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26 September, 22:36

Find the solution of the initial value problem ˙~x = F~ (~x) with ~x (0) = ~x0 for F~ (~x) = - µx1 - x2 + x 2 1, ~x = x1 x2, ~x0 = a b for arbitrarily chosen (a, b, µ) ∈ R 3. Is the solution ~u (t; a, b, µ) a continuous function of all variables (t, a, b, µ) ∈ R 4? Give an interval t ∈ [-a, a] and an open neighbourhood (a, b, µ) ∈ U ⊂ R 3 such that ~u ∈ C ([-a, a] * U, R 3).

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  1. 26 September, 22:55
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    Answer:F is the answer to the question
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