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**Unformatted text preview: **Review Problems
that is, x1 = x2 , x2 = x3 , 1 5 + et x1 − 2x2 . x3 = 3 11. This system is equivalent to x =t−y −y , y =x −x . Next, we introduce, as additional unknowns, derivatives of x(t) and y (t): x1 (t) := x(t), x4 (t) := y (t), With new variables, the system becomes x = (x ) =: x3 = t − y − y =: t − x5 − x6 , y = (y ) =: x6 = x − x =: x2 − x3 . Also, we have four new equations connecting xj ’s: x1 = x =: x2 , x2 = (x ) = x =: x3 , x4 = y =: x5 , x5 = (y ) = y =: x6 . Therefore, the answer is x1 = x2 , x2 = x3 , x3 = t − x5 − x6 , 337 x2 (t) := x (t), x5 (t) := y (t), x3 (t) := x (t), x6 (t) := y (t). ...

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