51.Using the Law of Exponential Change in Section 6.4 withappropriate changes of variables, the solution to thedifferential equation is L(x)5L0e2kx, where L05L(0) isthe surface intensity. We know 0.55e218k, so k5}l2n 01.85}and our equation becomes L(x)5L0e(ln 0.5)(x/18)5L01}12}2x/18. We now find the depthwhere the intensity is one-tenth of the surface value.0.151}12}2x/18ln 0.15}1x8}ln 1}12}2x5}18lnln0.05.1}<59.8 ftYou can work without artificial light to a depth of about59.8 feet.52. (a)}ddyt}5}kVA}(c2y)E}cd2yy}5E}kVA}dt2ln )c2y)5}kVA}t1Cln )c2y)5 2}kVA}t2C)c2y)5e2(kA/V)t2Cc2y5 6e2(kA/V)t2Cy5c 6e2(kA/V)t2Cy5c1De2(kA/V)tInitial condition y5y0when t50y05c1Dy02c5DSolution:y5c1(y02c)e2(kA/V)t(b)limt→‘y(t)5limt→‘[c1(y02c)e2(kA/V)t]5c53. (a)P(t)5}1115e04.32t}5}111e540.3e2t}This is P5}11MAe2kt}where M5150,A5e4.3, and k51. Therefore, it is a solution of the logisticdifferential equation.
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