Question

In each exercise, find the general solution of the homogeneous linear system and then solve the given initial value problem. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} 3 & 1 & 2 \\ 0 & 8 & 15 \\ 0 & -6 & -11 \end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r} 2 \\ 5 \\ -2 \end{array}\right] $$

Short answer

Question: Find the specific solution to the given initial value problem represented by the matrix $$ A = \left[\begin{array}{rrr} 3 & 1 & 2 \\\ 0 & 8 & 15 \\\ 0 & -6 & -11 \end{array}\right] $$ and the initial value \(\mathbf{y}(0) = \begin{bmatrix}2 \\ 5 \\ -2\end{bmatrix}\). Answer: The specific solution to the given initial value problem is: $$ \mathbf{y}(t) = \frac{5}{3}e^{t} \begin{bmatrix}1 \\-2 \\ 3\end{bmatrix} + \frac{1}{3}e^{3t} \begin{bmatrix}2 \\1 \\ 0\end{bmatrix} + \frac{34}{45} e^{-25t} \begin{bmatrix}1 \\ 5 \\ 15\end{bmatrix} $$

Step-by-step solution

Find the eigenvalues of the matrix A

First, we must find the eigenvalues of the matrix A associated with the differential equation. The matrix A is given as $$ A = \left[\begin{array}{rrr} 3 & 1 & 2 \\\ 0 & 8 & 15 \\\ 0 & -6 & -11 \end{array}\right] $$ To find the eigenvalues, we compute the characteristic equation as \(det(A - \lambda I) = 0\) where \(I\) is the identity matrix and \(\lambda\) is the eigenvalues. In this case, $$ det(A - \lambda I) = \left|\begin{array}{ccc} 3-\lambda & 1 & 2 \\\ 0 & 8-\lambda & 15 \\\ 0 & -6 & -11-\lambda \end{array}\right| $$ Compute the determinant, we get \(\lambda^3 - 3\lambda^2 - 92\lambda - 75 = 0\). Solving this equation, we get the eigenvalues \(\lambda_1 = 1, \lambda_2 = 3, \lambda_3 = -25\)

Find the eigenvectors of the matrix A

Now that we have the eigenvalues, we need to compute the eigenvectors associated with each eigenvalue. For an eigenvalue \(\lambda\), we need to solve the equation \((A - \lambda I) \mathbf{v} = \mathbf{0}\) for the eigenvector \(\mathbf{v}\). For \(\lambda_1 = 1\), solve \((A - I) \mathbf{v} = \mathbf{0}\): $$ \left[\begin{array}{rrr} 2 & 1 & 2 \\\ 0 & 7 & 15 \\\ 0 & -6 & -12 \end{array}\right] \left[\begin{array}{r} x \\\ y \\\ z \end{array}\right] = \left[\begin{array}{r} 0 \\\ 0 \\\ 0 \end{array}\right] $$ This gives us the eigenvector \(v_1 = \begin{bmatrix}1 \\-2 \\ 3\end{bmatrix}\). For \(\lambda_2=3\), solve \((A - 3I) \mathbf{v}=\mathbf{0}\): $$ \left[\begin{array}{rrr} 0 & 1 & 2 \\\ 0 & 5 & 15 \\\ 0 & -6 & -14 \end{array}\right] \left[\begin{array}{r} x \\\ y \\\ z \end{array}\right] = \left[\begin{array}{r} 0 \\\ 0 \\\ 0 \end{array}\right] $$ This gives us the eigenvector \(v_2 = \begin{bmatrix}2 \\1 \\ 0\end{bmatrix}\). For \(\lambda_3=-25\), solve \((A + 25I) \mathbf{v}=\mathbf{0}\): $$ \left[\begin{array}{rrr} 28 & 1 & 2 \\\ 0 & 33 & 15 \\\ 0 & -6 & 14 \end{array}\right] \left[\begin{array}{r} x \\\ y \\\ z \end{array}\right] = \left[\begin{array}{r} 0 \\\ 0 \\\ 0 \end{array}\right] $$ This gives us the eigenvector \(v_3 = \begin{bmatrix}1 \\ 5 \\ 15\end{bmatrix}\).

Find the general solution of the homogeneous linear system

With the eigenvalues and eigenvectors, we can now write the general solution of the homogeneous linear system given by $$ \mathbf{y}(t) = c_1 e^{\lambda_1 t} \mathbf{v}_1 + c_2 e^{\lambda_2 t} \mathbf{v}_2 + c_3 e^{\lambda_3 t} \mathbf{v}_3 $$ Substituting the eigenvalues and eigenvectors, we get $$ \mathbf{y}(t) = c_1 e^{t} \begin{bmatrix}1 \\-2 \\ 3\end{bmatrix} + c_2 e^{3t} \begin{bmatrix}2 \\1 \\ 0\end{bmatrix} + c_3 e^{-25t} \begin{bmatrix}1 \\ 5 \\ 15\end{bmatrix} $$

Solve the initial value problem

Now we will solve the initial value problem by using the given initial value \(\mathbf{y}(0) = \begin{bmatrix}2 \\ 5 \\ -2\end{bmatrix}\). We substitute \(t = 0\) into the general solution and set it equal to the initial value: $$ \begin{bmatrix}2 \\ 5 \\ -2\end{bmatrix} = c_1 \begin{bmatrix}1 \\-2 \\ 3\end{bmatrix} + c_2 \begin{bmatrix}2 \\1 \\ 0\end{bmatrix} + c_3 \begin{bmatrix}1 \\ 5 \\ 15\end{bmatrix} $$ Solving the above system of equations, we find that \(c_1 = \frac{5}{3}, c_2 = \frac{1}{3}\), and \(c_3 = \frac{34}{45}\).

Find the specific solution

Finally, we substitute the coefficients \(c_1\), \(c_2\), and \(c_3\) into the general solution to obtain the specific solution for the given initial value problem: $$ \mathbf{y}(t) = \frac{5}{3}e^{t} \begin{bmatrix}1 \\-2 \\ 3\end{bmatrix} + \frac{1}{3}e^{3t} \begin{bmatrix}2 \\1 \\ 0\end{bmatrix} + \frac{34}{45} e^{-25t} \begin{bmatrix}1 \\ 5 \\ 15\end{bmatrix} $$ This is the solution to the given initial value problem.

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are essential in solving differential equations, especially homogeneous linear systems. They help us understand how a system evolves over time. An eigenvalue is a special number associated with a square matrix that, when multiplied by its corresponding eigenvector, does not change its direction. Instead, it scales the eigenvector. This scaling is what allows us to analyze the behavior of solutions in differential equations.
To find the eigenvalues of a matrix, we solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \), where \( A \) is the matrix, \( \lambda \) represents the eigenvalues, and \( I \) is the identity matrix. Once we determine the eigenvalues, the next step is to find the eigenvectors, which are the directions in which the scaling occurs.
Eigenvectors are found by solving the equation \( (A - \lambda I) \mathbf{v} = \mathbf{0} \) for each eigenvalue. These calculations provide a foundational tool to decompose and understand complex systems, which is crucial for forming the general solution in differential equations.

General Solution of Differential Equations

In the context of homogeneous linear systems with constant coefficients, the general solution is constructed using the eigenvalues and eigenvectors we derived earlier. This solution is a combination of exponential functions multiplied by the corresponding eigenvectors, a form that reflects the system's behavior over time. The general solution encapsulates all possible states of the system, given any set of initial conditions.
The general solution can be expressed as:
\[\mathbf{y}(t) = c_1 e^{\lambda_1t} \mathbf{v}_1 + c_2 e^{\lambda_2t} \mathbf{v}_2 + c_3 e^{\lambda_3t} \mathbf{v}_3\]
Where \( c_1, c_2, \) and \( c_3 \) are constants determined by the initial values of the problem, and \( \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \) are the eigenvectors. The exponential functions \( e^{\lambda_it} \) show how each component evolves, dictated by the corresponding eigenvalue \( \lambda_i \).
This approach to finding the general solution streamlines the process of analyzing the dynamical behavior of equations, offering a way to predict how each factor will influence the system over time.

Initial Value Problems

Initial value problems add an extra layer to solving differential equations. Here, after determining the general solution, we apply specific initial conditions to find a particular solution that meets these conditions. This step is crucial because it tailors the general solution to match real-world scenarios specified at the beginning time \( t=0 \).
To solve an initial value problem, we substitute the given initial conditions into the general solution. In our case, we had \( \mathbf{y}(0) = \begin{bmatrix}2 \ 5 \ -2\end{bmatrix} \). By inserting these into the general solution, we create a system of equations that allows us to solve for the unknown constants \( c_1, c_2, \) and \( c_3 \).
Unlike the infinite solutions offered by the general system, the initial value problem pinpoints one specific solution that reflects the initial state of the system. This precision is what makes initial value problems applicable and significant in practical and diverse scientific and engineering contexts.