This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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,....
View
Full
Document
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.
 Spring '09
 unknown

Click to edit the document details