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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)  ˙ x = k x/ ( t + 1). (b)  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)  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)  2in (b)  0 . 72 < y < . 74. (c)  The curve leaves the window at approximately x =- 2 . 10....
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- Spring '09
- Constant of integration, Part II Solutions, xk yk Ak, steps. yk Ak, yk − yk