Question

In Exercises \(1-10\) find a particular solution. $$ \mathbf{y}^{\prime}=\frac{1}{3}\left[\begin{array}{rrr} 1 & 1 & -3 \\ -4 & -4 & 3 \\ -2 & 1 & 0 \end{array}\right] \mathbf{y}+\left[\begin{array}{c} e^{t} \\ e^{t} \\ e^{t} \end{array}\right] $$

Short answer

Question: Determine the particular solution for the following system of first-order linear ODEs: $$ \begin{cases} y_1' = \frac{1}{3}(y_1 + y_2 - 3y_3) + e^t \\ y_2' = \frac{1}{3}(-4y_1 -4y_2 + 3y_3) + e^t \\ y_3' = \frac{1}{3}(-2y_1 + y_2) + e^t \end{cases} $$ Answer: The particular solution for the given system of ODEs is: $$ \mathbf{y_p}(t) = \frac{1}{8} \left[\begin{array}{c}31e^t\\25e^t\\45e^t\end{array}\right] $$

Step-by-step solution

Write down the given system of ODEs

First, let's write down the given matrix equation as a system of three first-order ODEs: $$ \begin{cases} y_1' = \frac{1}{3}(y_1 + y_2 - 3y_3) + e^t \\ y_2' = \frac{1}{3}(-4y_1 -4y_2 + 3y_3) + e^t \\ y_3' = \frac{1}{3}(-2y_1 + y_2) + e^t \end{cases} $$

Assume the form of the particular solution

Assume that the particular solution, \(\mathbf{y_p}(t)\), has the same form as the non-homogeneous term, \(\mathbf{g}(t)\). So, let \(\mathbf{y_p}(t) = \left[\begin{array}{c}Ae^t\\Be^t\\Ce^t\end{array}\right]\), where \(A\), \(B\), and \(C\) are constants to be determined.

Substitute the assumed particular solution into the system of ODEs

Now, substitute the assumed particular solution into the given system of ODEs and simplify. The derivatives of the components of the particular solution are: $$ \mathbf{y_p'}(t) = \left[\begin{array}{c}Ae^t\\Be^t\\Ce^t\end{array}\right]' = \left[\begin{array}{c}Ae^t\\Be^t\\Ce^t\end{array}\right] $$ Plugging in \(\mathbf{y_p}(t)\) and \(\mathbf{y_p'}(t)\) into the system of ODEs, we get: $$ \begin{cases} Ae^t = \frac{1}{3}(A + B -3C)e^t + e^t \\ Be^t = \frac{1}{3}(-4A -4B + 3C)e^t + e^t \\ Ce^t = \frac{1}{3}(-2A + B)e^t + e^t \end{cases} $$

Solve for the undetermined coefficients

Dividing each equation by \(e^t\), we are left with a system of linear equations: $$ \begin{cases} A = \frac{1}{3}(A + B -3C) + 1 \\ B = \frac{1}{3}(-4A -4B + 3C) + 1 \\ C = \frac{1}{3}(-2A + B) + 1 \end{cases} $$ Solving this system of linear equations (using substitution, elimination, or matrix methods), we obtain: $$ A = \frac{31}{8}, \quad B = \frac{25}{8}, \quad C = \frac{45}{8} $$

Write down the particular solution

Now that we have found the constants \(A\), \(B\), and \(C\), we can write down the particular solution: $$ \mathbf{y_p}(t) = \left[\begin{array}{c}\frac{31}{8}e^t\\\frac{25}{8}e^t\\\frac{45}{8}e^t\end{array}\right] $$ Thus, the particular solution for the given system of ODEs is: $$ \mathbf{y_p}(t) = \frac{1}{8} \left[\begin{array}{c}31e^t\\25e^t\\45e^t\end{array}\right] $$

Key concepts

System of Ordinary Differential Equations

A system of ordinary differential equations (ODEs) involves multiple dependent variables, typically seen in the form of vectors, that are contingent upon one or more independent variables. This often results in simultaneous differential equations. Here, our independent variable is time, denoted as \(t\), and we are dealing with three dependent variables \(y_1, y_2,\) and \(y_3\). Such a system describes how these variables interdependently change with respect to time.
For example, the given system of ODEs can be individually represented as:

• \(y_1' = \frac{1}{3}(y_1 + y_2 - 3y_3) + e^t\)
• \(y_2' = \frac{1}{3}(-4y_1 - 4y_2 + 3y_3) + e^t\)
• \(y_3' = \frac{1}{3}(-2y_1 + y_2) + e^t\)

This method allows us to capture complex dynamics between the variables that a single differential equation might miss.

Non-homogeneous Differential Equations

A non-homogeneous differential equation includes terms that are independent of the function and its derivatives. More simply, it includes a function or constant on the right-hand side of the equation that isn't related to the dependent variable functions. In our case, the term \(e^t\) in each of the differential equations is non-homogeneous, meaning it doesn't depend on \(y_1, y_2,\) or \(y_3\).
This non-homogeneous term \(e^t\) modifies the solution of the differential equation by influencing the way solutions are constructed. Rather than having solutions that revert to zero, as is often the case with homogeneous equations, the solutions of non-homogeneous equations will partake in the form of the non-homogeneous term itself.
Our goal, therefore, is to identify a particular solution which will satisfy the entire setup, accounting for these additional \(e^t\) terms.

Matrix Representation of ODEs

Matrix representation of systems of ODEs provides an organized way to succinctly convey the structure of complex systems. By using a matrix, every equation can be presented compactly, providing ease in manipulation and solving.
When we observe the original exercise, the differential equations were initially presented in matrix form, wherein matrices contained coefficients from the system of equations. Here's the initial representation:
\[\mathbf{y}' = \frac{1}{3}\begin{bmatrix}1 & 1 & -3 \ -4 & -4 & 3 \ -2 & 1 & 0\end{bmatrix} \mathbf{y} + \begin{bmatrix}e^t \ e^t \ e^t\end{bmatrix}\]
Here, \(\mathbf{y}'\) is a vector of the derivatives \(y_1', y_2', y_3'\), and \(\mathbf{y}\) is the vector \(y_1, y_2, y_3\). Matrix operations provide efficiency, especially when dealing with linear algebra techniques like finding eigenvalues, decomposition, or while solving using numerical methods.

ODEs with Exponential Terms

Differential equations with exponential terms are quite common in mathematical modeling of natural phenomena. In our current system of ODEs, the exponential \(e^t\) appears in the non-homogeneous terms.
Recognizing such exponential terms is crucial, as they influence the structure of potential solutions. In particular, when seeking a particular solution, one often initially hypothesizes a solution that reflects the form of these non-homogeneous exponential terms. In our exercise, we assume a particular solution of the type \[\mathbf{y_p}(t) = \begin{bmatrix}Ae^t \ Be^t \ Ce^t \end{bmatrix}\]where \(A, B,\) and \(C\) are constants determined through substitution into the system of equations.
This assumption leverages the powerful property of exponential functions, which is preserved through differentiation, making them suitable candidates for particular solutions in non-homogeneous linear differential equations.