Question

In Exercises $1-16$ find the general solution. $${\mathbf{y}}^{\prime }=\left[\begin{array}{rrr}6& 0& -3\\ -3& 3& 3\\ 1& -2& 6\end{array}\right]{\mathbf{y}}^{\prime }$$

Short answer

Answer: The eigenvalues are λ₁ = 0, λ₂ = 9, and λ₃ = 8. The corresponding eigenvectors are v₁ = [1, 1, 2], v₂ = [1, 0, 1], and v₃ = [3, -2, 2]. The general solution is given by y(t) = C₁[1, 1, 2]e^(0t) + C₂[1, 0, 1]e^(9t) + C₃[3, -2, 2]e^(8t), where C₁, C₂, and C₃ are constants determined by initial conditions.

Step-by-step solution

Find the eigenvalues

To find the eigenvalues, we need to solve the characteristic equation given by the determinant of the matrix minus lambda times the identity matrix. $det(A - \lambda I) = \left|$$\begin{array}{rrr}6-\lambda & 0& -3\\ -3& 3-\lambda & 3\\ 1& -2& 6-\lambda \end{array}$$\right|$ We will compute the determinant: $det(A-\lambda I)=(6-\lambda )((3-\lambda )(6-\lambda )+6)+3(3+6(-1)+2\lambda )={\lambda }^3-15{\lambda }^2+72\lambda$ Now, we will solve the characteristic equation for eigenvalues: ${\lambda }^3-15{\lambda }^2+72\lambda =0$ This equation has a clear factor of $\lambda$ which can be factored out: $\lambda ({\lambda }^2-15\lambda +72)=0$ The new quadratic equation can be factored further: $\lambda (\lambda -9)(\lambda -8)=0$ This gives us the eigenvalues: ${\lambda }_1=0,{\lambda }_2=9,{\lambda }_3=8$.

Find the eigenvectors

Now we'll find the eigenvectors corresponding to each eigenvalue. We will substitute each eigenvalue back into the equation $(A-\lambda I)\mathbf{v}=0$ and solve for the eigenvector $\mathbf{v}$. 1. For ${\lambda }_1=0$, $(A-0I){\mathbf{v}}_1=\left[\begin{array}{rrr}6& 0& -3\\ -3& 3& 3\\ 1& -2& 6\end{array}\right]{\mathbf{v}}_1=0$ By row reduction, we find the eigenvector to be ${\mathbf{v}}_1=\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right]$. 2. For ${\lambda }_2=9$, $(A-9I){\mathbf{v}}_2=\left[\begin{array}{rrr}-3& 0& -3\\ -3& -6& 3\\ 1& -2& -3\end{array}\right]{\mathbf{v}}_2=0$ By row reduction, we find the eigenvector to be ${\mathbf{v}}_2=\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$. 3. For ${\lambda }_3=8$, $(A-8I){\mathbf{v}}_3=\left[\begin{array}{rrr}-2& 0& -3\\ -3& -5& 3\\ 1& -2& -2\end{array}\right]{\mathbf{v}}_3=0$ By row reduction, we find the eigenvector to be ${\mathbf{v}}_3=\left[\begin{array}{c}3\\ -2\\ 2\end{array}\right]$.

Write the general solution

Now we can write down the general solution in terms of the eigenvalues and eigenvectors: $\mathbf{y}(t)=C_1{\mathbf{v}}_1{e}^{{\lambda }_{1}t}+C_2{\mathbf{v}}_2{e}^{{\lambda }_{2}t}+C_3{\mathbf{v}}_3{e}^{{\lambda }_{3}t}$ Where $C_1,C_2,$ and $C_3$ are constants determined by initial conditions. Substituting our derived values of eigenvalues and eigenvectors: $\mathbf{y}(t)=C_1\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right]e^{0t}+C_2\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]e^{9t}+C_3\left[\begin{array}{c}3\\ -2\\ 2\end{array}\right]e^{8t}$ This is the general solution for the given system of differential equations.

Key concepts

Eigenvalues and Eigenvectors

Understanding the eigenvalues and eigenvectors of a matrix is crucial when solving systems of differential equations. These concepts are frequently encountered in various fields of science and engineering, such as when analyzing mechanical vibrations, electric circuits, or population models.

Eigenvalues are special scalars associated with a linear system of equations. In the context of a matrix, they are the values of $\lambda$ that make the following equation true for nonzero vectors \textbf{v}:
$$(A-\lambda I)\mathbf{\text{v}}=0$$
where \textbf{A} is our matrix and \textbf{I} is the identity matrix. The corresponding vectors \textbf{v} that satisfy this equation are called eigenvectors. They represent directions in which the action of the matrix \textbf{A} can be reduced to simple scaling by the eigenvalue.

When we have a set of eigenvalues and eigenvectors for a matrix, they provide fundamental insights into the behavior of the system described by the matrix. In our textbook solution, finding the eigenvalues and eigenvectors is the first step toward expressing the general solution for the given differential equation.

Characteristic Equation

The characteristic equation is a tool for finding the eigenvalues of a matrix. It is a polynomial whose roots are the eigenvalues. Specifically, the characteristic equation for a matrix $A$ is obtained from the determinant:
$$det(A-\lambda I)=0$$
By calculating this determinant, we create a polynomial equation in terms of $\lambda$. The solutions to this equation are our sought-after eigenvalues.

In the sample exercise, we calculated the determinant of matrix \textbf{A} minus $\lambda$ times the identity matrix and set it equal to zero. This gave us a third-degree polynomial, which we then factored to find the roots. These roots were the eigenvalues necessary for constructing the general solution of our system of differential equations.

The characteristic equation provides a bridge between an abstract matrix and the actual values (eigenvalues) that allow us to interpret and solve the system of equations it represents.

System of Differential Equations

A system of differential equations consists of multiple equations involving derivatives of several unknown functions. Solving such a system involves finding the functions that satisfy all equations simultaneously.

In the context of our exercise, the given system is described by a matrix equation involving the derivative of a vector ${\mathbf{\text{y}}}^{\prime }$. The general solution to this kind of system often involves exponential functions of time, scaled by the eigenvectors of the matrix, and exponentially modulated by the eigenvalues. As demonstrated in the solution steps, after finding the eigenvalues and eigenvectors, the general solution is written in the form:
$$\mathbf{\text{y}}(t)=C_1{\mathbf{\text{v}}}_1{e}^{{\lambda }_{1}t}+C_2{\mathbf{\text{v}}}_2{e}^{{\lambda }_{2}t}+C_3{\mathbf{\text{v}}}_3{e}^{{\lambda }_{3}t}$$
where $C_1,C_2,$ and $C_3$ are constants determined by the initial conditions of the problem. This compact form of the general solution encapsulates the dynamic behavior of the entire system and is a powerful result of linear algebra applied to differential equations.