Question

In each exercise, find the general solution of the homogeneous linear system and then solve the given initial value problem. $$ \mathbf{y}^{\prime}=\left[\begin{array}{ll} 1 & 2 \\ 0 & 3 \end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{l} 4 \\ 1 \end{array}\right] $$

Short answer

#Answer# The specific solution to the given initial value problem is: $$ \mathbf{y}(t) = 4 e^{t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$

Step-by-step solution

Calculate the eigenvalues of the matrix

To calculate the eigenvalues, we need to find the roots of the characteristic equation, which is obtained by subtracting the eigenvalue (denoted by λ) from the diagonals of the coefficient matrix and taking the determinant. First, subtract λ from the diagonal elements and take the determinant of the resulting matrix: $$ \begin{vmatrix} 1-\lambda & 2 \\ 0 & 3-\lambda \end{vmatrix} = (1-\lambda)(3-\lambda) \\ = \lambda^2 - 4\lambda + 3 $$ Now, factor the above quadratic equation: $$ \lambda^2 - 4\lambda + 3 = (\lambda-3)(\lambda-1) $$ Thus, the eigenvalues are λ₁ = 1 and λ₂ = 3.

Calculate the eigenvectors associated with the eigenvalues

Now, we need to find the eigenvectors corresponding to each eigenvalue. First, we will find the eigenvector associated with λ₁ = 1: Plugging the eigenvalue into the equation (A - λI)v = 0 gives us: $$ \begin{bmatrix} 1-1 & 2 \\ 0 & 3-1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ Simplifying the equation, we get: $$ \begin{bmatrix} 0 & 2 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ From the above equation: 2v₂ = 0, which gives us v₂ = 0. Thus, the eigenvector corresponding to λ₁ = 1 is: $$ \mathbf{v}_{1} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} $$ Next, find the eigenvector associated with λ₂ = 3: $$ \begin{bmatrix} 1-3 & 2 \\ 0 & 3-3 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ Simplifying the equation, we get: $$ \begin{bmatrix} -2 & 2 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$ From the above equation: -2w₁ + 2w₂ = 0, which gives us w₁ = w₂. Thus, the eigenvector corresponding to λ₂ = 3 is: $$ \mathbf{v}_{2} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$

Compute the general solution

Now that we have calculated the eigenvalues and the corresponding eigenvectors, we can find the general solution of the homogeneous linear system using these values: $$ \mathbf{y}(t) = c_1 e^{1t} \mathbf{v}_{1} + c_2 e^{3t} \mathbf{v}_{2} \\ = c_1 e^{t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$

Solve the initial value problem

Now let's solve the initial value problem by applying the given initial condition, which is: $$ \mathbf{y}(0) = \begin{bmatrix} 4 \\ 1 \end{bmatrix} $$ Plugging t = 0 into the general solution equation, we get: $$ \begin{bmatrix} 4 \\ 1 \end{bmatrix} = c_1 e^{0} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{0} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \\ \Rightarrow \begin{bmatrix} 4 \\ 1 \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$ From the above equation, we see that c₁ = 4 and c₂ = 1. Therefore, we can find the specific solution to the given initial value problem: $$ \mathbf{y}(t) = 4 e^{t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + e^{3t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} $$

Key concepts

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is crucial in solving a homogeneous linear system of differential equations. The process starts by representing the system using a matrix. In our example, this matrix is \( \begin{bmatrix} 1 & 2 \ 0 & 3 \end{bmatrix} \).

To find eigenvalues, we solve the characteristic equation, which arises from \( \det(A - \lambda I) = 0 \), where \( A \) is our matrix, \( \lambda \) represents the eigenvalue, and \( I \) is the identity matrix. After calculating, we find that the eigenvalues are \( \lambda_1 = 1 \) and \( \lambda_2 = 3 \).

Once the eigenvalues are determined, finding the corresponding eigenvectors gives us direction vectors that inherently point to the same orientation as our system. For each eigenvalue, solve \( (A - \lambda I)\mathbf{v} = 0 \) for vector \( \mathbf{v} \). This yields the eigenvectors \( \mathbf{v}_1 = \begin{bmatrix} 1 \ 0 \end{bmatrix} \) for \( \lambda_1 \) and \( \mathbf{v}_2 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \) for \( \lambda_2 \).

Eigenvalues and eigenvectors form the foundation of solving linear systems, allowing us to express any vector as a combination of eigenvectors, thereby simplifying the matrix operations involved.

Initial Value Problem

An initial value problem involves finding a specific solution to a differential equation, given an initial condition. This initial condition sets the starting point of the system. In this exercise, we are given \( \mathbf{y}(0) = \begin{bmatrix} 4 \ 1 \end{bmatrix} \), which pinpoints the system's state at time \( t = 0 \).

To solve the initial value problem, we use the general solution of the system. At \( t = 0 \), the coefficients associated with our eigenvectors need to satisfy the initial state.

Here’s what happens step-by-step:

• We evaluate the general solution at \( t = 0 \).
• This simplifies the general solution to the form \( c_1 \begin{bmatrix} 1 \ 0 \end{bmatrix} + c_2 \begin{bmatrix} 1 \ 1 \end{bmatrix} = \begin{bmatrix} 4 \ 1 \end{bmatrix} \).
• With simple substitution and solving, we find \( c_1 = 4 \) and \( c_2 = 1 \).

This ensures that the specific solution reflects the system dynamics observed at the start. The solution is tailored to not just satisfy the differential equation, but also align with the initial conditions imposed.

General Solution of Differential Equations

The concept of a general solution to a system of differential equations is central when dealing with linear algebra and dynamics. For homogeneous linear systems, like the one in the exercise, solutions are expressed using the system's eigenvalues and eigenvectors. These solutions describe the behavior of the entire system over time.

The general solution for our system is derived as follows:

• Understand the influence of each eigenvalue, where each contributes uniquely to the system behavior through exponential growth or decay events, shown as terms like \( e^{\lambda t} \).
• Combine these influences into a comprehensive expression that involves constants \( c_1 \) and \( c_2 \), which tune the solution to initial conditions.

The solution appears as: \( \mathbf{y}(t) = c_1 e^{t} \begin{bmatrix} 1 \ 0 \end{bmatrix} + c_2 e^{3t} \begin{bmatrix} 1 \ 1 \end{bmatrix} \). These exponentials mirror how the initial system setup evolves over time.

This general solution serves as a powerful insight, not only offering a way express the solution mathematically but also aiding in prediction of future behaviors of the system, enhancing one's ability to analyze and infer deeper insights into the system dynamics.