Consider the linear autonomous first-order systems i(t) = Ax(t), x(0) x(0) = xo 3 4 4 6 6 1 -12 -1 -6 14 -1 2 -8 -4 -4 w

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answerhappygod
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Consider the linear autonomous first-order systems i(t) = Ax(t), x(0) x(0) = xo 3 4 4 6 6 1 -12 -1 -6 14 -1 2 -8 -4 -4 w

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Consider The Linear Autonomous First Order Systems I T Ax T X 0 X 0 Xo 3 4 4 6 6 1 12 1 6 14 1 2 8 4 4 W 1
Consider The Linear Autonomous First Order Systems I T Ax T X 0 X 0 Xo 3 4 4 6 6 1 12 1 6 14 1 2 8 4 4 W 1 (19.59 KiB) Viewed 67 times
Consider The Linear Autonomous First Order Systems I T Ax T X 0 X 0 Xo 3 4 4 6 6 1 12 1 6 14 1 2 8 4 4 W 2
Consider The Linear Autonomous First Order Systems I T Ax T X 0 X 0 Xo 3 4 4 6 6 1 12 1 6 14 1 2 8 4 4 W 2 (62.13 KiB) Viewed 67 times
Consider the linear autonomous first-order systems i(t) = Ax(t), x(0) x(0) = xo 3 4 4 6 6 1 -12 -1 -6 14 -1 2 -8 -4 -4 with A = -6 -1 -4 4 3 0 5 -12 15 -5 and 4 -5 -8 0 Xo = 2 -5
E. This is now the main part of the question: Write down the solution to the differential equation i(t) = Ax(t), x(0) = xo along the lines of Equation (3.35) with special cases in Equations (3.36) and (3.41) of the book, or in the corresponding lecture notes. (You will have to generalize the method from 2 × 2 matrices to 5 × 5 matrices, and you might have to do some simple matrix-vector multiplications either by hand or by MATLAB.) You have to write down the vector-valued solution (t) component- by-component. ● F. What is the stability property of the origin? Is the origin source (solutions get away from the origin as t→∞); sink (solutions go toward the origin as t→ ∞); saddle (some solutions get away from, while others go toward the origin as t→∞); or center (solutions are going around the origin cyclically)? Explain your answer.
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