Question

Find the general solution. \(\mathbf{y}^{\prime}=\left[\begin{array}{rrr}-6 & -4 & -8 \\ -4 & 0 & -4 \\\ -8 & -4 & -6\end{array}\right] \mathbf{y}\)

Short answer

Question: Find the general solution of the first-order linear homogeneous system of ordinary differential equations (ODEs) given by \(\mathbf{y}'=A\mathbf{y}\), where \(A =\left[\begin{array}{rrr}-6 & -4 & -8 \\\ -4 & 0 & -4 \\\ -8 & -4 & -6\end{array}\right]\). Answer: The general solution for the given ODE system is \(\mathbf{y}(t)=c_1 e^{-4t}\left[\begin{array}{c}2 \\\ 1 \\\ 2\end{array}\right] + c_2 e^{-4t}\left[\begin{array}{c}2t+1 \\\ t \\\ 2t+2\end{array}\right]+c_3 e^{-4t}\left[\begin{array}{c}2t^2+2t \\\ 2t^2 \\\ 2t^2+4t\end{array}\right]\), where \(c_1, c_2\), and \(c_3\) are arbitrary constants.

Step-by-step solution

Find the eigenvalues of matrix A

To find the eigenvalues of the given matrix \(A=\left[\begin{array}{rrr}-6 & -4 & -8 \\\ -4 & 0 & -4 \\\ -8 & -4 & -6\end{array}\right]\), we need to calculate the characteristic polynomial. To do this, find the determinant of the matrix \((A-\lambda I)\) as shown below: \(\det(A-\lambda I) = \det\left(\left[\begin{array}{ccc}-6-\lambda & -4 & -8 \\\ -4 & -\lambda & -4 \\\ -8 & -4 & -6-\lambda\end{array}\right]\right)\)

Calculate the eigenvalues

After calculating the determinant, find the eigenvalues by finding the values of the variable \(\lambda\) for which the determinant equals zero: \(\det(A-\lambda I)=\lambda^3+12\lambda^2+56\lambda+64=(\lambda+4)^3=0,\) \(\Rightarrow \lambda_1=\lambda_2=\lambda_3=-4.\) Since we have only one unique eigenvalue (-4) with multiplicity 3, we proceed to find the eigenvectors and the generalized eigenvectors.

Solve for eigenvectors

To find the eigenvectors corresponding to the eigenvalue -4, solve the system \((A+\lambda I)\mathbf{x}=\mathbf{0}\): \(\left[\begin{array}{ccc}-2 & -4 & -8 \\\ -4 & 4 & -4 \\\ -8 & -4 & -2 \end{array}\right]\mathbf{x}=\mathbf{0}\) Here, we find an eigenvector as \(\mathbf{x}_1=\left[\begin{array}{c}2 \\\ 1 \\\ 2\end{array}\right]\)

Solve for generalized eigenvectors

To find the generalized eigenvectors, we need to solve the system \((A-\lambda I)^2\mathbf{x}=\mathbf{0}\) and \((A-\lambda I)^3\mathbf{x}=\mathbf{0}\) as shown below: \((A-\lambda I)^2\mathbf{x}=\mathbf{0}\) \(\left[\begin{array}{ccc}-2 & -4 & -8 \\\ -4 & 4 & -4 \\\ -8 & -4 & -2 \end{array}\right]^2\mathbf{x}=\mathbf{0}\) and we find a generalized eigenvector as \(\mathbf{x}_2=\left[\begin{array}{c}1 \\\ 0 \\\ 1\end{array}\right]\) For \((A-\lambda I)^3\mathbf{x}=\mathbf{0}\), since the rank of \((A-\lambda I)^3\) equals the rank of \((A-\lambda I)^2\), there is no need to solve this system, as the obtained eigenvectors and generalized eigenvectors are sufficient.

Construct the general solution

To construct the general solution of the ODE system, use the eigenvalues, eigenvectors, and generalized eigenvectors: \(\mathbf{y}(t)=c_1 e^{\lambda_1 t}\mathbf{x}_1 + c_2 e^{\lambda_2 t}(t\mathbf{x}_1+\mathbf{x}_2) + c_3 e^{\lambda_3 t}(t^2\mathbf{x}_1+2t\mathbf{x}_2)\) Substitute the values for eigenvalues and eigenvectors to obtain the general solution: \(\mathbf{y}(t)=c_1 e^{-4t}\left[\begin{array}{c}2 \\\ 1 \\\ 2\end{array}\right] + c_2 e^{-4t}\left[\begin{array}{c}2t+1 \\\ t \\\ 2t+2\end{array}\right]+c_3 e^{-4t}\left[\begin{array}{c}2t^2+2t \\\ 2t^2 \\\ 2t^2+4t\end{array}\right]\)

Key concepts

Eigenvalues and Eigenvectors

In the realm of linear algebra, eigenvalues and eigenvectors are essential components when analyzing linear transformations. They're particularly vital in solving systems of linear differential equations, as they help reveal the system's underlying structure.

Eigenvalues, denoted usually by \(\lambda\), are scalars that satisfy the equation \(A\mathbf{v} = \lambda\mathbf{v}\), where \(A\) is a square matrix, \(\mathbf{v}\) is an eigenvector, and the operation performed is matrix multiplication. This equation essentially states that when a linear transformation represented by \(A\) is applied to \(\mathbf{v}\), the output is a scalar multiple of \(\mathbf{v}\).

To find these values for a given matrix, we solve the characteristic polynomial, which leads us to the eigenvalues, and consequently, their corresponding eigenvectors. In the context of differential equations, each eigenvalue corresponds to a particular solution, and when an eigenvalue is repeated, as in our example with a unique eigenvalue of multiplicity three, generalized eigenvectors are also needed to construct the complete solution.

The process of finding eigenvalues and eigenvectors typically involves:

• Creating the matrix \(A - \lambda I\) from the given matrix \(A\)
• Computing the determinant of \(A - \lambda I\), which gives us the characteristic polynomial
• Solving the equation \(\det(A - \lambda I) = 0\) for \(\lambda\) to obtain eigenvalues
• Substituting each eigenvalue back into \(A - \lambda I\) to solve for the corresponding eigenvectors

Characteristic Polynomial

The characteristic polynomial plays a pivotal role in determining the eigenvalues of a matrix. It is a polynomial which, when set to zero, gives us the values of \lambda that are the eigenvalues of the matrix. You can think of it as a bridge between a matrix and its eigenvalues.

To compute the characteristic polynomial of matrix \(A\), one takes the determinant of the matrix \(A - \lambda I\), where \(I\) is the identity matrix of the same dimension as \(A\) and \(\lambda\) represents a scalar value. This polynomial will be in terms of \lambda and by finding its roots, we determine the matrix's eigenvalues.

In our exercise, the characteristic polynomial was found to be \(\lambda^3+12\lambda^2+56\lambda+64\). It simplifies to \(\lambda+4)^3 = 0\), indicating that \(\lambda = -4\) is a root with multiplicity three. In practice, finding the characteristic polynomial is a fundamental step in analyzing matrix behavior and solving differential equations.

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are fundamental in modeling a wide array of real-world phenomena from physics, engineering, economics, and biology, describing how physical quantities change over time or space.

In our example, we're dealing with a system of linear ordinary differential equations (ODEs), which is represented in matrix form. The solution to this system is a vector function that describes the state of the system at any time. To solve these equations, we often use the matrix's eigenvalues and eigenvectors. For a system \(\mathbf{y}' = A\mathbf{y}\), where \(A\) is a matrix and \(\mathbf{y}\) is a vector of functions, the solution is constructed using the eigenvalues \lambda and eigenvectors \mathbf{v}, leading to solutions of the form \(\mathbf{y}(t) = c e^{\lambda t}\mathbf{v}\).

If the eigenvalue \lambda has multiplicity greater than one, we must also find generalized eigenvectors to create a complete set of linearly independent solutions. This is exactly what was demonstrated in the exercise with the unique eigenvalue of -4. The final solution combines these vectors into a general solution that includes constants \(c_1\), \(c_2\), and \(c_3\) to account for initial conditions or further specifications of the problem.