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**Unformatted text preview: **t 3 + (-1 + 2 3 a ) te-2 t 3 (2) y ( t ) contains two terms and the rst term is ae-2 t 3 and the second term is (-1 + 2 3 a ) te-2 t 3 . And we also know that the rst term is positive as long as a > 0 and second term is positive as long as-1 + 2 3 a > a > 3 2 . Hence we know that y ( t ) will be positive as long as a > 0 and a > 3 2 So the critical value is a = 0 or a = 3 2 . But for this problem when t is really big, the second term will dominates the solution. So the critical value is a = 3 2 ....

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