Question

Use the operator method described in this section to find the general solution of each of the linear systems in Exercises 1-26. $$ \begin{aligned} &x^{\prime \prime}+y^{\prime}=e^{2 t} \\ &x^{\prime}+y^{\prime}-x-y=0 \end{aligned} $$

Short answer

The general solution for the given linear system is: \[ \begin{aligned} x(t) &= k_1e^{r_1t} + k_2e^{r_2t} \\ y(t) &= ct + d - (k_1e^{r_1t} + k_2e^{r_2t}) \end{aligned} \]

Step-by-step solution

1. Identifying the Operator

First, observe that the first equation says \(x'' + y' = e^{2t}\), while the second equation says \(x' + y' -x - y = 0\). We can rewrite the second equation as \(x' - x = -y' + y\). Now let's use the differential operator D, where D is the derivative with respect to t. We can rewrite our system using the operator: \[ \begin{aligned} &(D^2 + D)x + Dy = e^{2t} \\ &(D - 1)x + (D - 1)y = 0 \end{aligned} \] In the second equation, we can notice that there is a common factor of (D - 1), which gives us an operator:

2. Applying the Operator

Now, we will factor the operator (D - 1) in the second equation: \((D - 1)(x + y) = 0\) This equation states that the sum of x and y has constant derivative. Let's set \(z = x + y\), then \(z' = \text{constant}\). Let this constant be c. So, we have: \(z' = c\) Integrating both sides with respect to t, we get: \(z = ct + d\) Where d is an integration constant.

3. Solving for x and y

Now, we have the equation for z, which is the sum of x and y: \(x + y = ct + d\) We will use the first equation to solve for x: \((D^2 + D)x + Dy = e^{2t}\) Since \(Dy = z' - Dx = c - Dx\), substitute this expression into the equation: \((D^2 + D)x + (c - Dx) = e^{2t}\) \(D^2x + 2Dx - c = e^{2t}\) To solve this differential equation for x(t), assume that \(x = e^{rt}\) and substitute: \((r^2 + 2r - c)e^{rt} = e^{2t}\) We can solve this characteristic equation for r: \(r^2 + 2r - c = 2\) Taking r1 and r2 as roots, we can write the general solution for x(t) as follows: \(x(t) = k_1e^{r_1t} + k_2e^{r_2t}\) Now, to find y(t), use the relation: \(y(t) = ct + d - x(t) = ct + d - (k_1e^{r_1t} + k_2e^{r_2t})\)

4. General Solution

Finally, we have the general solution for the given linear system as: \[ \begin{aligned} x(t) &= k_1e^{r_1t} + k_2e^{r_2t} \\ y(t) &= ct + d - (k_1e^{r_1t} + k_2e^{r_2t}) \end{aligned} \] In this way, the operator method was used to find the general solution of the given linear system.

Key concepts

Linear Systems of Differential Equations

When solving linear systems of differential equations, understanding the interaction between multiple functions and their derivatives is crucial. In the given exercise, we encounter such a system with two equations involving the functions x(t) and y(t), and their derivatives.

In the realm of mathematics, these systems showcase the interconnected nature of physical quantities in applied problems, like in engineering, physics, and other sciences. The solution pathway often involves identifying shared characteristics, or 'operators', in the differential equations that allow us to simplify and solve the problem.

For instance, recognizing that an operator, such as (D - 1), is common to terms in different equations helps us create a new equation involving a new variable z, which is a linear combination of x and y. This simplification is a key strategy that turns a system of equations into something more manageable, eventually leading us to find the particular solutions for x(t) and y(t) individually.

Characteristic Equation

The characteristic equation is a pivotal concept when dealing with linear differential equations. It is a polynomial whose roots are essential in determining the solutions to differential equations.

In our exercise, the characteristic equation arises from an assumption: assuming a solution of the form x(t) = e^{rt} for some constant r, we substitute this form into the differentials of x to obtain a polynomial equation in r. Solving this equation gives us the values of r that satisfy the original differential equation.

Why 'Characteristic'?

These values are 'characteristic' of the differential operator defined by the equation, representing the natural frequencies or modes of the system described by the linear differential equation. Knowing how to derive and solve this characteristic equation is a foundational skill for any student delving into the study of ordinary differential equations and their applications in various fields.

Integration of Functions

Integration is a fundamental mathematical operation, often viewed as the inverse of differentiation, and it plays a significant role in solving differential equations.

In the context of our system of equations, integration is utilized to find a general solution for a new function z defined as a sum of x and y. Once we knew that z' is a constant, integrating with respect to t gave us z as a linear function of t, uncovering part of the system's behavior over time.

The Power of Integration

Through integration, we obtain functions that describe cumulative quantities or the accumulation of change over time. For students, mastering the technique of integrating functions—whether by recognizing antiderivatives directly or applying methods like substitution and integration by parts—is critical for solving a wide range of problems in calculus and beyond.