Question

Let $$ A=\left[\begin{array}{ll} 2 & 4 \\ 4 & 2 \end{array}\right], \quad \mathbf{y}_{1}=\left[\begin{array}{c} e^{6 t} \\ e^{6 t} \end{array}\right], \quad \mathbf{y}_{2}=\left[\begin{array}{r} e^{-2 t} \\ -e^{-2 t} \end{array}\right], \quad \mathbf{k}=\left[\begin{array}{r} -3 \\ 9 \end{array}\right] $$ (a) Verify that \(\left\\{\mathbf{y}_{1}, \mathbf{y}_{2}\right\\}\) is a fundamental set of solutions for \(\mathbf{y}^{\prime}=A \mathbf{y}\). (b) Solve the initial value problem $$ \mathbf{y}^{\prime}=A \mathbf{y}, \quad \mathbf{y}(0)=\mathbf{k} $$ (c) Use the result of Exercise \(6(\mathbf{b})\) to find a formula for the solution of (A) for an arbitrary initial vector \(\mathbf{k}\)

Short answer

Question: Verify that the given vectors form a fundamental set of solutions for the differential equation and find the formula for solving the initial value problem with an arbitrary initial vector 𝑘. Answer: The vectors 𝑦1 and 𝑦2 form a fundamental set of solutions for the differential equation 𝑦′=𝐴𝑦, and the formula for solving the initial value problem with an arbitrary initial vector 𝑘 is given by: $$ \mathbf{y}(t) = \frac{k_{1}-k_{2}}{2}\begin{bmatrix} e^{6t}\\ e^{6t} \end{bmatrix} + \frac{k_{1}+k_{2}}{2}\begin{bmatrix} e^{-2t} \\ -e^{-2t} \end{bmatrix}. $$

Step-by-step solution

Verification of the fundamental set

First, let's calculate the derivatives 𝑦₁' and 𝑦₂' as follows: $$ \mathbf{y}_{1}^{\prime}=\begin{bmatrix} 6e^{6t}\\ 6e^{6t} \end{bmatrix} $$ and $$ \mathbf{y}_{2}^{\prime}=\begin{bmatrix} 2e^{-2t}\\ 2e^{-2t} \end{bmatrix}. $$ Now, let's compute 𝐴𝑦₁ and 𝐴𝑦₂: $$ A\mathbf{y}_{1}= \begin{bmatrix} 2 & 4\\ 4 & 2 \end{bmatrix}\begin{bmatrix} e^{6t}\\ e^{6t} \end{bmatrix} =\begin{bmatrix} 6e^{6t} \\ 6e^{6t} \end{bmatrix} =\mathbf{y}_{1}^{\prime}, $$ and $$ A\mathbf{y}_{2}= \begin{bmatrix} 2 & 4\\ 4 & 2 \end{bmatrix}\begin{bmatrix} e^{-2t}\\ -e^{-2t} \end{bmatrix} =\begin{bmatrix} 2e^{-2t}\\ 2e^{-2t} \end{bmatrix} =\mathbf{y}_{2}^{\prime}. $$ Since both 𝑦₁' = 𝐴𝑦₁ and 𝑦₂' = 𝐴𝑦₂, it verifies that 𝑦₁ and 𝑦₂ form a fundamental set of solutions for the differential equation 𝑦′=𝐴𝑦.

Solving the initial value problem

The general solution of the homogeneous linear system 𝑦′ = 𝐴𝑦 is given by a linear combination of the fundamental set of solutions: 𝑦(𝑡) = 𝑐₁𝑦₁ + 𝑐₂𝑦₂. Then, we find the values of 𝑐₁ and 𝑐₂ given the initial state 𝑦(0) = 𝑘: $$ \begin{bmatrix} -3\\ 9 \end{bmatrix}= \begin{bmatrix} c_{1} \\ c_{1} \end{bmatrix}+ \begin{bmatrix} c_{2} \\ -c_{2} \end{bmatrix}. $$ From the equation above, we can see that \(c_1=-3\) and \(c_2=6\). Thus, the initial value problem solution is: $$ \mathbf{y}(t) = -3\begin{bmatrix} e^{6t}\\ e^{6t} \end{bmatrix} + 6 \begin{bmatrix} e^{-2t} \\ -e^{-2t} \end{bmatrix} =R(t) $$

Formula for arbitrary 𝑘

We can express the general solution we found in step 2 in terms of an arbitrary initial state vector 𝑘 = [𝑘₁, 𝑘₂]𝑇 as follows: $$ \mathbf{y}(t) = \frac{k_{1}-k_{2}}{2}\begin{bmatrix} e^{6t}\\ e^{6t} \end{bmatrix} + \frac{k_{1}+k_{2}}{2}\begin{bmatrix} e^{-2t} \\ -e^{-2t} \end{bmatrix}. $$ This is the formula for solving the initial value problem with an arbitrary initial vector 𝑘.

Key concepts

Fundamental Set of Solutions

Understanding the concept of a fundamental set of solutions is essential for solving systems of linear differential equations. When we have a system of linear differential equations represented in matrix form as \[\begin{equation}\mathbf{y}^{\text{}}=A \mathbf{y}\end{equation}\], a fundamental set of solutions is a group of solution vectors that form a basis for the solution space -- this means that any solution to the system can be expressed as a linear combination of these basis vectors.

To verify that a given set of solutions is fundamental, each solution in the set must be a solution to the system, and the set must be linearly independent. Linear independence ensures that no solution in the set is a linear combination of the others, providing the 'building blocks' needed to construct any possible solution. In the textbook example, verification was completed by showing that each solution vector satisfies the differential equation when multiplied by the matrix A, confirming that both \[\begin{equation}\mathbf{y}_1\end{equation}\] and \[\begin{equation}\mathbf{y}_2\end{equation}\] are indeed solutions to the system.

Moreover, the property that \[\begin{equation}A\mathbf{y}_i = \mathbf{y}_i^{\prime}\end{equation}\] for each solution vector indicates the set is fundamental. The fact that these vectors are also linearly independent (due to their distinct exponential components) confirms their status as a fundamental set of solutions.

Initial Value Problem

An initial value problem (IVP) occurs when a differential equation is supplemented by initial conditions, that is, the values of the unknown functions at a specific point. Solving an initial value problem means finding a function, or in the case of systems, a set of functions that satisfy both the differential equation and the initial conditions.

In our textbook problem, the initial condition is provided as the vector \[\begin{equation}\mathbf{k}\end{equation}\], representing the value of the solution at time t=0. Solving the IVP entails determining constants so that the solution fits this condition perfectly. When we tackled the IVP in the exercise, we did this by expressing the general solution as a linear combination of our fundamental set of solutions, and then finding the coefficients that align with the initial vector \[\begin{equation}\mathbf{k}\end{equation}\]. This process showed that specific coefficients indeed satisfy the initial conditions, giving us a unique solution that fits the given initial state.Understanding initial value problems is crucial because many real-world applications involve determining the progress of a system from a known starting point. It's like knowing your starting position on a map and figuring out the path you need to take.

Homogeneous Linear System

A homogeneous linear system is a set of linear equations where each equation sums up to zero. In matrix terms, it's where we have \[\begin{equation}\mathbf{y}^{\text{'}} = A \mathbf{y}\end{equation}\]and we're looking for vectors \[\begin{equation}\mathbf{y}\end{equation}\] that satisfy this relationship. A key feature of such systems is that if \[\begin{equation}\mathbf{y}_1\end{equation}\] and\[\begin{equation}\mathbf{y}_2\end{equation}\] are solutions, then any linear combination of these solutions is also a solution. This property is known as the principle of superposition.

Our exercise example demonstrated this by solving for \[\begin{equation}\mathbf{y}(t)\end{equation}\] which represents the general solution to the homogeneous linear system. This system is important to study because homogeneous linear systems often represent steady state or equilibrium conditions in various physical, biological, and economic models. Moreover, understanding these systems paves the way for dealing with non-homogeneous systems, where an external force or influence is introduced, by employing methods such as the superposition principle.