Question

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

Short answer

The general solutions for \(x(t)\) and \(y(t)\) are: \( x(t) = \frac{1}{2}\left(C_1 + \frac{1}{6}t^2 + t - \frac{5}{6}\right) \) \( y(t) = \frac{1}{2}\left(C_1 - \frac{1}{6}t^2 - t + \frac{5}{6}\right) \)

Step-by-step solution

Rewrite the ODEs with operators

Rewrite the given ODEs in terms of operators \(D_x\) and \(D_y\), where \(D_x\) is the derivative \( \frac{d}{dt} \) applied to the function \(x(t)\) and \(D_y\) is the derivative \( \frac{d}{dt} \) applied to the function \(y(t)\). Our ODEs can be rewritten as: \( (2D_x + D_y - 1)(x) = -2t \) \( (D_x + D_y + 1)(y) = t^2 \)

Add or subtract the equations to obtain a simplified equation

We can add both equations to obtain a simplified equation: \( (2D_x + D_y - 1)(x) + (D_x + D_y + 1)(y) = -2t + t^2 \)

Solve the simplified equation

Now, we are left with the following simplified equation: \( (3D_x + 2D_y)(x + y) = t^2 - 2t \) To solve this equation, let \(z(t) = x(t) + y(t)\). Then, apply the operators on the left-hand side: \( 3D_x(z) + 2D_y(z) = 3z' + 2z'' = t^2 - 2t \) This is a second-order linear ODE, which can be solved using integrating factors or other methods. In this case, we will use the method of undetermined coefficients. First, find the complementary solution by solving the homogeneous equation: \( 3z_c' + 2z_c'' = 0 \) Guess a solution of the form \(z_c(t) = C_1e^{\lambda t}\), and plug it in: \( (3\lambda C_1e^{\lambda t} + 2\lambda^2 C_1e^{\lambda t}) = 0 \) Divide by \(C_1e^{\lambda t}\): \( 3\lambda + 2\lambda^2 = 0 \) Solve for \(\lambda\) and get the complementary solution: \( \lambda_1 = 0 \Rightarrow z_c(t) = C_1 \) Now we need to find a particular solution of the non-homogeneous equation by guessing a function of the same form as the inhomogeneous term; \(t^2 - 2t\). Hence, we assume a particular solution of the form: \( z_p(t) = A t^2 + Bt + C \) Differentiate and substitute the derivatives in the non-homogeneous equation: \( z_p'(t) = 2At + B \) \( z_p''(t) = 2A \) Substitute these expressions into the original ODE: \( 3(2At + B) + 2(2A) = t^2 - 2t \) Solve for the coefficients: \( A = 1/6 \) \( B = -1 \) \( C = -11/6 \) So our particular solution is: \( z_p(t) = \frac{1}{6}t^2 - t - \frac{11}{6} \) Next, we combine the complementary and particular solutions to obtain the general solution for \(z(t)\): \( z(t) = C_1 + \frac{1}{6}t^2 - t - \frac{11}{6}\)

Use the operators in the initial ODEs to find the general solution of the linear system

Now that we have the general solution for \(z(t)\), we can go back to the original ODEs. Recall that \(z(t) = x(t) + y(t)\). Since: \( (2D_x + D_y -1)(x) = -2t \) \( x(t) = \frac{1}{2}(z(t) + 1 + 2t) \) \( x(t) = \frac{1}{2}(C_1 + \frac{1}{6}t^2 - t - \frac{11}{6} + 1 + 2t) \) \( x(t) = \frac{1}{2}\left(C_1 + \frac{1}{6}t^2 + t - \frac{5}{6}\right) \) Similarly, since: \( (D_x + D_y +1)(y) = t^2 \) \( y(t) = -x(t) + z(t) - 1 \) \( y(t) = -\frac{1}{2}\left(C_1 + \frac{1}{6}t^2 + t - \frac{5}{6}\right) + \left(C_1 + \frac{1}{6}t^2 - t - \frac{11}{6} - 1\right) \) \( y(t) = \frac{1}{2}\left(C_1 - \frac{1}{6}t^2 - t + \frac{5}{6}\right) \) Finally, we have our general solutions for \(x(t)\) and \(y(t)\): \( x(t) = \frac{1}{2}\left(C_1 + \frac{1}{6}t^2 + t - \frac{5}{6}\right) \) \( y(t) = \frac{1}{2}\left(C_1 - \frac{1}{6}t^2 - t + \frac{5}{6}\right) \)

Key concepts

Linear Systems of Differential Equations

Linear systems of differential equations involve multiple functions and their derivatives that are linked together. In the context of the exercise, we encounter a pair of equations that represent a system where the rate of change of two variables, in this case, functions represented as x(t) and y(t), depends not only on time t but also on each other.

In an attempt to solve such systems, we often aim to simplify and decouple the equations to handle one function at a time. Techniques such as adding or subtracting the equations, as seen in the step-by-step solution, are employed to reduce the complexity of the system. The eventual goal is to express each variable in terms of t alone, which yields the individual functions that solve the system comprehensively.

Operator Method

The operator method, used in our example, transforms differential equations into a more manageable algebraic form by replacing derivatives with operators. These operators, like D_x and D_y, act upon the functions x(t) and y(t), simplifying the expressions and highlighting the relationships between the differentials.

In the context of the exercise, the use of operators helps combine the equations, which then allows for the application of algebraic techniques to find the solution to the system. The operator method is a powerful tool as it provides clarity in manipulation and can often reveal solutions that are not immediately apparent.

Method of Undetermined Coefficients

The method of undetermined coefficients is an approach to find particular solutions to non-homogeneous linear differential equations. In the given solution, it is used to determine a specific form of solution that satisfies the non-homogeneous part of our equation, which is represented as a polynomial in t.

The key idea is to guess a solution with undetermined coefficients and then determine those coefficients by substituting back into the original equation. It requires an understanding of the form of the non-homogeneous term to choose a correct 'guess' for the particular solution. The coefficients A, B, and C are then calibrated to fit the equation. The final particular solution augments the complementary solution derived from the homogeneous part, resulting in the general solution for the system.

Second-order Linear ODE

A second-order linear ordinary differential equation (ODE) contains derivatives up to the second order and can be expressed in a standard form. The solution process involves finding two parts: the complementary solution, which is the solution of the homogeneous equation (where the right-hand side is zero), and the particular solution, which accounts for the non-homogeneous part.

In the exercise, once simplified, we obtain a second-order linear ODE in the form of a new variable z(t). Even though the original problem started with a first-order system, combining the equations led us to face this second-order challenge. Traditionally solved by either the characteristic equation method or by using integrating factors, this class of ODEs can also be approached with the method of undetermined coefficients when the non-homogeneous term has a regular, easily predictable format, as we saw with the polynomial term in the exercise.