Question

Use the operator method described in this section to find the general solution of each of the linear systems in Exercises 1-26. $$ \begin{aligned} &2 x^{\prime}+y^{\prime}-4 x+y=e^{2 t} \\ &x^{\prime}+y^{\prime}-5 x-2 y=e^{-t} \end{aligned} $$

Short answer

The general solution of the given system of linear differential equations is: $$ \begin{aligned} x(t) &= \frac{1}{4}e^{-t}(-t + 2) + \frac{1}{2}C_1e^{5t} + C_3e^{5t} - 2C_2 \\ y(t) &= \frac{1}{4}e^{-t}(t+2) + C_3e^{5t} + 2C_2 \end{aligned} $$

Step-by-step solution

Define the operators

Let's define the operators D and P for these differential equations: For the first equation: $$D_1 = 2\frac{d}{dt} - 4$$ $$P_1 = 1$$ So, we have \( D_1(x) + P_1(y) = e^{2t} \). For the second equation: $$D_2 = \frac{d}{dt} - 5$$ $$P_2 = -2$$ So, we have \( D_2(x) + P_2(y) = e^{-t} \).

Eliminate one of the operators

Now, let's eliminate one of the operators using substitution. We will do this by solving one equation for x and substitute it into the other equation. We will solve the second equation for x: $$x = D_2^{-1}(e^{-t} - P_2(y)) + C_1$$ Now substitute this expression for x into the first equation: $$D_1(D_2^{-1}(e^{-t} - P_2(y)) + C_1) + P_1(y) = e^{2t}$$

Solve the resulting differential equation

We have now simplified the given system into a single differential equation. We can solve this equation using separation of variables or an integrating factor (if the equation is non-homogeneous). Here, the equation is non-homogeneous, so we will use the method of integrating factors. $$D_1 \circ D_2^{-1}(e^{-t} - P_2(y)) + P_1 \circ D_2^{-1}(P_2(y) - e^{-t}) = e^{2t} - P_1(C_1)$$ First, find the integrating factor for the given equation, which is given by: $$v(t) = e^{\int P_1 dt} = e^{t}$$ Now, multiply through by the integrating factor: $$e^tD_1 \circ D_2^{-1}(e^{-t} - P_2(y)) + e^{t}P_1\circ D_2^{-1}(P_2(y) - e^{-t}) = e^{3t} - P_1(C_1)e^{t}$$ Now, we can integrate this equation with respect to t: $$e^{t}D_2^{-1}(e^{-t} - P_2(y)) + P_1(C_1)te^{t} = \frac{1}{4}e^{3t} + C_2e^{t}$$ Now, we can solve this equation for \(D_2^{-1}(e^{-t} - P_2(y))\): $$D_2^{-1}(e^{-t} - P_2(y)) = \frac{1}{4}te^{-t} + C_2$$ Now, to find y, we can apply the inverse operator \(D_2\) to this expression to obtain: $$y = D_2(\frac{1}{4}te^{-t} + C_2) = \frac{1}{4}e^{-t}(t+2) + C_3e^{5t} + 2C_2$$ Now substituting this expression for y into the equation we found for x (in step 2), we can find the expression for x: $$x = \frac{1}{4}e^{-t}(-t + 2) + \frac{1}{2}C_1e^{5t} + C_3e^{5t} - 2C_2$$ Now combining the expressions for x and y, we have found the general solution of the given system of linear differential equations: $$ \begin{aligned} x(t) &= \frac{1}{4}e^{-t}(-t + 2) + \frac{1}{2}C_1e^{5t} + C_3e^{5t} - 2C_2 \\ y(t) &= \frac{1}{4}e^{-t}(t+2) + C_3e^{5t} + 2C_2 \end{aligned} $$

Key concepts

linear differential systems

Linear differential systems are a set of differential equations where the variables are interdependent. This means that the equations involve derivatives of two or more unknown functions. In the context of this exercise, we are looking at a system with equations involving variables \(x\) and \(y\), their derivatives, and some given functions.

Linear differential systems often appear in mathematical models that describe real-world situations, like population dynamics or electrical circuits. The coefficients of these systems might be constant or functions of an independent variable like time \(t\).

To solve these systems, we use various methods, including the operator method described here. The goal is to find functions \(x(t)\) and \(y(t)\) that satisfy both differential equations in the system.

general solution

The general solution of a differential system encompasses all possible solutions that satisfy the system of equations. It includes particular solutions which satisfy the non-homogeneous part of the equations and the general solution of the associated homogeneous system.

In the context of our exercise:

• The general solution has constants \(C_1, C_2,\) and \(C_3\) to incorporate all potential constants of integration stemming from solving the differential equations.
• For each function \(x(t)\) and \(y(t)\), these constants represent how initial conditions can change the shape of the solution in real applications.

Thus, the general solution here integrates both arbitrary constants and specific functions of \(t\), providing a full depiction of the solution space of the system.

inverse operator

The inverse operator method is a robust technique for solving linear differential equations. When dealing with operators like \(D_1\) and \(D_2\), defined in the solution, the inverse operator is used to isolate and solve for one of the variables in the system.

Here's how it works:

• Given a differential operator, the inverse essentially "undoes" the effect of that operator.
• In your system, applying \(D_2^{-1}\) was crucial for expressing \(x\) in terms of \(y\) or vice versa.

By converting these complex equations into simpler ones where variables can be easily isolated, the inverse operator allows us to make progress towards finding the general solution.

integrating factors

Integrating factors are essential tools for solving linear differential equations, especially when they are non-homogeneous.

An integrating factor is chosen to simplify a differential equation into a form where direct integration is possible. For our system, the integrating factor is determined by:

• Calculating \( e^{\int P_1 dt} \) to transform the equation.
• The integrating factor \(e^t\) here effectively simplifies the handling of the term \(P_2(y)\).

By multiplying the original differential equation by this factor, we ensure it becomes simpler to integrate, ultimately leading us to find expressions for both \(x(t)\) and \(y(t)\) in the general solution.