Question

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

Short answer

Question: Solve the given initial value problem involving a first-order linear system of differential equations using the Laplace transforms method. Initial value problem: \(y'(t) = \begin{bmatrix} 5 & -4 \\ 5 & -4 \end{bmatrix}y(t) + \begin{bmatrix} 0 \\ 1 \end{bmatrix}\), \(y(0) = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\) Solution: \(\mathbf{y}(t) = \left[\begin{array}{l}y_1(t) \\ y_2(t)\end{array}\right] = \left[\begin{array}{l}-\frac{1}{9}(e^{5t}-e^{-4t}) \\ \frac{1}{9}(e^{5t}-e^{-4t})\end{array}\right]\)

Step-by-step solution

Apply the Laplace transform to the system

First, we need to apply the Laplace transform to the given initial value problem. The Laplace transform of the derivative of a function is given by the product of s and the Laplace transform of the function, minus the initial value of the function. In this case, the initial values of the functions are zero. So, applying the Laplace transform to the given initial value problem, we get: \(L\{ \mathbf{y'}\} = sL\{\mathbf{y}\} - \mathbf{y}(0)\) which in matrix form becomes: \(s\left[\begin{array}{l}L\{y_1\} \\ L\{y_2\}\end{array}\right] - \left[\begin{array}{l}0 \\ 0\end{array}\right] = \left[\begin{array}{ll}5 & -4 \\ 5 & -4\end{array}\right]\left[\begin{array}{l}L\{ y_1\} \\ L\{ y_2\}\end{array}\right] + \left[\begin{array}{l}0 \\ 1\end{array}\right]\)

Solve for the Laplace transforms of the solution

Now, we need to solve the transformed system of equations for \(L\{y_1\}\) and \(L\{y_2\}\). We can rewrite the system as: \(\left[\begin{array}{cc}s-5 & 4 \\ -5 & s+4\end{array}\right]\left[\begin{array}{l}L\{y_1\} \\ L\{y_2\}\end{array}\right] = \left[\begin{array}{l}0 \\ 1\end{array}\right]\) To solve it, we find the inverse of the matrix on the left-hand side and multiply it by the right-hand side vector. The inverse of a 2x2 matrix \(\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]\) is given by \(\frac{1}{ad-bc}\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]\). So in our case: \(\left[\begin{array}{cc}s-5 & 4 \\ -5 & s+4\end{array}\right]^{-1} = \frac{1}{(s-5)(s+4)-(-20)}\left[\begin{array}{cc}s+4 & -4 \\ 5 & s-5\end{array}\right]\) Let's denote \(M = (s-5)(s+4)+20\). So we get: \(\left[\begin{array}{l}L\{y_1\} \\ L\{y_2\}\end{array}\right] = \frac{1}{M}\left[\begin{array}{cc}s+4 & -4 \\ 5 & s-5\end{array}\right]\left[\begin{array}{l}0 \\ 1\end{array}\right] = \frac{1}{M}\left[\begin{array}{l}-4 \\ s-5\end{array}\right]\)

Apply the inverse Laplace transform

Finally, we apply the inverse Laplace transform to \(L\{y_1\}\) and \(L\{y_2\}\) to get the solution in the time domain: \(\mathbf{y}(t) = \left[\begin{array}{l}y_1(t) \\ y_2(t)\end{array}\right] = \mathcal{L}^{-1}\left\{\frac{1}{M}\left[\begin{array}{l}-4 \\ s-5\end{array}\right]\right\}\) Upon using the inverse Laplace transform table, we find that \(y_1(t) = \mathcal{L}^{-1} \left\{\frac{-4}{M}\right\} = -4\frac{e^{5t}-e^{-4t}}{(5+4)(e^{5t}-e^{-4t})} = -\frac{1}{9}(e^{5t}-e^{-4t})\) and \(y_2(t) = \mathcal{L}^{-1} \left\{\frac{s-5}{M}\right\} = \frac{(s-5)(e^{5t}-e^{-4t})}{(5+4)(e^{5t}-e^{-4t})} = \frac{1}{9}(e^{5t}-e^{-4t})\) Thus, the solution of the given initial value problem is: \(\mathbf{y}(t) = \left[\begin{array}{l}y_1(t) \\ y_2(t)\end{array}\right] = \left[\begin{array}{l}-\frac{1}{9}(e^{5t}-e^{-4t}) \\ \frac{1}{9}(e^{5t}-e^{-4t})\end{array}\right]\)

Key concepts

Initial Value Problems

Initial value problems (IVPs) are common in differential equations where the goal is to find a solution given specific initial conditions. As in the exercise, we have a system of differential equations and initial conditions at time zero:

• These initial conditions are usually expressed as specific values the solution must satisfy at the initial time.
• In our exercise, the initial condition is given as \( \mathbf{y}(0) = \begin{bmatrix} 0 \ 0 \end{bmatrix} \). This provides the starting point or 'initial state' for solving the problem.
• IVPs are crucial in modeling real-world situations where you know the starting point and need to determine how the system evolves over time.

Handling IVPs involves applying techniques like the Laplace Transform, which simplifies complex differential equations by transforming them into an algebraic form that is easier to solve.

Matrix Inversion

Matrix inversion plays a vital role in solving systems of equations, especially when dealing with Laplace transforms, as seen in the original exercise.

• Matrix inversion essentially finds a matrix that, when multiplied by the original matrix, yields the identity matrix.
• For a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the inverse is given by \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \), provided the determinant \( ad-bc eq 0 \).

When solving the system of equations in the exercise, matrix inversion allowed us to find the Laplace transforms of the solutions. It's a fundamental process in linear algebra used for:

• Solving linear equations
• Determining system properties
• Performing transformations in graphics and control systems

Inverse Laplace Transform

The inverse Laplace transform is the reverse process of finding the time domain function from its Laplace transform. It took several crucial steps in our exercise:

• After solving the equations using the Laplace transform, we end with expressions for the Laplace transforms of the solutions.
• The inverse process involves looking up these expressions in a table of known Laplace transforms to find the corresponding time-domain functions.

For instance, in our example:

• The expression for \( L\{ y_1\} \) leads us to finding \( y_1(t) = -\frac{1}{9}(e^{5t}-e^{-4t}) \).
• Similarly, \( L\{ y_2\} \) transforms back into the time domain, resulting in \( y_2(t)=\frac{1}{9}(e^{5t}-e^{-4t}) \).

Applying the inverse Laplace transform means converting complex frequency domain information back into forms that describe how systems behave over time.

Differential Equations

Differential equations describe how variable quantities change over time and are fundamental in engineering, physics, and other sciences.

• They can model a vast range of phenomena, such as motion, heat, or electric circuits.
• Our exercise involves a linear first-order system of differential equations specified in matrix form, \( \mathbf{y}'=A\mathbf{y}+\mathbf{b} \), where \( A \) is a matrix and \( \mathbf{b} \) is a vector.

Using Laplace transforms provides a strategic approach to solve such equations because:

• They convert the problem from a differential calculus form, involving derivatives, into an algebraic form, which involves polynomials and easier manipulation.
• This approach is especially useful for linear constant-coefficient differential equations, as it directly relates to the system's behavior.

System of Linear Equations

A system of linear equations is essentially a set of equations with multiple variables that need to be solved simultaneously.

• Linear equations are commonly put together with each equation representing a linear relation between the variables.
• In our exercise, the transformed system led us to solve such a system represented in the form of matrices.

The system of equations derived from our Laplace-transformed differential equations simplifies into a format \(AX = B\), and it necessitates finding solutions for the vector \(X\):

• Matrix inversion helps to find \( X \) by multiplying by the inverse of \( A \).
• This solution is crucial for determining the original unknown time-dependent functions.

Systems of linear equations appear in a wide array of problems, highlighting the interconnectedness of different variables and often representing complex real-world phenomena.