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Chapter 5: Problem 20

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

Expert verified
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

01

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 \(\mathbf{G}(s) = \mathscr{L}\{\mathbf{g}(t)\}\).
02

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, $$\mathbf{Y}(s) = (sI - A)^{-1}\mathbf{G}(s)$$
03

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 $$\mathbf{y}(t) = \left[\begin{array}{c} e^t \\\ t(e^t - 1) \\\ 2(e^t -1 - t) \end{array}\right]$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Give the form of the partial fraction expansion for the given rational function \(F(s)\). You need not evaluate the constants in the expansion. However, if the denominator of \(F(s)\) contains irreducible quadratic factors of the form \(s^{2}+2 \alpha s+\beta^{2}, \beta^{2}>\alpha^{2}\), complete the square and rewrite this factor in the form \((s+\alpha)^{2}+\omega^{2}\). $$F(s)=\frac{s^{3}-1}{\left(s^{2}+1\right)^{2}(s+4)^{2}}$$

Give the form of the partial fraction expansion for the given rational function \(F(s)\). You need not evaluate the constants in the expansion. However, if the denominator of \(F(s)\) contains irreducible quadratic factors of the form \(s^{2}+2 \alpha s+\beta^{2}, \beta^{2}>\alpha^{2}\), complete the square and rewrite this factor in the form \((s+\alpha)^{2}+\omega^{2}\). $$F(s)=\frac{s^{2}+1}{s^{2}\left(s^{2}+2 s+10\right)}$$

Solve the given initial value problem, in which inputs of large amplitude and short duration have been idealized as delta functions. Graph the solution that you obtain on the indicated interval. \(\frac{d}{d t}\left[\begin{array}{l}y_{1} \\\ y_{2}\end{array}\right]=\left[\begin{array}{ll}2 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{l}y_{1} \\\ y_{2}\end{array}\right]+\left[\begin{array}{l}0 \\\ 1\end{array}\right]-\delta(t-1)\left[\begin{array}{l}1 \\\ 0\end{array}\right], \quad\left[\begin{array}{l}y_{1}(0) \\\ y_{2}(0)\end{array}\right]=\left[\begin{array}{l}0 \\ 0\end{array}\right], \quad 0 \leq t \leq 2\)

Use Laplace transforms to solve the given initial value problem. \(\mathbf{y}^{\prime}=\left[\begin{array}{ll}5 & -4 \\ 3 & -2\end{array}\right] \mathbf{y}+\left[\begin{array}{l}t \\\ 1\end{array}\right], \quad \mathbf{y}(0)=\left[\begin{array}{l}0 \\\ 0\end{array}\right]\)

Suppose that \(\mathcal{L}\left\\{f_{1}(t)\right\\}=F_{1}(s)\) and \(\mathcal{L}\left\\{f_{2}(t)\right\\}=F_{2}(s), s>a\). Use the fact that $$\mathcal{L}^{-1}\left\\{c_{1} F_{1}(s)+c_{2} F_{2}(s)\right\\}=c_{1} \mathcal{L}^{-1}\left\\{F_{1}(s)\right\\}+c_{2} \mathcal{L}^{-1}\left\\{F_{2}(s)\right\\}, \quad a
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