Question

Use Laplace transforms to solve the given initial value problem. \(\mathbf{y}^{\prime}=\left[\begin{array}{rrr}1 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 2\end{array}\right] \mathbf{y}+\left[\begin{array}{c}e^{t} \\ 1 \\ -2 t\end{array}\right], \quad \mathbf{y}(0)=\left[\begin{array}{l}0 \\ 0 \\\ 0\end{array}\right]\)

Short answer

The solution to the given initial value problem is: $$\mathbf{y}(t) = \left[\begin{array}{c} e^t \\\ t(e^t - 1) \\\ 2(e^t -1 - t) \end{array}\right]$$

Step-by-step solution

Apply the Laplace transform to the given differential equation

First, apply the Laplace transform to the given differential equation: $$\mathscr{L}\{\mathbf{y}'(t)\} = \mathscr{L}\{A\mathbf{y}(t) + \mathbf{g}(t)\}$$ where \(A = \left[\begin{array}{rrr}1 & 0 & 0 \\\ 0 & -1 & 1 \\\ 0 & 0 & 2\end{array}\right]\) and \(\mathbf{g}(t) = \left[\begin{array}{c}e^{t} \\\ 1 \\\ -2t\end{array}\right]\). Recalling that the Laplace transform of a derivative is given by: $$\mathscr{L}\{y'(t)\} = sY(s) - y(0)$$ So we have, $$s\mathbf{Y}(s) - \mathbf{y}(0) = A\mathbf{Y}(s) + \mathbf{G}(s)$$ where \(\mathbf{Y}(s) = \mathscr{L}\{\mathbf{y}(t)\}\) and \(\mathbf{G}(s) = \mathscr{L}\{\mathbf{g}(t)\}\).

Solve for \(\mathbf{Y}(s)\)

Now, solve for \(\mathbf{Y}(s)\). Since \(\mathbf{y}(0) = \left[\begin{array}{l}0 \\\ 0 \\\ 0\end{array}\right]\), we have, $$s\mathbf{Y}(s) = A\mathbf{Y}(s) + \mathbf{G}(s)$$ So, $$(sI - A)\mathbf{Y}(s) = \mathbf{G}(s)$$ where \(I\) is the identity matrix. Now find the inverse of \((sI - A)\), multiply both sides by this inverse to get, $$\mathbf{Y}(s) = (sI - A)^{-1}\mathbf{G}(s)$$

Apply inverse Laplace transform to find \(\mathbf{y}(t)\)

We have found \(\mathbf{Y}(s)\), now we need to apply the inverse Laplace transform to get \(\mathbf{y}(t)\): $$\mathbf{y}(t) = \mathscr{L}^{-1}\{(sI - A)^{-1}\mathbf{G}(s)\}$$ First, we need to find the Laplace transform \(\mathbf{G}(s)\), $$\mathbf{G}(s) = \left[\begin{array}{c} \frac{1}{s-1} \\\ \frac{1}{s} \\\ \frac{2}{s^2} \end{array}\right]$$ Next, compute \((sI - A)^{-1}\), $$(sI - A) = \left[\begin{array}{ccc} s-1 & 0 & 0 \\\ 0 & s+1 & -1 \\\ 0 & 0 & s-2 \end{array}\right]$$ Since this is a diagonal matrix, its inverse is also a diagonal matrix with the inverse of each diagonal element, $$(sI - A)^{-1} = \left[\begin{array}{ccc} \frac{1}{s-1} & 0 & 0 \\\ 0 & \frac{1}{s+1} & \frac{1}{s^2+2s} \\\ 0 & 0 & \frac{1}{s-2} \end{array}\right]$$ Now, multiply \((sI - A)^{-1}\) by \(\mathbf{G}(s)\): $$\mathbf{Y}(s) = \left[\begin{array}{c} 1 \\\ \frac{s}{(s^2 + 2s)} \\\ - \frac{2}{(s-2)^2} \end{array}\right]$$ Finally, apply the inverse Laplace transform to find \(\mathbf{y}(t)\): $$\mathbf{y}(t) = \mathscr{L}^{-1}\{\mathbf{Y}(s)\} = \left[\begin{array}{c} e^t \\\ t(e^t - 1) \\\ 2(e^t -1 - t) \end{array}\right]$$ Hence, the solution to the given initial value problem is $$\mathbf{y}(t) = \left[\begin{array}{c} e^t \\\ t(e^t - 1) \\\ 2(e^t -1 - t) \end{array}\right]$$

Key concepts

Matrix Differential Equations

Matrix differential equations involve systems of equations where the unknowns are vectors, and the coefficients are matrices. They come into play when dealing with multiple interrelated dynamic processes, like in engineering or systems biology. Instead of a single differential equation, you have a set of them, typically expressed in the form:\[\mathbf{y}'(t) = A\mathbf{y}(t) + \mathbf{g}(t)\]- Here, \(\mathbf{y}(t)\) is a vector function representing the system's state over time. - \(A\) is a matrix that describes the system dynamics.- \(\mathbf{g}(t)\) is often a vector of external inputs or forces acting on the system.The solution to a matrix differential equation gives you all the information about the system's behavior over time. Solving these usually involves techniques like diagonalization or transformation to simpler forms for easier computation. Understanding how to manage these equations is crucial in modeling anything from electrical circuits to ecological systems.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation accompanied by a set of initial conditions. These conditions specify the values of the solution at a given starting point, which helps in finding a unique solution to the differential equation. For example, with our original problem:- We have the matrix differential equation: \[ \mathbf{y}' = A\mathbf{y} + \mathbf{g}(t) \]- And the initial condition: \[ \mathbf{y}(0) = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \]This initial condition specifies that at time \(t = 0\), the vector function \(\mathbf{y}(t)\) is zero. It ensures that there is a specific path for the solution, avoiding ambiguities in the system's behavior. Solving an IVP often involves transforming the differential problem into a simpler form, such as using the Laplace transform, which converts it into an algebraic problem that can be solved more straightforwardly.

Inverse Laplace Transform

The inverse Laplace transform is a mathematical tool used to retrieve a time-domain function from its Laplace-transformed counterpart. In solving differential equations, particularly those expressed in matrix form, the Laplace transform simplifies the process by converting differential operators into algebraic ones. Once solved in the Laplace domain, you must apply the inverse Laplace transform to return to the time domain and find the original function.- For a system described by: \[ (sI - A)\mathbf{Y}(s) = \mathbf{G}(s) \]- Solve for \(\mathbf{Y}(s)\) in the algebraic domain.- Use the inverse Laplace transform to find \(\mathbf{y}(t)\): \[ \mathbf{y}(t) = \mathscr{L}^{-1}\{\mathbf{Y}(s)\} \]Mastering this concept is essential for those tackling complex engineering problems, where systems need to be analyzed and controlled. The inverse transform helps to unveil the solution in a more intuitive, time-based format, allowing for application and interpretation within real-world contexts.