Question

Use Laplace transforms to convert the following system of differential equations into an algebraic system, and find the solution of the differential equations. $$ \begin{array}{ll} x^{\prime \prime}=3 x-6 y, & x(0)=1, & x^{\prime}(0)=0 \\ y^{\prime \prime}=x+y, & y(0)=0, & y^{\prime}(0)=0 \end{array} $$

Short answer

Applying the Laplace transform to the system of differential equations and solving the resulting algebraic equations yield the solutions for \(x(t)\) and \(y(t)\).

Step-by-step solution

Laplace Transform

First, apply the Laplace transform to each differential equation in the system. The Laplace transform of \(x^\prime\prime\) is \(s^2X(s)-sx(0)-x^\prime(0)\), and the transform of \(y^\prime\prime\) is \(s^2Y(s)-sy(0)-y^\prime(0)\). In this case, \(x(0)=1\), \(x^\prime(0)=0\), \(y(0)=0\), and \(y^\prime(0)=0\). Therefore, given \(x^\prime\prime=3x-6y\), the transformed equation is \(s^2X(s) - s = 3X(s) - 6Y(s)\). Secondly, for \(y^\prime\prime=x+y\), the transformed equation is \(s^2Y(s) = X(s) + Y(s)\).

Algebraic System

After applying the Laplace transform, rearrange the equations to form an algebraic system. The equation for \(x\) yields \(s^2X(s) - 3X(s) = s + 6Y(s)\), and the equation for \(y\) yields \(s^2Y(s) - Y(s) = X(s)\). These two equations form the algebraic system.

Solving the Algebraic System

Solve the algebraic system of equations for \(X(s)\) and \(Y(s)\). Substitute \(X(s)\) from the first equation into the second equation, get \(Y(s)\). Substitute this \(Y(s)\) into the first equation, get \(X(s)\)

Inverse Laplace Transform

Move back to the time domain by applying the inverse Laplace transform. The inverse Laplace transform of \(X(s)\) gives the solution \(x(t)\), and the transform of \(Y(s)\) gives the solution \(y(t)\).

Key concepts

System of Differential Equations

In the realm of mathematics, specifically within the study of dynamics and systems, a system of differential equations consists of multiple equations that relate to the rates at which quantities change. These systems are crucial in modeling the behavior of interacting components in fields like physics, biology, and engineering.

A typical system includes several unknown functions and their derivatives, all interconnected. The system given in our exercise, \[\left\{\begin{array}{l} x^{\text{\prime\prime}} = 3x - 6y, \ y^{\text{\prime\prime}} = x + y\end{array}\right.\]
comprises of two second-order differential equations where the primes denote differentiation with respect to time, and involves two dependent variables, \(x(t)\) and \(y(t)\), which are functions of time. The initial conditions provided help in determining a unique solution to the system.

These equations describe the rates of change of \(x\) and \(y\) in response to themselves and each other, reflecting an interconnected dynamic system where the behavior of each variable influences, and is influenced by, the other.

Algebraic System

After transforming the system of differential equations using Laplace transforms, the result is an algebraic system. Unlike differential equations, algebraic equations involve functions and their coefficients only, without derivatives. The beauty of Laplace transforms lies in their ability to convert complex differential equations into easier-to-manage algebraic equations.

In the context of the given exercise, we utilize the Laplace transform to convert the original differential system into:

Transformed Equations

\[\begin{aligned} s^2X(s) - 3X(s) &= s + 6Y(s), \ s^2Y(s) - Y(s) &= X(s).\end{aligned}\]
This new system consists of linear equations with the variables \(X(s)\) and \(Y(s)\) representing the Laplace transforms of the functions \(x(t)\) and \(y(t)\). The transformed equations are less daunting to handle as they only require algebraic manipulations for solving, leading to solutions that can be transformed back into the time domain.

Inverse Laplace Transform

The final phase in solving a system of differential equations with Laplace transforms is to apply the inverse Laplace transform. This process translates the algebraic solutions back into their original time-domain functions, which describe the behavior of the system as it evolves over time.

In practice, once we have the solutions \(X(s)\) and \(Y(s)\) to the algebraic system, we look up or compute their corresponding inverse transforms. This step requires familiarity with common Laplace pairs and the properties of the transform. For example, the inverse Laplace transform of a function \(F(s)\) is denoted by \(\mathcal{L}^{-1}\{F(s)\}\).

Returning to the time domain implies converting the functions from a form that is dependent on 's' (a complex frequency domain), back to a function of 't' (time), such as:
\[\mathcal{L}^{-1}\left\{X(s)\right\} = x(t) \quad \text{and} \quad \mathcal{L}^{-1}\left\{Y(s)\right\} = y(t)\].
The transformed expressions \(X(s)\) and \(Y(s)\) from our exercise would be inverted to yield the time-domain solutions to the original second-order differential equations, hence completing the task at hand.