1 1 2 0 2 2 x; x(0) = | 0 -1 1 3 6. The solution of the initial-value problem x' = is: (a) x = e* 2 +e 2 1 -1 (b) x = 2e2 1 -1 2 (c) x = 2e3t 2 - 3e2t +et 2 1 (d) x= 3e" | 1 3t 2 (e) None of the above.

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