Question

In Exercises \(1-10\) find a particular solution. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} -6 & -3 \\ 1 & -2 \end{array}\right] \mathbf{y}+\left[\begin{array}{c} 4 e^{-3 t} \\ 4 e^{-5 t} \end{array}\right] $$

Short answer

The particular solution to this differential equation is given by: $$\mathbf{y}(t) = \left[\begin{array}{c} e^{-3t}\left(-\frac{4}{3} - 2t\right) \\\ e^{-3t}(8 + 2t) \end{array}\right]$$

Step-by-step solution

Extracting the system of linear differential equations

Since the matrix form is given, we can read off each of these equations in the form:$$y_1^{\prime}=-6y_1-3y_2+4e^{-3t}$$$$y_2^{\prime}=y_1 - 2y_2 + 4e^{-5t}$$The task now is to solve this system of linear differential equations and find the particular solution.

Solve for \(y_1(t)\) and \(y_2(t)\)

We start with the first equation:$$y_1^{\prime}=-6y_1-3y_2+4e^{-3t}$$Let \(y_1=u\) and \(y_2=v\), then the equation becomes$$u^{\prime}=-6u-3v+4e^{-3t}$$To find function \(u(t)\) and \(v(t)\), first we assume a solution in the form of \(u(t)=e^{-3t}(A+Bt)\). We assume this form as it would lead to getting similar terms after differentiating. Similarly, we assume the solution for \(v(t) = e^{-3t}(C+Dt)\). Now, we need to differentiate u(t) and replace it in the equation above to find the values of constants A, B, C, and D.

Differentiate the functions and plug them into the equation

Differentiating \(u(t)\) and \(v(t)\) gives us$$u^{\prime}=-3Ae^{-3t}+(\text{-}3 B-6A)t e^{-3t}+\text{-}3 B t^2 e^{-3t}$$$$v^{\prime}=-3Ce^{-3t}+(\text{-}3D-6C)t e^{-3t}+\text{-}3 D t^2 e^{-3t}$$Now, plugging these back into the first equation:$$-3Ae^{-3t}+(\text{-}3 B-6A)t e^{-3t}+\text{-}3 B t^2 e^{-3t}=-6u - 3v + 4e^{-3t}$$We can now find the values of A, B, C, and D by equating the coefficients of like terms.

Find the values of constants A, B, C, and D

Equating coefficients of the similar terms, we get the system of equations$$-3A = 4$$$$-6A - 3B = -3C$$$$-3B = -3D$$Solve the equations to obtain A=-4/3, B=-2, C=8,and D=2. Now we can write the found functions u(t) and v(t): $$u(t)=e^{-3t}\left(-\frac{4}{3} - 2t\right)$$$$v(t)=e^{-3t}(8 + 2t)$$

Write the final solution for \(\mathbf{y}\)

Since \(y_1(t)=u(t)\) and \(y_2(t)=v(t)\), we now have the solutions for both components of the vector y(t). The particular solution can be written in vector form $$\mathbf{y}(t) = \left[\begin{array}{c} y_1(t) \\\ y_2(t) \end{array}\right] = \left[\begin{array}{c} e^{-3t}\left(-\frac{4}{3} - 2t\right) \\\ e^{-3t}(8 + 2t) \end{array}\right]$$So, this is the particular solution to the given exercise.

Key concepts

System of Linear Differential Equations

A system of linear differential equations comprises multiple equations that relate two or more functions and their derivatives. Solving such systems is vital in modeling real-world phenomena where multiple factors are dynamically interdependent, such as in engineering and the natural sciences.

In the exercise, we tackle a system involving two functions, \(y_1(t)\) and \(y_2(t)\), and their derivatives. The system is represented in matrix form, indicating a strong interconnectedness between the two functions where their rates of change are influenced by one another and by external functions, in this case, \(4 e^{-3 t}\) and \(4 e^{-5 t}\).

The initial step in finding a solution is to extract individual equations for \(y_1^{\prime}\) and \(y_2^{\prime}\) from the matrix, leading to two coupled first-order linear differential equations with variable coefficients. These equations describe the rate at which each function changes over time, influenced by both its current state and the state of the other function.

Particular Solution

Within the context of differential equations, there are typically two types of solutions: the 'general solution' which includes arbitrary constants to encompass all possible solutions, and the 'particular solution' which relates to a specific instance that satisfies both the equation and initial or boundary conditions presented in a problem.

In the given exercise, the goal was to find a particular solution for the given system of linear differential equations. To achieve this, specific forms for \(y_1(t)\) and \(y_2(t)\) are posited based on the form of the non-homogeneous part of the differential equations (i.e., \(4 e^{-3 t}\) and \(4 e^{-5 t}\)). The chosen forms are then differentiated and substituted back into the original equations, enabling the calculation of the constants that define the particular solution.

This methodology transforms the problem from finding an arbitrary function to determining fixed values for the coefficients within the assumed structure of the functions, which are ultimately used to construct the specific functions that solve the given system under the constraints imposed by the non-homogeneous components.

Differential Equations with Constant Coefficients

When dealing with linear differential equations, the presence of constant coefficients simplifies the process of finding a solution. These coefficients, being constant, imply that the associated rates of change are uniform, not dependent on the independent variable (often time). This feature allows us to apply standard methods such as characteristic equations and exponential trial solutions.

The exercise illustrates the process in a context where the functions' coefficients in the system are constant. This constancy allows the assumption of an exponential solution form \(e^{-3t}(A+Bt)\), which helps in reducing the complexity of the problem. The derivatives of these exponential forms still yield exponential terms, enabling easy comparison of coefficients on either side of the equation to find the unknowns A, B, and so forth.

The constancy of coefficients throughout the equations simplifies the steps required to reach a solution, as it ensures the structure of the differential equations remains manageable after applying the required operations, such as differentiation and equating coefficients, to find the particular solution.