differential equations involve functions and their derivatives, describing how one quantity changes with respect to another.
Linear systems like \(\mathbf{x}' = A \mathbf{x} + \mathbf{b}\) involve linear transformation of vectors, using matrices to represent these transformations.
Key ideas include:

• Homogeneous differential equations, where \(\mathbf{b}=0\).
• Non-homogeneous differential equations, where there is an external input or forcing term \(\mathbf{b}\).
• Solving such systems requires leveraging matrix operations and calculus.

Differential equations model real-world processes, allowing mathematicians to predict behaviors over time.

General Solution

The general solution to a differential equation encompasses all possible solutions, including homogeneous and particular solutions.
For our matrix-driven system, it is represented as:

• \(\mathbf{x}(t) = X(t) \mathbf{c} + \mathbf{x}_p(t)\)
• Here, \(\mathbf{c}\) represents arbitrary constants, allowing for a variety of solutions based on initial conditions.
• \(\mathbf{x}_p(t)\) is the particular solution, bringing in the influences of \(\mathbf{b}\).

This solution package wraps up all dynamics from the matrix into a single expression, demonstrating how starting conditions and external forces work hand-in-hand in shaping the system's behavior.