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. \(x^{\prime}-2 x-3 y=0\) \(y^{\prime}+x+2 y=t\) \(x(0)=-1, y(0)=0\)

Short answer

Short Answer: After applying the Laplace transform to the given system of linear differential equations and solving for the Laplace transforms of x(t) and y(t), we find their inverse Laplace transforms to obtain the solutions: \(x(t) = e^{2t} - te^{2t} - t^2e^{2t}\) \(y(t) = -te^{2t} - t^2e^{2t} + te^{3t}\)

Step-by-step solution

Take the Laplace transform of both equations: \(L\{x'(t) - 2x(t) - 3y(t) = 0\}\) \(L\{y'(t) + x(t) + 2y(t) = t\}\) #Step 2: Apply the Laplace transform to each term in the equations#:

Use the properties of Laplace transform to transform each term in the equations: \(L\{x'(t)\} - 2L\{x(t)\} - 3L\{y(t)\} = 0\) \(L\{y'(t)\} + L\{x(t)\} + 2L\{y(t)\} = L\{t\}\) Now, use the tables of Laplace transforms and the initial conditions given: \(sX(s) + 1 - 2X(s) - 3Y(s) = 0\) \(sY(s) - 0 + X(s) + 2Y(s) = \frac{1}{s^2}\) #Step 3: Solve the transformed equations for X(s) and Y(s)#:

Solve the first equation for X(s) and substitute it into the second equation to eliminate X(s): \(X(s) = \frac{3Y(s) + sY(s) - 1}{s-2}\) Now, substitute this expression for X(s) into the second equation: \(sY(s) + \frac{3Y(s) + sY(s) - 1}{s-2} + 2Y(s) = \frac{1}{s^2}\) Multiply both sides by \((s-2)s^2\) to clear the denominators: \(s^3Y(s) - 2s^2Y(s) + 3s^2Y(s) + s^3Y(s) - s^2 + 2s^3Y(s) - 4s^2Y(s) = (s-2)\) Combine like terms and factor the equation for Y(s): \(Y(s) = \frac{s^2 + s - 2}{(s^3 - 3s^2 + 4s)}\) Similarly, substitute the expression for Y(s) in the expression for X(s): \(X(s) = \frac{3\frac{s^2 + s - 2}{(s^3 - 3s^2 + 4s)} + s\frac{s^2 + s - 2}{(s^3 - 3s^2 + 4s)} - 1}{s-2}\) Simplify and factor the equation for X(s): \(X(s) = \frac{(s-1)(s^2 + s)}{(s-2)(s^3 - 3s^2 + 4s)}\) #Step 4: Find the inverse Laplace transform of X(s) and Y(s)#:

To find the inverse Laplace transform of both X(s) and Y(s), perform partial fraction decomposition, then use the tables of inverse Laplace transforms: \(L^{-1}\{X(s)\} = x(t) = L^{-1}\left\{\frac{(s-1)(s^2 + s)}{(s-2)(s^3 - 3s^2 + 4s)}\right\}\) \(L^{-1}\{Y(s)\} = y(t) = L^{-1}\left\{\frac{s^2+s-2}{(s^3 - 3s^2 + 4s)}\right\}\) After performing partial fraction decomposition and using the table of inverse Laplace transforms, we get the solutions for x(t) and y(t) in terms of t: \(x(t) = e^{2t} - te^{2t} - t^2e^{2t}\) \(y(t) = -te^{2t} - t^2e^{2t} + te^{3t}\)

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with specific initial conditions. These conditions are used to find a unique solution to the differential equations. Typically, initial conditions specify the value of the function (and possibly its derivatives) at a particular point, often at time zero.

In the exercise, we are given a system of differential equations along with initial conditions: \( x(0) = -1 \) and \( y(0) = 0 \). These initial conditions play a crucial role in determining the specific solutions for \( x(t) \) and \( y(t) \). Without these, we would only have a general solution, which might contain arbitrary constants.

To solve an initial value problem using the Laplace Transform, we first apply the transform to the differential equation, incorporate the initial conditions, and then solve for the transformed variable, which in this case is in terms of \( X(s) \) and \( Y(s) \).

Linear Systems of Differential Equations

Linear systems of differential equations consist of multiple linear differential equations working together. Each equation describes the behavior of one part of the system and is linked to others through shared variables.

Our task is to solve such a system:

• \( x'(t) - 2x(t) - 3y(t) = 0 \)
• \( y'(t) + x(t) + 2y(t) = t \)

These equations are interconnected since \( x(t) \) from the first equation appears in the second, and \( y(t) \) from the second is in the first.

This necessitates the solving of both equations simultaneously. We do this using the Laplace Transform, which transforms the problem from the time domain into the complex plane where differentiation becomes algebraic manipulation. This allows us to solve the equations more straightforwardly.

Inverse Laplace Transform

The Inverse Laplace Transform is the process used to convert a function back from its Laplace-transformed form into the time domain. It is the last step in finding the solution to an initial value problem using the Laplace Transform.

For our solution, after obtaining \( X(s) \) and \( Y(s) \) in terms of \( s \), we have to determine \( x(t) \) and \( y(t) \). The inverse involves transforming each term in the Laplace domain back to its original function form as time-dependent variables.

This often requires **Partial Fraction Decomposition** to simplify expressions, followed by using the standardized Laplace transform tables that relate common forms like \( \frac{1}{s} \) to corresponding time domain functions, providing the solution explicitly in terms of \( t \).

This process enables us to interpret our results meaningfully in terms of dynamical behavior over time, which wasn't possible in the complex \( s \) domain.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions, making them easier to handle. In solving differential equations using the Laplace Transform, this step is crucial for finding inverse transforms.

When something like \( X(s) \) or \( Y(s) \) is an expression involving polynomials over polynomials, the partial fraction decomposition splits it into a sum of simpler fractions. This helps in finding the inverse, as each simpler fraction typically corresponds to a basic function in the Laplace transform table.

For example, suppose you have \( \frac{s^2 + s - 2}{(s^3 - 3s^2 + 4s)} \). Partial fraction decomposition simplifies this into manageable terms. Once decomposed, each term can be looked up directly or easily manipulated to find the inverse Laplace transform. Hence, partial fraction decomposition forms an integral bridge between the algebraic solution of \( X(s) \), \( Y(s) \), and their respective time-domain functions \( x(t) \), \( y(t) \).