Question

Use Laplace transforms to convert the following nonhomogeneous systems of differential equations into an algebraic system, and find the solutions of the differential equations. a. $$ \begin{aligned} &x^{\prime}=2 x+3 y+2 \sin 2 t, \quad x(0)=1 \\ &y^{\prime}=-3 x+2 y, \quad y(0)=0 \end{aligned} $$ \(\mathrm{b}\) $$ \begin{aligned} &x^{\prime}=-4 x-y+e^{-1}, \quad x(0)=2 \\ &y^{\prime}=x-2 y+2 e^{-3 t}, \quad y(0)=-1 \end{aligned} $$ c. $$ \begin{array}{ll} x^{\prime} & =x-y+2 \cos t, & x(0)=3 \\ y^{\prime} & =x+y-3 \sin t, & y(0)=2 \end{array} $$

Short answer

To solve these types of nonhomogeneous systems of differential equations, apply Laplace transforms, simplify the transformed equations, solve the algebraic system for \(X(s)\) and \(Y(s)\), then apply inverse Laplace transformation. The detailed solutions would depend on the specifics of the equations in the exercise.

Step-by-step solution

Apply the Laplace Transform

Take the Laplace transform on both sides of the equation. Use the properties of Laplace transform, namely linearity and the shifting property. For instance, the Laplace transform of \(x^{\prime}\) is \(sX(s) - x(0)\), and the Laplace transform of \(y^{\prime}\) is \(sY(s) - y(0)\). Using these, transform the given equations.

Simplify the Transformed Equations

Write the transformed equations in the standard form by grouping terms. This will simplify further calculations.

Solve the Transformed Equations

Now solve the transformed equations to find \(X(s)\) and \(Y(s)\). In order to do it, you can use the technique of substitution or elimination, which is familiar from solving systems of algebraic equations. From the solved equations, you get the Laplace transforms of the solutions of the original differential equations.

Apply Inverse Laplace Transform

Using the solutions obtained in the previous step, apply the inverse Laplace transformation. Note that to perform inverse Laplace transformation, the solution should be simplified into a form listed in the standard table of Laplace transforms. If the differential equations were written correctly and all calculations were made without mistakes, the result will meet the initial conditions specified in the exercise.

Key concepts

Nonhomogeneous Systems of Differential Equations

When faced with nonhomogeneous systems of differential equations, like the ones presented in our exercise, students encounter a set of linked equations that involve both the derivative(s) of the unknown function(s) and an additional non-zero term, often understood as an external force or input.
To comprehend these systems, it's critical to recognize that they represent interconnected dynamic phenomena. Each equation implies how one variable's change rate is related to itself and others, while the nonhomogeneous term injects complexity and external influence into the system.

Characteristic vs. Nonhomogeneous Parts

Understanding the solution comprises two parts: one that is associated with the characteristic equation (homogeneous solution), and another that arises due to the nonhomogeneous part. The latter is special in each problem and can involve constants, functions like exponentials, or trigonometric inputs, such as the \(2 \sin 2t\) and \(e^{-3t}\) seen in our given exercise.

Superposition Principle

An interesting aspect is the principle of superposition, which asserts that the total solution is a sum of the homogeneous and particular solutions. This property enables us to tackle each part separately before merging them for the full answer, simplifying the overall process.
In practice, the nonhomogeneous systems require careful examination. Every term contributes to the character and behavior of the overall system, and missing any connection or influence could lead to misinterpretation of the system's dynamics.

Algebraic System Conversion

Algebraic system conversion is a pivotal step in handling differential equations using the Laplace transform. It primarily serves as a bridge between the time domain, where the original differentials live, and a sort of frequency or complex domain that allows for a more manageable manipulation of the equations.

Transforming Nonhomogeneous Systems

In our exercise, applying the Laplace transform to each term in the differential equations converts the problem into an algebraic system. This system appears less intimidating since it usually consists of polynomials and algebraic equations. It's important to emphasize that the accuracy of this transformation depends on the correct application of Laplace properties, such as linearity and the initial value theorem akin to the shifting property.

Advantages of Conversion

One key advantage of this conversion is that solving an algebraic system is a process familiar to most students. Using methods like substitution or elimination - techniques common in solving algebraic equations - they can systematically find solutions for the variable expressions in the Laplace domain. Moreover, this translations allow students to bypass the complexity of directly solving differential equations, particularly when they involve intricate or non-standard nonhomogeneous terms.
However, remember that at this stage, our solutions represent the system's behavior in an abstract form. For complete solutions, we will need to revert to the time domain, showing the true evolution of our system over time.

Inverse Laplace Transform

After successfully transforming and solving for the Laplace domain expressions, it's time for the crucial reversal: the inverse Laplace transform. This mathematical operation converts the algebraic solutions back to the time domain, unveiling the original differential system's dynamics.
In the exercise, having found the expressions for \(X(s)\) and \(Y(s)\), we revert to \(x(t)\) and \(y(t)\) by applying the inverse Laplace transform. This requires a keen understanding of the standard forms and properties of Laplace transforms to ensure that the algebraic expressions can indeed be reverted.

Standard Forms and Lookup Tables

It is often necessary to manipulate the transformed solutions algebraically into standard forms for which the inverse transform is known and tabulated. The challenge here is to recognize and apply partial fraction decomposition, convolution theorem, or other algebraic manipulation techniques to the transformed functions, preparing them for a straightforward inverse process.

Checking the Initial Conditions

The final step is to check that our transformed functions satisfy the initial conditions given in the problem statement. This is an essential verification act that ensures the validity and correctness of our solutions. If discrepancies arise, revisiting the transformation steps for potential errors is warranted. The beauty of the inverse transform lies not only in its power to reveal the time-based solutions but also in validating the entire process - much like closing a loop that started with our initial application of the Laplace transform.