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(Solved): Problem 3: Consider the Linear Systems dtdY=AY where Y=(y1y2) and A is given by: (24 ...



Problem 3: Consider the Linear Systems \( \frac{d \vec{Y}}{d t}=A \vec{Y} \) where \( \vec{Y}=\left(\begin{array}{l}y_{1} \\
Problem 3: Consider the Linear Systems where and is given by: (a) Determine the type of the system, i.e., sink (node), source, saddle, spiral source, spiral sink, center. (b) Draw the phase portrait of the system. If the eigenvalues are real you need to compute the eigenvectors and indicate them clearly on the phase portrait. (c) Compute the General Solution of


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In order to determine the type of the system described by the matrix A, we need to analyze its eigenvalues.

The matrix A is   

To find the eigenvalues, we solve the characteristic equation: det(A - ?I) = 0 where ? is the eigenvalue and I is the identity matrix. The characteristic equation becomes:    Expanding the determinant, we have: (-2-?)(-2-?) - (4)(4) = 0 Simplifying further: (?+2)(?+2) - 16 = 0 (?+2)² - 16 = 0 Expanding and rearranging: ?² + 4? + 4 - 16 = 0 ?² + 4? - 12 = 0 Factoring the quadratic equation: (? - 2)(? + 6) = 0 This equation has two eigenvalues: ? = 2 and ? = -6.

Now, let's analyze the types of the system based on the eigenvalues: 1. If both eigenvalues have negative real parts, the system is a sink (node). 2. If both eigenvalues have positive real parts, the system is a source. 3. If the eigenvalues have opposite signs, the system is a saddle. 4. If the eigenvalues have positive real parts and a non-zero imaginary part, the system is a spiral source. 5. If the eigenvalues have negative real parts and a non-zero imaginary part, the system is a spiral sink. 6. If the eigenvalues have zero real parts, the system is a center.

For the eigenvalues ? = 2 and ? = -6, both eigenvalues have negative real parts.
Therefore, the system is a saddle.
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