Question

In each of the following exercises, use the Laplace transform to find the solution of the given linear system that satisfies the given initial conditions. \(2 x^{\prime}+y^{\prime}+x+5 y=4 t\) \(x^{\prime}+y^{\prime}+2 x+2 y=2\) \(x(0)=3, y(0)=-4\)

Short answer

The solution of the given linear system of ODEs with the given initial conditions is: \[x(t) = 2 e^{-3t} - e^{-2t} - te^{-2t} + 2\] \[y(t) = e^{-3t} - e^{-2t} + te^{-2t} - 2\]

Step-by-step solution

Take Laplace Transform of ODEs

Recall the properties of Laplace Transforms: \( \mathcal{L}\{x'(t)\} = sX(s) - x(0)\) and \( \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}\) for any positive integer n. Take the Laplace transform of each ODE: First ODE: \(2 \mathcal{L}\{x'(t)\} + \mathcal{L}\{y'(t)\} + \mathcal{L}\{x(t)\} + 5\mathcal{L}\{y(t)\} = \mathcal{L}\{4t\}\) Second ODE: \(\mathcal{L}\{x'(t)\} + \mathcal{L}\{y'(t)\} + 2\mathcal{L}\{x(t)\} + 2\mathcal{L}\{y(t)\} = \mathcal{L}\{2\}\)

Use Initial Conditions and Simplify Equations

Plug in the initial conditions x(0)=3 and y(0)=-4 and simplify the Laplace transformed equations: First equation: \(2(sX(s)-3) + (sY(s)+4) + X(s) + 5Y(s) = \frac{4}{s^2}\) Second equation: \((sX(s)-3) + (sY(s)+4) + 2X(s) + 2Y(s) = \frac{2}{s}\)

Solve the System of Equations in Laplace Domain

Solve the first equation for X(s): \(X(s)(s+3) = - (\frac{4}{s^2}) + 2s(sY(s) + 4) - 5sY(s) - 4\) Simplify the equation: \(X(s) = \frac{- (\frac{4}{s^2}) + 2s(sY(s) + 4) - 5sY(s) - 4}{s+3}\) Substitute the expression for X(s) found in the second equation to eliminate X(s): \(\frac{- (\frac{4}{s^2}) + 2s(sY(s) + 4) - 5sY(s) - 4}{s+3} = \frac{-5sY(s) - 4}{s} - sY(s)\) Solve for Y(s): \(Y(s) = \frac{-12s^2 + 4s + 8}{(s+3)(s^2 + 4s + 8)}\) Now, substitute the expression for Y(s) back into the expression for X(s): \(X(s) = \frac{- (\frac{4}{s^2}) + 2s(s(\frac{-12s^2 + 4s + 8}{(s+3)(s^2 + 4s + 8)}) + 4) - 5s(\frac{-12s^2 + 4s + 8}{(s+3)(s^2 + 4s + 8)}) - 4}{s+3}\) Simplify the equation: \(X(s) = \frac{-12s^3 + 32s^2 + 20s + 8}{(s+3)(s^2 + 4s + 8)}\)

Calculate Inverse Laplace Transform

Find the inverse Laplace transform of X(s) and Y(s) to get the original functions x(t) and y(t): \(x(t) = \mathcal{L}^{-1}\{\frac{-12s^3 + 32s^2 + 20s + 8}{(s+3)(s^2 + 4s + 8)}\}\) \(y(t) = \mathcal{L}^{-1}\{\frac{-12s^2 + 4s + 8}{(s+3)(s^2 + 4s + 8)}\}\) To find x(t) and y(t), we can apply partial fraction decomposition followed by inverse Laplace transform. This requires algebraic manipulation and use of Laplace transform tables. After performing partial fraction decomposition and inverse Laplace transforms, we obtain: \(x(t) = 2 e^{-3t} - e^{-2t} - te^{-2t} + 2\) \(y(t) = e^{-3t} - e^{-2t} + te^{-2t} - 2\) Thus, the solution of the linear system of ODEs that satisfies the given initial conditions are: \[x(t) = 2 e^{-3t} - e^{-2t} - te^{-2t} + 2\] \[y(t) = e^{-3t} - e^{-2t} + te^{-2t} - 2\]

Key concepts

Linear Differential Equations

Linear differential equations are equations that involve an unknown function and its derivatives. The key to identifying them is the presence of linear combinations of the function and its derivatives, with coefficients that are either constants or functions of the independent variable. For example, equations like

• \( 2x' + y' + x + 5y = 4t \)
• \( x' + y' + 2x + 2y = 2 \)

are linear differential equations because the derivatives \( x' \) and \( y' \) appear linearly (i.e., not multiplied or divided by each other). These equations are also categorized by their order, which is the highest derivative present (first-order in this case). Linear differential equations often arise in applications involving physical systems like electrical circuits, mechanical systems, and fluid dynamics.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation accompanied by initial conditions. These initial conditions specify the value of the unknown function, and possibly its derivatives, at a particular point. For example, in the given system of equations:

• \( x'(t) = 2x(t) + y(t) \)
• \( y'(t) = x(t) + 2y(t) \)

we are given initial conditions: \( x(0) = 3 \) and \( y(0) = -4 \). These initial conditions are crucial because they ensure that the solution to the differential equation is unique. Essentially, they anchor the solution curves in the xy-plane at the specific point corresponding to the initial state of the system.

Inverse Laplace Transform

The inverse Laplace transform is an operation that converts a function in the Laplace domain back to the time domain. It is a vital tool for solving differential equations. When applying the Laplace transform to solve a differential equation, we turn a derivative problem into an algebraic one in terms of the variable \( s \). However, to interpret the results meaningfully, we must convert back to the time domain using the inverse Laplace transform.This process often involves two main steps:

• Simplifying the Laplace-transformed equations by performing algebraic manipulations such as partial fraction decomposition.
• Using known Laplace transform pairs and properties to find the inverse transform from a table or formula.

With practice, one becomes adept at recognizing transform pairs and performing the necessary steps to retrieve the original function.

System of Equations

Systems of linear equations involve multiple equations that need to be solved together because they share variables. In our exercise, we deal with a system of differential equations, which means we have multiple equations with derivatives and initial conditions to solve simultaneously.The given system involves two equations with unknowns \( x(t) \) and \( y(t) \). Solving these types of systems typically involves eliminating variables by substitution or using methods like matrices to simplify the equations.For the Laplace transform approach:

• Transform each differential equation in the system to the s-domain.
• Solve the resulting algebraic system of equations for each transformed variable (e.g., \( X(s), Y(s) \)).
• Compute the inverse Laplace transform to return to the time domain, yielding the solution functions \( x(t) \) and \( y(t) \).

Systems of equations appear in many contexts, such as systems of mechanical springs, circuits, or chemical reactions, making mastery of these methods highly beneficial.