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Chapter 9: Problem 33

\(\mathbf{x}^{\prime}(t)=\left[ \begin{array}{rrr}{1} & {-2} & {2} \\ {-2} & {1} & {-2} \\ {2} & {-2} & {1}\end{array}\right] \mathbf{x}(t), \quad \mathbf{x}(0)=\left[ \begin{array}{r}{-2} \\ {-3} \\ {2}\end{array}\right]\)

Short Answer

Expert verified
The solution of the system is given by \(\mathbf{x}(t) = e^{At} \mathbf{x}(0)\), where \(e^{At}\) is computed as \(e^{At} = Pe^{Dt}P^{-1}\) using the eigenvalues and eigenvectors of A, and \(\mathbf{x}(0)\) is the given initial condition.

Step by step solution

01

Compute the eigenvalues of the matrix A

The eigenvalues of the matrix A can be found by solving the characteristic equation, which is given by \(det(A - \lambda I) = 0\). So, the first step is to compute the determinant of the matrix \(\left[ \begin{array}{rrr}{1-\lambda} & {-2} & {2} \ {-2} & {1-\lambda} & {-2} \ {2} & {-2} & {1-\lambda}\end{array}\right]\) and solve it for \(\lambda\).
02

Compute the eigenvectors associated with each eigenvalue

Once the eigenvalues are found, the next step is to find the corresponding eigenvectors by solving the homogenous system of equations \((A - \lambda I)v = 0\), where \(v\) are the eigenvectors. For each eigenvalue, you substitute it into the equation and find the corresponding eigenvector.
03

Diagonalize the matrix A

We can diagonalize the matrix A as \(A=PDP^{-1}\), where D is a diagonal matrix whose diagonal elements are the eigenvalues of A and P is the matrix whose columns are the eigenvectors of A. So, the third step is to form the matrices P and D using the obtained eigenvalues and eigenvectors, and then compute the inverse of P.
04

Compute the matrix exponential

The matrix exponential \(e^{At}\) can be computed as \(e^{At} = Pe^{Dt}P^{-1}\), where \(e^{Dt}\) is the matrix with \(e^{\lambda t}\) on the diagonal. To compute \(e^{Dt}\), we get the matrix with the elements \(\exp{(\lambda_i t)}\) on the diagonal, where \(\lambda_i\) are the eigenvalues of A.
05

Compute the solution vector

Now that we have \(e^{At}\), we can compute \(\mathbf{x}(t) = e^{At} \mathbf{x}(0)\), which is the solution of the system.

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Most popular questions from this chapter

\(\mathbf{X}(t)=\left[ \begin{array}{ccc}{e^{t}} & {e^{-t}} & {e^{2 t}} \\\ {e^{t}} & {-e^{-t}} & {2 e^{2 t}} \\ {e^{t}} & {e^{-t}} & {4 e^{2 t}}\end{array}\right]\)

\(\mathbf{A}=\left[ \begin{array}{rrr}{0} & {1} & {0} \\ {0} & {0} & {1} \\\ {1} & {-1} & {1}\end{array}\right]\)

For the coupled mass-spring system governed by system \((10),\) assume \(m_{1}=m_{2}=1 \mathrm{kg}, k_{1}=k_{2}=2 \mathrm{N} / \mathrm{m},\) and \(k_{3}=3 \mathrm{N} / \mathrm{m} .\) Determine the normal frequencies for this coupled mass-spring system.

\(\mathbf{x}^{\prime}(t)=\left[ \begin{array}{rrr}{1} & {-2} & {2} \\ {-2} & {1} & {-2} \\ {2} & {-2} & {1}\end{array}\right] \mathbf{x}(t), \quad \mathbf{x}(0)=\left[ \begin{array}{r}{-2} \\ {-3} \\ {2}\end{array}\right]\)

Prove that if \(\mathbf{x}_{p}\) satisfies \(\mathbf{A x}_{p}=\mathbf{b},\) then every solution to the nonhomogeneous system \(\mathbf{A x}=\mathbf{b}\) is of the form \(\mathbf{x}=\mathbf{x}_{p}+\mathbf{x}_{h},\) where \(\mathbf{x}_{h}\) is a solution to the corresponding homogeneous system \(\mathbf{A} \mathbf{x}=\mathbf{0}\) .
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