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ps2sol

# ps2sol - 18.03 Problem Set 2 Solutions 4(W 15 Feb[Part I...

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18.03 Problem Set 2: Solutions 4. (W 15 Feb) [Part I problems: 6 pts.] (a) [4] x decreases exponentially to zero. y grows linearly (with slope 1 / 2) at first, then levels off and finally decays exponentially to zero. z grows quadratically at first, then levels off and converges exponentially to value 1 as t → ∞ . (b) [5] The decay constant l of an exponential decay is related to half life t 1 / 2 by 1 2 = e - lt 1 / 2 , or ln 2 = lt 1 / 2 . Thus the decay constant of Lu is 1 2 . I’ll write l for the decay constant of Lu because the formulas are clearer with this symbol than with 1 2 . So ˙ x + lx = 0, ˙ y + ky = lx , ˙ z = ky . (c) [5] x = x (0) e - lt . Using x (0) = 1 we find x = e - lt . Thus y + ky = le - lt . Solve this using the integrating factor e kt : d dt ( e kt y ) = le ( k - l ) t , so e kt y = le ( k - l ) t / ( k - l )+ c (assuming k 6 = l , at least), or y = le - lt / ( k - l ) + ce - kt . To find c , evaluate at t = 0: 0 = l k - l + c , so y = l e - lt - e - kt k - l . Thus z = kl k - l e - lt - l - e - kt - k + c . To find the constant of integration here, evaluate at t = 0 again: 0 = kl k - l ( 1 - l - 1 - k ) + c , or c = 1: z = kl k - l e - kt k - e - lt l + 1.

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