ps2sol

Ps2sol - 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

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Unformatted text preview: 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,....
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This note was uploaded on 02/06/2009 for the course 18 18.03 taught by Professor Unknown during the Spring '09 term at MIT.

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Ps2sol - 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

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