Question

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

Short answer

Answer: The solution vector for the given initial value problem using Laplace transforms is \(\mathbf{y}(t) = \left[\begin{array}{c}2\cos(2t) - 2\sin(2t)\\2\cos(2t) - 2\sin(2t)\end{array}\right]\).

Step-by-step solution

Take the Laplace transform of the ODE

Taking the Laplace transform of the ODE will give us: \(\mathcal{L}\{\mathbf{y'}(t)\} = \mathcal{L}\{\left[\begin{array}{rr}1 & 4 \\\ -1 & 1\end{array}\right]\mathbf{y}(t)\}\) By linearity of the Laplace transform, we can write this as: \(p\mathcal{L}\{\mathbf{y}(t)\} - \mathbf{y}(0) = \left[\begin{array}{rr}1 & 4 \\\ -1 & 1\end{array}\right]\mathcal{L}\{\mathbf{y}(t)\}\), where \(p\) is the Laplace transform variable.

Solve for the Laplace transform of the solution vector

We will first plug in the initial conditions, \(\mathbf{y}(0) = \left[\begin{array}{l}2\\0\end{array}\right]\), which gives: \(p\mathcal{L}\{\mathbf{y}(t)\} - \left[\begin{array}{l}2\\0\end{array}\right] = \left[\begin{array}{rr}1 & 4 \\\ -1 & 1\end{array}\right]\mathcal{L}\{\mathbf{y}(t)\}\) Now we will solve for the Laplace transform of the solution vector, \(\mathcal{L}\{\mathbf{y}(t)\}\): \((pI - A) \mathcal{L}\{\mathbf{y}(t)\} = \left[\begin{array}{l}2\\0\end{array}\right]\), where \(I\) is the identity matrix and \(A = \left[\begin{array}{rr}1 & 4 \\\ -1 & 1\end{array}\right]\). The matrix \((pI - A)\) is given by: \(\left[\begin{array}{cc}p-1 & -4 \\\ 1 & p-1\end{array}\right]\). To find its inverse, we compute the determinant and the adjugate. The determinant is: \((p-1)^2 - (-4)(1) = p^2 - 2p + 5\). The adjugate of \((pI - A)\) is: \(\left[\begin{array}{cc}p-1 & 4 \\\ -1 & p-1\end{array}\right]\). Then, \((pI - A)^{-1} = \frac{1}{p^2 - 2p + 5}\left[\begin{array}{cc}p-1 & 4 \\\ -1 & p-1\end{array}\right]\). Multiplying \((pI - A)^{-1}\) by \(\left[\begin{array}{l}2\\0\end{array}\right]\), we get: \(\mathcal{L}\{\mathbf{y}(t)\} = \frac{1}{p^2 - 2p + 5}\left[\begin{array}{c}(p-1)(2)\\(-1)(2)\end{array}\right] = \left[\begin{array}{c}\frac{2(p-1)}{p^2 - 2p + 5} \\\ -\frac{2}{p^2 - 2p + 5}\end{array}\right]\).

Apply the inverse Laplace transform

Finally, we apply the inverse Laplace transform to find the solution vector \(\mathbf{y}(t)\): \(\mathbf{y}(t) = \mathcal{L}^{-1}\left\{\left[\begin{array}{c}\frac{2(p-1)}{p^2 - 2p + 5} \\\ -\frac{2}{p^2 - 2p + 5}\end{array}\right]\right\}\). By using tables or partial fraction decomposition to find the inverse Laplace transforms, we obtain: \(\mathbf{y}(t) = \left[\begin{array}{c}2\cos(2t) - 2\sin(2t)\\2\cos(2t) - 2\sin(2t)\end{array}\right]\). Thus, the solution to the initial value problem is: \(\mathbf{y}(t) = \left[\begin{array}{c}2\cos(2t) - 2\sin(2t)\\2\cos(2t) - 2\sin(2t)\end{array}\right]\).

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between functions and their derivatives. They are fundamental in modeling situations where a certain quantity's rate of change is dependent on the quantity itself. For instance, they are crucial in physics for modeling dynamics and in engineering for portraying how systems evolve over time.

When it comes to systems of differential equations with multiple variables, as in our example, we express the relationships using matrix notation. Systems of linear differential equations can be very complex, but with the help of linear algebra and operations like matrix multiplication, we can manipulate these systems into more manageable forms. Solving such a system involves finding a vector-valued function \( \mathbf{y}(t) \) that satisfies the given differential equation with the initial condition specified.

Matrix Exponential

Understanding the Matrix Exponential

Matrix exponential is a concept from linear algebra, crucial for solving systems of linear differential equations. It extends the idea of exponential functions to matrix-valued functions. Just as the exponential function can be defined by a power series, matrix exponentials are defined via a series expansion using matrix powers. In theory, this method allows us to solve linear differential equations without explicitly calculating the inverse Laplace transform, by simply evaluating the matrix exponential function at a particular point.

For the given exercise, the matrix exponential helps express the solution to the linear system in terms of \( e^{At} \) where \( A \) is the coefficients matrix from our differential equation. It serves as a powerful tool when the given matrix \( A \) has complex eigenvalues or does not have a full set of eigenvectors.

Inverse Laplace Transform

Applying the Inverse Laplace Transform

The inverse Laplace transform is a method used to revert Laplace transforms and recover the original time-domain function. It's widely employed to solve differential equations that have been transformed to the s-domain using Laplace transform. By expressing the differential equations in the s-domain, we often simplify the complex interdependencies between variables and their derivatives, allowing for algebraic manipulation.

In our exercise, after transforming the system of equations and applying initial conditions, we've obtained a resultant matrix in the s-domain. The inverse transform is then applied element-wise to each entry of this matrix using known transformations or methods like partial fraction decomposition when necessary. This step is crucial because it takes us back from the algebraically convenient s-domain to the original domain where the problem was stated and where the solution is practically applicable.

Initial Conditions

Significance of Initial Conditions

Initial conditions provide required information to uniquely determine the solution to a differential equation. They specify the state of the system at a starting time, often denoted as \( t = 0 \). Without initial conditions, a differential equation might have infinitely many solutions.

In the context of our problem, the initial condition \( \mathbf{y}(0) \) is used right after the Laplace transform is applied to the system of equations. It helps in finding the specific solution to the equation that corresponds to our initial state, guiding us towards the unique solution that not only fits the equation but also matches the given starting scenario. Initial conditions bridge the gap between abstract mathematical solutions and real-world scenarios where the initial state of the system is known.