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Unformatted text preview: 18.03 Problem Set 1: Part II Solutions 0. (T 5 Feb) Part I problems: 12 pts. (a) [4] ˙ x = k x/ ( t + 1). (b) [4] Separate: dx/x = k dt/ ( t + 1). Integrate: ln  x  + c 1 = k ln  t + 1  + c 2 = ln(  t + 1  k ) + c 2 . Amalgamate constants and exponentiate:  x  = e c  t + 1  k . Eliminate the absolute value: x = C  t + 1  k , where C = ± e c . Reintroduce the solution we lost by dividing by x in the first step: allow C = 0. So the general solution is x = C  t + 1  k . (c) [4] If x (0) > 0 then C > 0. The function C  t + 1  k grows without bound: for any number L > C we can solve for L = C  t + 1  k :  t + 1  = ( L/C ) 1 /k . [The function C  t + 1  k grows more slowly than does Ae kt for any k > 0, however: the virus causes exponential growth to be replaced by a slower growth rate.] 1. (W 6 Feb) Part I problems: 6 pts. (a) [3] 2in (b) [3] 0 . 72 < y < . 74. (c) [3] The curve leaves the window at approximately x = 2 . 10....
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 Spring '09
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 Constant of integration, Part II Solutions, xk yk Ak, steps. yk Ak, yk − yk

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