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Flashcards: Homogeneous Linear Systems

Solve the initial value problem $\mathbf{x'} = \begin{bmatrix} 3&-2 \\ -1&2 \end{bmatrix}\mathbf{x}$ . Where $\mathbf{x}_0 = \begin{bmatrix} 1\\-1 \end{bmatrix}$

$\mathbf{x} = e^t\begin{bmatrix} 3\\1 \end{bmatrix} +e^{4t}\begin{bmatrix} 5\\6 \end{bmatrix}$

$\mathbf{x} = e^{-t}\begin{bmatrix} -2\\1 \end{bmatrix} +e^{3t}\begin{bmatrix} 2\\2 \end{bmatrix}$

$\mathbf{x} = e^{-t}\begin{bmatrix} 5\\2 \end{bmatrix} +e^{3t}\begin{bmatrix} -1\\3 \end{bmatrix}$

$\mathbf{x} = e^t\begin{bmatrix} 1\\4 \end{bmatrix} +e^{4t}\begin{bmatrix} 5\\2 \end{bmatrix}$

$\mathbf{x} = e^t\begin{bmatrix} -3\\-3 \end{bmatrix} +e^{4t}\begin{bmatrix} 4\\2 \end{bmatrix}$

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Flashcards: Homogeneous Linear Systems

Solve the initial value problem $\mathbf{x'} = \begin{bmatrix} 3&-2 \\ -1&2 \end{bmatrix}\mathbf{x}$ . Where $\mathbf{x}_0 = \begin{bmatrix} 1\\-1 \end{bmatrix}$

Your answer:

$\mathbf{x} = e^t\begin{bmatrix} 3\\1 \end{bmatrix} +e^{4t}\begin{bmatrix} 5\\6 \end{bmatrix}$

Your answer:

$\mathbf{x} = e^{-t}\begin{bmatrix} -2\\1 \end{bmatrix} +e^{3t}\begin{bmatrix} 2\\2 \end{bmatrix}$

Your answer:

$\mathbf{x} = e^{-t}\begin{bmatrix} 5\\2 \end{bmatrix} +e^{3t}\begin{bmatrix} -1\\3 \end{bmatrix}$

Your answer:

$\mathbf{x} = e^t\begin{bmatrix} 1\\4 \end{bmatrix} +e^{4t}\begin{bmatrix} 5\\2 \end{bmatrix}$

Correct answer:

$\mathbf{x} = e^t\begin{bmatrix} -3\\-3 \end{bmatrix} +e^{4t}\begin{bmatrix} 4\\2 \end{bmatrix}$

Explanation:

To solve the homogeneous system, we will need a fundamental matrix. Specifically, it will help to get the matrix exponential. To do this, we will diagonalize the matrix. First, we will find the eigenvalues which we can do by calculating the determinant of $\lambda I - A$ .

$|\lambda I - A| = \begin{vmatrix} \lambda - 3&2 \\ 1&\lambda - 2 \end{vmatrix} = (\lambda -3)(\lambda -2) - 2 = \lambda^2 - 5\lambda +4 = (\lambda -4)(\lambda -1)$

Finding the eigenspaces, for lambda = 1, we have

$\begin{bmatrix} -2&2 &\vline &0 \\ 1&-1 &\vline & 0 \end{bmatrix}$

Adding -1/2 Row 1 to Row 2 and dividing by -1/2, we have $\begin{bmatrix} 1&-1 &\vline &0 \\ 0&0 &\vline & 0 \end{bmatrix}$ which means x_1 + x_2 = 0

Thus, we have an eigenvector of $\begin{bmatrix} 1\\1 \end{bmatrix}$ .

For lambda = 4

$\begin{bmatrix} 1&-2 &\vline &0 \\ -1&2 &\vline & 0 \end{bmatrix}$

Adding Row 1 to Row 2, we have

$\begin{bmatrix} 1&-2 &\vline &0 \\ 0&0 &\vline & 0 \end{bmatrix}$

So x_1 - 2x_2 = 0 with an eigenvector $\begin{bmatrix} 2\\1 \end{bmatrix}$ .

Thus, we have $P = \begin{bmatrix} 1&2 \\ 1&1 \end{bmatrix}$ and $D = \begin{bmatrix} 1&0 \\ 0&4 \end{bmatrix}$ . Using the inverse formula for 2x2 matrices, we have that $P^{-1} = \begin{bmatrix} -1&2 \\ 1&-1 \end{bmatrix}$ . As we know that $e^{tA} = Pe^{tD}P^{-1}$ , we have $e^{tA} = \begin{bmatrix} 1&2 \\ 1&1 \end{bmatrix}\begin{bmatrix} e^t&0 \\ 0&e^{4t} \end{bmatrix}\begin{bmatrix} -1&2 \\ 1&-1 \end{bmatrix}= \begin{bmatrix} -e^t + 2e^{4t}&2e^{t} -2e^{4t} \\ -e^t + e^{4t}&2e^t-e^{4t} \end{bmatrix}$

The solution to a homogenous system of linear equations is simply to multiply the matrix exponential by the intial condition. For other fundamental matrices, the matrix inverse is needed as well.

Thus, our final answer is $\begin{bmatrix} -e^t + 2e^{4t}&2e^{t} -2e^{4t} \\ -e^t + e^{4t}&2e^t-e^{4t} \end{bmatrix}\begin{bmatrix} 1\\-1 \end{bmatrix} = \begin{bmatrix} -3e^t + 4e^{4t}\\-3e^t + 2e^{4t} \end{bmatrix}= e^t\begin{bmatrix} -3\\-3 \end{bmatrix} +e^{4t}\begin{bmatrix} 4\\2 \end{bmatrix}$

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All Differential Equations Resources

Flashcards: Homogeneous Linear Systems

Flashcards: Homogeneous Linear Systems

Flashcards: Homogeneous Linear Systems

Flashcards: Homogeneous Linear Systems

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All Differential Equations Resources

Flashcards: Homogeneous Linear Systems

Flashcards: Homogeneous Linear Systems

Flashcards: Homogeneous Linear Systems

Flashcards: Homogeneous Linear Systems

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