Problem 2 - Differential Equations and Linear Algebra

Question

Show that the given functions are solutions of the system \(\mathbf{x}^{\prime}(t)=A(x) \mathbf{x}(t)\) for the given matrix \(A,\) and hence, find the general solution to the system (remember to check linear independence). If auxiliary conditions are given, find the particular solution that satisfies these conditions. $$\begin{aligned} &\mathbf{x}_{1}(t)=\left[\begin{array}{c} e^{4 t} \\ 2 e^{4 t} \end{array}\right], \quad \mathbf{x}_{2}(t)=\left[\begin{array}{c} 3 e^{-t} \\ e^{-t} \end{array}\right],\\\ &A=\left[\begin{array}{ll} -2 & 3 \\\ -2 & 5 \end{array}\right], \quad \mathbf{x}(0)=\left[\begin{array}{r} -2 \\\ 1 \end{array}\right]. \end{aligned}$$

Step-by-Step Solution

Answer

The general solution of the system is \(\mathbf{x}(t) = c_{1}\left[\begin{array}{c} e^{4 t} \\ 2 e^{4t} \end{array}\right] + c_{2}\left[\begin{array}{c} 3 e^{-t} \\ e^{-t} \end{array}\right]\). Given the initial condition \(\mathbf{x}(0)=\left[\begin{array}{r} -2 \\ 1 \end{array}\right]\), we find the particular solution as \(\mathbf{x}(t) =\left[\begin{array}{c} -e^{4 t} + 3 e^{-t} \\ -2 e^{4t} + e^{-t} \end{array}\right]\).

Step 1: We are given the functions \(\mathbf{x}_{1}(t)=\left[\begin{array}{c} e^{4 t} \\ 2e^{4t} \end{array}\right]\) and \(\mathbf{x}_{2}(t)=\left[\begin{array}{c} 3e^{-t} \\ e^{-t} \end{array}\right]\), and we have to show that they are solutions to the differential equation system \(\mathbf{x}^{\prime}(t)=A(x) \mathbf{x}(t)\) with matrix \(A = \left[\begin{array}{ll} -2 & 3 \\ -2 & 5 \end{array}\right]\). To show this, we have to compute the derivatives of \(\mathbf{x}_{1}(t)\) and \(\mathbf{x}_{2}(t)\), and check if they are equal to the matrix multiplication \(A \mathbf{x}_{1}(t)\) and \(A \mathbf{x}_{2}(t)\), respectively. Let's first compute the derivatives of \(\mathbf{x}_{1}(t)\) and \(\mathbf{x}_{2}(t)\): \[\mathbf{x}_{1}^{\prime}(t)=\left[\begin{array}{c} 4 e^{4 t} \\ 8 e^{4t} \end{array}\right],\quad \mathbf{x}_{2}^{\prime}(t)=\left[\begin{array}{c} -3 e^{-t} \\ - e^{-t} \end{array}\right]\] Now compute \(A \mathbf{x}_{1}(t)\) and \(A \mathbf{x}_{2}(t)\): \[\begin{aligned} A \mathbf{x}_{1}(t) &= \left[\begin{array}{ll} -2 & 3 \\ -2 & 5 \end{array}\right] \left[\begin{array}{c} e^{4 t} \\ 2e^{4t} \end{array}\right] = \left[\begin{array}{c} 4 e^{4 t} \\ 8 e^{4t} \end{array}\right] \\ A \mathbf{x}_{2}(t) &= \left[\begin{array}{ll} -2 & 3 \\ -2 & 5 \end{array}\right] \left[\begin{array}{c} 3e^{-t} \\ e^{-t} \end{array}\right] = \left[\begin{array}{c} -3 e^{-t} \\ - e^{-t} \end{array}\right] \end{aligned}\] As we can see, \(\mathbf{x}_{1}^{\prime}(t) = A \mathbf{x}_{1}(t)\) and \(\mathbf{x}_{2}^{\prime}(t) = A \mathbf{x}_{2}(t)\), so the given functions are solutions of the system. #Step 2: Check the linear independence of \(\mathbf{x}_{1}(t)\) and \(\mathbf{x}_{2}(t)\)#

In order to find the general solution, we need to check the linear independence of the two given functions. The Wronskian of the two functions is given by: \[W[\mathbf{x}_{1}, \mathbf{x}_{2}](t) = \text{det}\begin{bmatrix} e^{4 t} & 3 e^{-t} \\ 2 e^{4 t} & e^{-t} \end{bmatrix} = (e^{4 t})(e^{-t}) - (3 e^{-t})(2 e^{4 t}) = e^{3t} - 6 e^{3t}\] Since the Wronskian is nonzero \((-5 e^{3t})\) for all \(t\), the given functions are linearly independent. #Step 3: Find the general solution#

Step 2: Since the given functions are linearly independent solutions of the system, the general solution can be expressed as a linear combination of \(\mathbf{x}_{1}(t)\) and \(\mathbf{x}_{2}(t)\): \[\mathbf{x}(t) = c_{1}\mathbf{x}_{1}(t) + c_{2}\mathbf{x}_{2}(t) = c_{1}\left[\begin{array}{c} e^{4 t} \\ 2 e^{4t} \end{array}\right] + c_{2}\left[\begin{array}{c} 3 e^{-t} \\ e^{-t} \end{array}\right]\] #Step 4: Find the particular solution using initial conditions#

Now, let's find the particular solution that satisfies the given initial conditions \(\mathbf{x}(0)=\left[\begin{array}{r} -2 \\ 1 \end{array}\right]\). Plug in \(t=0\) into the general solution: \[\mathbf{x}(0) = c_{1}\left[\begin{array}{c} 1 \\ 2 \end{array}\right] + c_{2}\left[\begin{array}{c} 3 \\ 1 \end{array}\right] = \left[\begin{array}{r} -2 \\ 1 \end{array}\right]\] This leads to the following system of equations: \[\begin{cases} c_{1} + 3c_{2} = -2 \\ 2c_{1} + c_{2} = 1 \end{cases}\] Solving this system, we get \(c_{1}=-1\) and \(c_{2}=1\). Thus, the particular solution is: \[\mathbf{x}(t) = -1\left[\begin{array}{c} e^{4 t} \\ 2 e^{4t} \end{array}\right] + 1\left[\begin{array}{c} 3 e^{-t} \\ e^{-t} \end{array}\right] = \left[\begin{array}{c} -e^{4 t} + 3 e^{-t} \\ -2 e^{4t} + e^{-t} \end{array}\right]\] So, we have found the general solution and the particular solution that satisfy the conditions given in the exercise.

Key Concepts

Linear Algebra and Differential Equations

Linear algebra is an essential branch of mathematics when it comes to understanding and solving differential equations systems. In our context, differential equations describe how certain functions change over time, and linear algebra provides the tools to handle and solve these systems efficiently.

For instance, when we are given a differential equation system like \(\mathbf{x}^\prime(t)=A\mathbf{x}(t)\), we are looking at a first-order linear system where \(\mathbf{x}(t)\) is a vector of functions and \(A\) is a matrix that affects how the system evolves over time. To solve such a system, we must ensure that our solution functions indeed satisfy the system's requirements which in this case, is verified through matrix multiplication.

Matrix Multiplication

Matrix multiplication is a central operation in linear algebra used to calculate the product of two matrices. This operation is not only fundamental in many areas of mathematics but also crucial for solving systems of linear differential equations. When we multiply matrix \(A\) with vector \(\mathbf{x}(t)\), we're essentially transforming \(\mathbf{x}(t)\) according to the rules defined by \(A\).

In our exercise, matrix multiplication is used to verify that the given functions \(\mathbf{x}_1(t)\) and \(\mathbf{x}_2(t)\) are solutions to the system by checking whether \(\mathbf{x}^\prime(t) = A\mathbf{x}(t)\). This step is critical because if the functions satisfy this condition, we can proceed to find general and particular solutions of the system.

Wronskian and Linear Independence

The Wronskian is a determinant used in the study of differential equations to test for linear independence among solutions. The concept is named after the Polish mathematician Hoene Wronski. In the context of our problem, linear independence ensures that the solutions form a basis set for the space of all solutions to the differential equation.

If the Wronskian is nonzero for at least one value of \(t\), then the functions are linearly independent. Moreover, the absence of linear independence would mean that the functions do not provide the complete solution set to the differential equation, which is a crucial aspect to check before establishing the general solution of the system.

Initial Conditions

Initial conditions help us determine a particular solution from a family of possible solutions to a differential equation. They are specific values assigned to the function (and possibly its derivatives) at a particular point, often when \(t=0\).

In our case, the initial condition \(\mathbf{x}(0)=\left[\begin{array}{r} -2 \ 1 \end{array}\right]\) allows us to find specific values for constants in the general solution, pinpointing the unique solution that fits the given starting scenario. This is a crucial step for translating a general solution into a particular one that corresponds to realistic initial conditions of the problem at hand.