Question

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

Short answer

Answer: The solution to the given initial value problem is \(\mathbf{y}(t) = \left[\begin{array}{l}\frac{1}{2}\sin t \\ \sin t\end{array}\right]\).

Step-by-step solution

Write down the given ODE system and initial values.

Given \(\mathbf{y}^{\prime \prime}=\left[\begin{array}{ll}1 & -1 \\ 1 & -1\end{array}\right] \mathbf{y}+\left[\begin{array}{l}2 \\ 1\end{array}\right]\), with \(\mathbf{y}(0)=\left[\begin{array}{l}0 \\ 1\end{array}\right]\), and \(\mathbf{y}^{\prime}(0)=\left[\begin{array}{l}0 \\ 0\end{array}\right]\) Let \(\mathbf{y}=\left[\begin{array}{l}y_1(t) \\ y_2(t)\end{array}\right].\) Then the given ODE system can be written as: \((1)\:\: y_1^{\prime\prime}(t) - y_2^{\prime\prime}(t) = y_1(t) - y_2(t) + 2 \) \((2)\:\: y_1^{\prime\prime}(t) - y_2^{\prime\prime}(t) = y_1(t) - y_2(t) + 1 \)

Apply the Laplace Transform

Applying the Laplace transform to both sides of equations (1) and (2) gives: \((1)\:\: s^2Y_1(s) - sy_1(0) - y_1^{\prime}(0) - s^2Y_2(s)+ sy_2(0) + y_2^{\prime}(0) = Y_1(s) - Y_2(s) + \frac{2}{s} \) \((2)\:\: s^2Y_1(s) - sy_1(0) - y_1^{\prime}(0) - s^2Y_2(s)+ sy_2(0) + y_2^{\prime}(0) = Y_1(s) - Y_2(s) + \frac{1}{s} \) Substitute the initial values \(y_1(0) = 0\), \(y_2(0) = 1\), \(y_1^{\prime}(0) = 0\), and \(y_2^{\prime}(0) = 0\): \((1)\:\: s^2Y_1(s) - s^2Y_2(s) = Y_1(s) - Y_2(s) + \frac{2}{s} \) \((2)\:\: s^2Y_1(s) - s^2Y_2(s) + s = Y_1(s) - Y_2(s) + \frac{1}{s} \)

Solve the algebraic system of equations for \(Y_1(s)\) and \(Y_2(s)\)

Subtract (1) from (2): \(s = \frac{1}{s} - \frac{2}{s} \) Multiplying by \(s\) gives \(s^2 = -1\), which implies \(s = \pm i\). Now we can solve for \(Y_1(s)\) and \(Y_2(s)\) for each value of \(s\): \(s = i\) \((1)\:\: -Y_1(i) + Y_2(i) = Y_1(i) - Y_2(i) + \frac{2}{i} \) \((2)\:\: -Y_1(i) + Y_2(i) + i = Y_1(i) - Y_2(i) + \frac{1}{i} \) Subtract (1) from (2): \(i - \frac{2}{i} = -\frac{1}{i} \) Multiplying by \(i\) gives \(i^2 = -1\): \(i^2 + 1 = 2 \) And substituting back: \(-Y_1(i) + Y_2(i) + i = -Y_1(i) + Y_2(i) + 1\) \(Y_1(i) = \frac{1}{1+i}\) \(Y_2(i) = \frac{2}{1+i}\) \(s = -i\) \((1)\:\: Y_1(-i) - Y_2(-i) = Y_1(-i) - Y_2(-i) + \frac{2}{-i} \) \((2)\:\: Y_1(-i) - Y_2(-i) - i = Y_1(-i) - Y_2(-i) + \frac{1}{-i} \) Subtract (1) from (2): \(i - \frac{2}{-i} = \frac{1}{-i} \) Multiplying by \(-i\) gives: \(i^2 + 1 = -2 \) And substituting back: \(Y_1(-i) - Y_2(-i) - i = Y_1(-i) - Y_2(-i) - 1\) \(Y_1(-i) = -\frac{1}{1-i}\) \(Y_2(-i) = -\frac{2}{1-i}\)

Apply the inverse Laplace transform

Apply the inverse Laplace transform to obtain the expressions for \(y_1(t)\) and \(y_2(t)\): \(y_1(t) = \mathcal{L}^{-1}\left(\frac{1}{s^2+1}\right) = \frac{1}{2}\sin t\) \(y_2(t) = \mathcal{L}^{-1}\left(\frac{2}{s^2+1}\right) = \sin t\) Therefore, the solution to the given initial value problem is: \(\mathbf{y}(t) = \left[\begin{array}{l}\frac{1}{2}\sin t \\ \sin t\end{array}\right]\)

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) involves finding a function that satisfies a differential equation along with specific conditions given at the start of the interval in question. In the problem, you have a second-order differential equation expressed as \[\mathbf{y}^{\prime \prime} = \left[\begin{array}{ll}1 & -1 \ 1 & -1\end{array}\right] \mathbf{y} + \left[\begin{array}{l}2 \ 1\end{array}\right]\] with initial conditions \[\mathbf{y}(0) = \left[\begin{array}{l}0 \ 1\end{array}\right]\] and \[\mathbf{y}^{\prime}(0) = \left[\begin{array}{l}0 \ 0\end{array}\right].\] The goal is to find the function \( \mathbf{y}(t) \) that fulfills both the equation and these initial conditions.

• "Initial value" refers to conditions given at the starting point \( t = 0 \).
• These provide necessary information for a unique solution.
• Without them, there would be infinitely many solutions due to the second-order nature of the differential equation.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In the context of the given exercise, the system of differential equations is presented in matrix form:\[y^{\prime \prime} = \left[\begin{array}{ll}1 & -1 \1 & -1\end{array}\right] y + \left[\begin{array}{l}2 \1\end{array}\right]\]This form helps in dealing with systems that have multiple interdependent varying quantities—here represented by \( y_1(t) \) and \( y_2(t) \).

• The order of a differential equation (in this case, second-order) indicates the highest derivative present.
• The matrix component \( \mathbf{y}^{\prime \prime} \) signifies a coupled system, meaning the derivatives of \( y_1 \) and \( y_2 \) affect each other.
• Such systems often appear in physics and engineering, modeling scenarios where forces or interactions between two or more states need understanding.

Recognizing and transforming these equations is crucial in predicting system behavior over time.

Inverse Laplace Transform

The inverse Laplace Transform is a mathematical operation used to convert a function from the frequency domain back into the time domain. This operation is crucial in finding the solution to the differential equations expressed in terms of Laplace transforms.In the final step of the exercise, after obtaining \( Y_1(s) \) and \( Y_2(s) \) in the Laplace domain, the inverse Laplace transform helps translate these results back into functions of time, \( y_1(t) \) and \( y_2(t) \):

• \[y_1(t) = \mathcal{L}^{-1}\left(\frac{1}{s^2+1}\right) = \frac{1}{2}\sin t\]
• \[y_2(t) = \mathcal{L}^{-1}\left(\frac{2}{s^2+1}\right) = \sin t\]

This transformation is done using known pairs from Laplace transform tables.

• The inverse Laplace transform is essential to return the solution to its original time-based form.
• It provides insights into how the dynamic behavior, like oscillations in this case, evolves over time.
• Mastering these transforms allows efficient resolution of initial value problems in engineering and science.