Question

Use Laplace transforms to solve the given initial value problem. \(\mathbf{y}^{\prime}=\left[\begin{array}{ll}5 & -4 \\ 5 & -4\end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l}5 \\\ 6\end{array}\right]\)

Short answer

Based on the step-by-step solution provided above, answer the following question: **Question:** Find the solution of the given initial value problem for the system of linear differential equations: $$\mathbf{y}^{\prime}(t) = \left[\begin{array}{ll}5 & -4 \\ 5 & -4\end{array}\right] \mathbf{y}(t), \hspace{15pt} \mathbf{y}(0) = \left[\begin{array}{l}5\\ 6\end{array}\right]$$ **Answer:** The solution of the given initial value problem is: $$\mathbf{y}(t) = \left[\begin{array}{l}-e^{-4t}+e^{5t}\\ 0\end{array}\right].$$

Step-by-step solution

1. Apply Laplace Transform to the Matrix Equation

Apply Laplace transform to both sides of the equation, remembering that the Laplace transform of a derivative is \(L\{y^{\prime}\}(s) = sY(s) - y(0)\), where \(Y(s)\) is the Laplace transform of \(y(t)\). We get: $$s\mathbf{Y}(s) - \mathbf{y}(0)=\left[\begin{array}{ll}5 & -4 \\ 5 &-4\end{array}\right] \mathbf{Y}(s)$$ With \(\mathbf{y}(0)=\left[\begin{array}{l}5\\ 6\end{array}\right]\).

2. Solve for the Laplace Transform of the Vector

Rearrange the matrix equation to isolate the Laplace transforms of the unknown vectors: $$\mathbf{Y}(s) = (s\mathbf{I}-\mathbf{A})^{-1}\mathbf{y}(0),$$ Where \(\mathbf{A} =\left[\begin{array}{ll}5 & -4 \\ 5 & -4\end{array}\right]\) and \(\mathbf{I}\) is the identity matrix. Thus, we can write $$\mathbf{Y}(s) = \left[\begin{array}{ll}s-5 & 4 \\ -5 & s+4\end{array}\right]^{-1}\left[\begin{array}{l}5\\ 6\end{array}\right].$$

3. Find the Inverse of the Matrix

To compute the inverse of \(\left[\begin{array}{ll}s-5 & 4 \\ -5 & s+4\end{array}\right]\), we'll use the formula: $$\mathbf{B}^{-1} = \frac{1}{\det(\mathbf{B})}\rm{adj}(\mathbf{B}),$$ Where \(\det(\mathbf{B})\) is the determinant of the matrix \(\mathbf{B}\) and adj\(\rm{(\mathbf{B})}\) is its adjugate. We find that $$\det(\mathbf{B}) = (s-5)(s+4) - (-20) = s^2 - s - 20,$$ And $$\rm{adj}(\mathbf{B}) = \left[\begin{array}{ll}s+4 & -4 \\ 5 & s-5\end{array}\right].$$ Therefore, the inverse of the given matrix is $$\left[\begin{array}{ll}s-5 & 4 \\ -5 & s+4\end{array}\right]^{-1} = \frac{1}{s^2-s-20}\left[\begin{array}{ll}s+4 & -4 \\ 5 & s-5\end{array}\right].$$

4. Multiply the Inverse Matrix by the Initial Conditions

Multiply the inverse matrix by the initial conditions vector: $$\mathbf{Y}(s) = \frac{1}{s^2-s-20}\left[\begin{array}{ll}s+4 & -4 \\ 5 & s-5\end{array}\right]\left[\begin{array}{l}5\\ 6\end{array}\right] = \frac{1}{s^2-s-20}\left[\begin{array}{l}30-4s\\ 5s\end{array}\right].$$

5. Apply Inverse Laplace Transform

Finally, apply the inverse Laplace transform to each component of the vector to obtain the solution of the initial value problem: $$\mathbf{y}(t) = L^{-1}\{\mathbf{Y}(s)\} = \left[\begin{array}{l}L^{-1}\{30-4s\}\\ L^{-1}\{5s\}\end{array}\right].$$ To simplify it, we need to perform a partial fraction decomposition on \(\frac{30-4s}{s^2-s-20}\): $$\frac{30-4s}{s^2-s-20} = \frac{A}{s-(-4)}+\frac{B}{s-5}$$ From here, we find that \(A=-1\) and \(B=1\). Thus, $$\mathbf{y}(t) = \left[\begin{array}{l}L^{-1}\{-\frac{1}{s+4}+\frac{1}{s-5}\}\\ L^{-1}\{5s\}\end{array}\right] = \left[\begin{array}{l}-e^{-4t}+e^{5t}\\ 0\end{array}\right].$$ So, the solution of the given initial value problem is: $$\mathbf{y}(t) = \left[\begin{array}{l}-e^{-4t}+e^{5t}\\ 0\end{array}\right].$$

Key concepts

Differential Equations

Differential equations are equations involving a function and its derivatives, often used in various scientific fields like physics and engineering to model real-world phenomena. In simple terms, differential equations relate a function to its rates of change. They come in various forms: ordinary differential equations (ODEs), where functions depend on a single variable, and partial differential equations (PDEs), involving multivariable functions.

When dealing with matrices, systems of differential equations can be represented using matrix algebra. This allows multiple governing equations to be expressed compactly, simplifying problem-solving, as demonstrated in our exercise scenario. By representing the differential system as a matrix equation, such as \[\mathbf{y}^{\prime} = \mathbf{A} \mathbf{y} \], we make complex equations easier to manage and solve.

Matrix Algebra

Matrix algebra is a branch of mathematics focusing on operations involving matrices. In the context of differential equations, using matrices allows us to handle complex systems of linear equations efficiently. Each column or row represents a linear equation or its coefficients.

In the present problem, matrix algebra is used to manipulate the matrix that represents the coefficients of the differential equation system. We use matrix inversion to find solutions by isolating the variable of interest. This involves steps like computing the determinant of a matrix, which is crucial for finding its inverse. With the inverse matrix, it becomes possible to solve the system of equations derived from our original differential equation by directly obtaining expressions for the Laplace transform of the unknown functions.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation accompanied by initial conditions, which specify values for the unknown function at a specific point. IVPs are essential in ensuring the uniqueness of a solution, helping us predict future behavior based on initial states.

For example, knowing the initial conditions \[\mathbf{y}(0) = \begin{bmatrix} 5 \ 6 \end{bmatrix} \] is vital for determining the specific paths the functions will take through time. These initial conditions were used in our problem to solve \[\mathbf{Y}(s)\]. Given these conditions and applying techniques like the Laplace transform, we can systematically break down the solution process. The initial states enable us to transform from the outcome of a system, such as function behavior, into an equation that represents the system.

Inverse Laplace Transform

The inverse Laplace transform is a mathematical operation that helps revert a function from its Laplace-transformed domain back to the original time domain. These transforms are particularly useful for solving differential equations because they convert differential operations into algebraic operations, which are simpler to handle.

To apply the inverse Laplace transform in our exercise, we first found the Laplace transform's algebraic form by solving the matrix equation. Once in this form, partial fraction decomposition aided in breaking down complex rational expressions, enabling the process of inversion back to time-domain functions. Our final solution involved reverting the expressions to standard time functions using known inverse transforms, such as recovering exponential forms from simple rational expressions. This process conclusively solved the initial value problem, yielding explicit time-based solutions for each component of the system.