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The matrix \(A\) has complex eigenvalues. Find a fundamental set of real solutions of the system \(\mathbf{y}^{\prime}=A \mathbf{y}\). \(A=\left(\begin{array}{rr}-1 & 1 \\ -5 & -5\end{array}\right)\)

Short Answer

Expert verified
The fundamental set of real solutions is \(\{e^{-3t} (\cos t, -2\cos t + \sin t), e^{-3t} (\sin t, -2\sin t - \cos t)\}\).

Step by step solution

01

Find the Eigenvalues

To find the eigenvalues of the matrix \(A\), we solve the characteristic equation \(\det(A - \lambda I) = 0\). For \(A = \begin{pmatrix} -1 & 1 \ -5 & -5 \end{pmatrix}\), the characteristic equation is \(|-1-\lambda & 1| ewline |-5 & -5-\lambda|\). This leads to the equation \((\lambda + 1)(\lambda + 5) + 5 = 0\), simplifying to \(\lambda^2 + 6\lambda + 10 = 0\). Using the quadratic formula, the eigenvalues are \(\lambda = -3 \pm i\).
02

Determine the Eigenvectors

Next, we find the eigenvectors corresponding to the eigenvalues \(-3 + i\) and \(-3 - i\). Substituting \(\lambda = -3 + i\) into \((A - \lambda I)\mathbf{v} = 0\), we simplify \((-1 + 3 - i & 1)|( -5 & -5 + 3 - i)|\) to \((2 - i & 1 ewline -5 & -8 - i)\). Solving the first row, \((2 - i)x + y = 0\), gives \(y = -(2 - i)x\). Choosing \(x = 1\), we have \(\mathbf{v} = \begin{pmatrix} 1 \ -(2 - i) \end{pmatrix}\). Similarly, \(\mathbf{v}_2\) for \(\lambda = -3 - i\) will be the complex conjugate, \(\mathbf{v}_2 = \begin{pmatrix} 1 \ -(2 + i) \end{pmatrix}\).
03

Construct the General Solution

The general solution uses the eigenvectors and eigenvalues to create real solutions. For the eigenvalue \(-3 + i\), the associated real solution vector involves \(e^{-3t}\)(\(\cos t + i\sin t\)) applied to the eigenvector. Writing this as \(e^{-3t}(\cos t \mathbf{v}_1^R - \sin t \mathbf{v}_1^I + i(\sin t \mathbf{v}_1^R + \cos t \mathbf{v}_1^I))\), where \(\mathbf{v}_1^R\) and \(\mathbf{v}_1^I\) are the real and imaginary parts of \(\mathbf{v}_1 = \begin{pmatrix} 1 \ -2 + i \end{pmatrix}\). This yields two fundamental solutions: \(\mathbf{y}_1 = e^{-3t} \begin{pmatrix} \cos t \ -2\cos t + \sin t \end{pmatrix}\) and \(\mathbf{y}_2 = e^{-3t} \begin{pmatrix} \sin t \ -2\sin t - \cos t \end{pmatrix}\).
04

Verify Independence

Check that \(\mathbf{y}_1\) and \(\mathbf{y}_2\) are linearly independent by calculating their Wronskian. Compute \(W(\mathbf{y}_1, \mathbf{y}_2) = \det \begin{pmatrix} e^{-3t} \cos t & e^{-3t} \sin t \ -2e^{-3t} \cos t + e^{-3t} \sin t & -2e^{-3t} \sin t - e^{-3t} \cos t \end{pmatrix}\). Simplifying, \(W\) reduces to \(e^{-6t}\), which is not zero, confirming independence.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Real Solutions
When dealing with complex eigenvalues in a differential equation, it's essential to express solutions in terms of real numbers. Starting with complex numbers can be intimidating, but rest assured, it's a standard procedure to transform these into real solutions. In the exercise, we faced eigenvalues of \( \lambda = -3 \pm i \). These lead to exponential expressions involving both sine and cosine functions after using Euler's formula. This way, we construct real solutions, which can be more intuitive and easier to interpret in the real world. The key takeaway is understanding that even with complex numbers, we aim to express our results in real forms.
Eigenvectors
Eigenvectors play a critical role when it comes to finding solutions for a system of differential equations. Once eigenvalues are found, eigenvectors help to shape the particular solutions of the differential equation system. In our case, for the eigenvalues \( \lambda = -3 + i \) and \( \lambda = -3 - i \), we derived complex eigenvectors. These vectors are inherently tied to the system of equations we're solving, and their real and imaginary parts directly contribute to forming the basis of the solution space. Recognizing the importance of choosing a suitable eigenvector is key to accurately describing the differential system.
General Solution
The general solution to a differential equation system uses both the eigenvector and eigenvalue information derived earlier. Combining these insights, the exercise's general solution involved forming two real-valued functions from the complex eigen pair. By applying trigonometric identities and combining real and imaginary parts of eigenvectors, we rendered pairs of sinusoidal functions multiplied by exponential decays, specifically: \( \mathbf{y}_1 = e^{-3t} \begin{pmatrix} \cos t \ -2\cos t + \sin t \end{pmatrix} \) and \( \mathbf{y}_2 = e^{-3t} \begin{pmatrix} \sin t \ -2\sin t - \cos t \end{pmatrix} \). These represent the fundamental real solutions satisfying the system, and together, they characterize the comprehensive behavior of the system over time.
Linear Independence
Verifying linear independence is the final key step when confirming our solutions are distinct and essential. If solutions are not linearly independent, they can't span the solution space properly, leading to incomplete results. In our context, using the Wronskian determinant ensures this independence. Here, the computed Wronskian, \(W = e^{-6t}\), stayed non-zero, validating that our constructed solutions, \( \mathbf{y}_1 \) and \( \mathbf{y}_2 \), do indeed form a valid basis for the solution space. This means they don't overlap in a way that would cause them to be redundant, hence capturing all potential solution behaviors.

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Most popular questions from this chapter

Find the general solution of the system \(\mathbf{y}^{\prime}=A \mathbf{y}\) for the given matrix \(A\). \(A=\left(\begin{array}{ll}-10 & 4 \\ -12 & 4\end{array}\right)\)

Consider the system $$ \mathbf{y}^{\prime}=\left(\begin{array}{rrr} 1 & -1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right) \mathbf{y} . $$ (a) Find the eigenvalues and eigenvectors. Use your numerical solver to sketch the half-lines generated by the exponential solutions \(\mathbf{y}(t)=C_{i} e^{\lambda_{i} t} \mathbf{v}_{i}\) for \(i=1,2\), and 3 . (b) The six half-line solutions found in part (a) come in pairs that form straight lines. These three lines, taken two at a time, generate three planes. In turn these planes divide phase space into eight octants. Use your numerical solver to add solution trajectories with initial conditions in each of the eight octants. (c) Based on your phase portrait, what would be an appropriate name for the equilibrium point in this case?

Verify that the equilibrium point at the origin is a center by showing that the real parts of the system's complex eigenvalues are zero. In each case, calculate and sketch the vector generated by the right-hand side of the system at the point \((1,0)\). Use this to help sketch the elliptic solution trajectory for the system passing through the point \((1,0)\). Draw arrows on the solution, indicating the direction of motion. Use your numerical solver to check your result. \(\mathbf{y}^{\prime}=\left(\begin{array}{rr}0 & 1 \\ -4 & 0\end{array}\right) \mathbf{y}\)

For each of the matrices, perform the following activities. (i) Determine where in the trace-determinate plane the system \(\mathbf{y}^{\prime}=A \mathbf{y}\) fits. (ii) Find all of the equilibrium points for the system \(\mathbf{y}^{\prime}=A \mathbf{y}\), and plot them in the phase plane. (iii) Find the general solution and use the result to create a phase portrait for this system, taking care to sketch enough solution curves to show what happens in every part of the phase plane.\(A=\left(\begin{array}{rr}2 & 1 \\\ -10 & -5\end{array}\right)\)

For the matrices, use a computer to help find a fundamental set of solutions to the system \(\mathbf{y}^{\prime}=A \mathbf{y}\). \(A=\left(\begin{array}{rrr}8 & 12 & -4 \\ -9 & -13 & 4 \\ -1 & -3 & 0\end{array}\right)\)

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