Problem 10
Question
Solve \(\mathbf{x}^{\prime}=A \mathbf{x}\) by determining \(n\) linearly independent solutions of the form \(\mathbf{x}(t)=e^{A t} \mathbf{v}\). \(A=\left[\begin{array}{rrrr}1 & 0 & 0 & 0 \\ 0 & 6 & -7 & 3 \\ 0 & 0 & 3 & -1 \\ 0 & -4 & 9 & -3\end{array}\right] .\) You may assume that \(p(\lambda)=(\lambda-1)(\lambda-2)^{3}.\)
Step-by-Step Solution
Verified Answer
The general solution of the given differential equation with 4 linearly independent solutions is:
\[\mathbf{x}(t)=\begin{bmatrix}c_1e^t\\c_2e^{2t}\\c_3te^{2t}\\c_2e^{2t}+c_3te^{2t}+c_4t^2e^{2t}\end{bmatrix}\]
1Step 1: Find the eigenvectors and generalized eigenvectors for eigenvalue 1.
First, we'll find an eigenvector corresponding to eigenvalue \(\lambda=1\). Solve the equation \((A-\lambda I)\mathbf{v}=0\), where \(\lambda=1\) and \(\mathbf{v}\) is the eigenvector we're looking for.
Substitute \(\lambda=1\) from our given characteristic equation:
\[(A-I)\mathbf{v}=\left[\begin{array}{rrrr}0 & 0 & 0 & 0 \\\ 0 & 5 & -7 & 3 \\\ 0 & 0 & 2 &-1 \\\ 0 & -4 & 9 & -4\end{array}\right]\mathbf{v}=0\]
Solving this system, we obtain that the eigenvector \(\mathbf{v_1}=\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0\end{bmatrix}\).
Since there's only one eigenvalue and the algebraic multiplicity is 1, we have no generalized eigenvectors corresponding to \(\lambda=1\).
2Step 2: Find the eigenvectors and generalized eigenvectors for eigenvalue 2.
Next, we'll find an eigenvector corresponding to eigenvalue \(\lambda=2\). Solve the equation \((A-\lambda I)\mathbf{v}=0\), where \(\lambda=2\) and \(\mathbf{v}\) is the eigenvector we're looking for.
Substitute \(\lambda=2\) from our given characteristic equation:
\[(A-2I)\mathbf{v}=\left[\begin{array}{rrrr}-1 & 0 & 0 & 0 \\\ 0 & 4 & -7 & 3 \\\ 0 & 0 & 1 &-1 \\\ 0 & -4 & 9 & -5\end{array}\right]\mathbf{v}=0\]
Solving this system, we obtain that the eigenvector \(\mathbf{v_2}=\begin{bmatrix} 0 \\ 1 \\ 0 \\ 1\end{bmatrix}\).
Now, we have an eigenvector corresponding to the eigenvalue \(\lambda=2\). We need to find 2 more linearly independent generalized eigenvectors to find the solution.
Next, we compute \((A-2I)^2\mathbf{v}=0\). Using the matrix \((A-2I)\) computed above, we have:
\[(A-2I)^2=\left[\begin{array}{rrrr}1 & 0 & 0 & 0 \\\ 0 & 1 & -1 & 0 \\\ 0 & 0 & 1 &-1 \\\ 0 & 0 & 1 & 1\end{array}\right]\]
Solving the system \((A-2I)^2\mathbf{v}=0\), we obtain that a linearly independent generalized eigenvector \(\mathbf{v_3}=\begin{bmatrix} 0 \\ 0 \\ 1 \\ 1\end{bmatrix}\).
To find the last generalized eigenvector, we will compute \((A-2I)^3\mathbf{v}=0\):
\[(A-2I)^3=\left[\begin{array}{rrrr}1 & 0 & 0 & 0 \\\ 0 & 1 & -1 & 1 \\\ 0 & 0 & 1 &-1 \\\ 0 & 0 & 0 & 1\end{array}\right]\]
Solving the system \((A-2I)^3\mathbf{v}=0\), we obtain that the linearly independent generalized eigenvector \(\mathbf{v_4}=\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1\end{bmatrix}\).
3Step 3: Write the general solution of the given differential equation
We have found 4 linearly independent eigenvectors and generalized eigenvectors of \(A\). Now, we can express the general solution for \(\mathbf{x}(t)\) using the formula \(e^{A t}\mathbf{v}\):
\[\mathbf{x}(t)=c_1 e^{1t}\mathbf{v_1} + c_2 e^{2t}\mathbf{v_2} + c_3 te^{2t}\mathbf{v_3} + c_4 t^2e^{2t}\mathbf{v_4}\]
Substituting the eigenvectors and generalized eigenvectors we found:
\[\mathbf{x}(t)=c_1 e^{t}\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0\end{bmatrix}+ c_2 e^{2t}\begin{bmatrix} 0 \\ 1 \\ 0 \\ 1\end{bmatrix} + c_3 te^{2t}\begin{bmatrix} 0 \\ 0 \\ 1 \\ 1\end{bmatrix} + c_4 t^2e^{2t}\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1\end{bmatrix}\]
This expression represents the general solution of the given differential equation with 4 linearly independent solutions:
\[\mathbf{x}(t)=\begin{bmatrix}c_1e^t\\c_2e^{2t}\\c_3te^{2t}\\c_2e^{2t}+c_3te^{2t}+c_4t^2e^{2t}\end{bmatrix}\]
Key Concepts
EigenvectorsEigenvaluesGeneral Solution
Eigenvectors
One of the key concepts when dealing with differential equations like the one given is eigenvectors. An eigenvector, in simple terms, is a non-zero vector that changes by a scalar factor when a linear transformation is applied. This means that its direction does not change, only its magnitude. When solving differential equations, identifying eigenvectors allows us to understand the structure of a system's transformation.
In this particular exercise, we discovered the eigenvector \(\mathbf{v_1} = \begin{bmatrix} 1 \ 0 \ 0 \ 0 \end{bmatrix}\) associated with the eigenvalue \(\lambda=1\), and another eigenvector \(\mathbf{v_2} = \begin{bmatrix} 0 \ 1 \ 0 \ 1 \end{bmatrix}\) associated with \(\lambda=2\). These vectors fit into the equation \((A-\lambda I)\mathbf{v}=0\), where \(I\) is the identity matrix.
Eigenvectors, especially when used in combination with their corresponding eigenvalues, provide crucial insight into the behavior of dynamic systems modelled by differential equations.
In this particular exercise, we discovered the eigenvector \(\mathbf{v_1} = \begin{bmatrix} 1 \ 0 \ 0 \ 0 \end{bmatrix}\) associated with the eigenvalue \(\lambda=1\), and another eigenvector \(\mathbf{v_2} = \begin{bmatrix} 0 \ 1 \ 0 \ 1 \end{bmatrix}\) associated with \(\lambda=2\). These vectors fit into the equation \((A-\lambda I)\mathbf{v}=0\), where \(I\) is the identity matrix.
Eigenvectors, especially when used in combination with their corresponding eigenvalues, provide crucial insight into the behavior of dynamic systems modelled by differential equations.
Eigenvalues
Eigenvalues are another fundamental part of understanding how linear systems behave over time. They are scalars that provide the factor by which the associated eigenvector is scaled during a linear transformation. To find the eigenvalues of a matrix, you solve the characteristic equation \(\det(A - \lambda I) = 0\). This exercise tells us the characteristic polynomial is \((\lambda-1)(\lambda-2)^3\).
In our exercise, the eigenvalues \(\lambda = 1\) and \(\lambda = 2\) are significant because they help us determine the long-term behavior of solutions to the system \( \mathbf{x}^{\prime}=A \mathbf{x} \). The eigenvalue \(\lambda = 1\) has an algebraic multiplicity of 1, while \(\lambda = 2\) has an algebraic multiplicity of 3, which signifies the complexity and potential for multiple solutions related to this eigenvalue.
By identifying eigenvalues, we can predict the stability and type of behavior the system will exhibit, such as whether solutions will converge, diverge, or oscillate.
In our exercise, the eigenvalues \(\lambda = 1\) and \(\lambda = 2\) are significant because they help us determine the long-term behavior of solutions to the system \( \mathbf{x}^{\prime}=A \mathbf{x} \). The eigenvalue \(\lambda = 1\) has an algebraic multiplicity of 1, while \(\lambda = 2\) has an algebraic multiplicity of 3, which signifies the complexity and potential for multiple solutions related to this eigenvalue.
By identifying eigenvalues, we can predict the stability and type of behavior the system will exhibit, such as whether solutions will converge, diverge, or oscillate.
General Solution
The general solution of a differential equation is a combination of several particular solutions and is crucial for understanding the overall behavior of the system. In the given exercise, we are tasked with finding linearly independent solutions to express the general solution of the system \( \mathbf{x}^{\prime}=A \mathbf{x} \).
The general solution formula \(\mathbf{x}(t) = c_1 e^{1t}\mathbf{v_1} + c_2 e^{2t}\mathbf{v_2} + c_3 t e^{2t}\mathbf{v_3} + c_4 t^2 e^{2t}\mathbf{v_4}\) ,
tells us how the system behaves depending on initial conditions provided. This includes contributions from the exponential solutions given by eigenvectors and their associated eigenvalues, and from generalized eigenvectors due to the repeated eigenvalue \(\lambda=2\).
Each term represents different behaviors or modes of the system, such as exponential growth or decay, and the presence of terms with \(t\) and \(t^2\) indicate more complex interactions because of higher algebraic multiplicity. Understanding the general solution allows one to tailor this model to specific conditions by plugging in the initial values, leading to practical applications in predicting real-world systems.
The general solution formula \(\mathbf{x}(t) = c_1 e^{1t}\mathbf{v_1} + c_2 e^{2t}\mathbf{v_2} + c_3 t e^{2t}\mathbf{v_3} + c_4 t^2 e^{2t}\mathbf{v_4}\) ,
tells us how the system behaves depending on initial conditions provided. This includes contributions from the exponential solutions given by eigenvectors and their associated eigenvalues, and from generalized eigenvectors due to the repeated eigenvalue \(\lambda=2\).
Each term represents different behaviors or modes of the system, such as exponential growth or decay, and the presence of terms with \(t\) and \(t^2\) indicate more complex interactions because of higher algebraic multiplicity. Understanding the general solution allows one to tailor this model to specific conditions by plugging in the initial values, leading to practical applications in predicting real-world systems.
Other exercises in this chapter
Problem 9
Solve the given initial-value problem. $$x_{1}^{\prime}=2 x_{2}, \quad x_{2}^{\prime}=x_{1}+x_{2}, \quad x_{1}(0)=3, \quad x_{2}(0)=0$$
View solution Problem 10
Determine the general solution to the linear system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\). $$\left[\begin{array}{rr} -3 & -10 \\ 5 & 1
View solution Problem 10
Determine the general solution to the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\) $$\left[\begin{array}{rll} 3 & 1 & 0 \\ -1 & 5 & 0
View solution Problem 10
Characterize the equilibrium point for the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) and sketch the phase portrait. $$A=\left[\begin{array}{rr} -2 & 1 \\ 1 &
View solution