121616949-math.244 - current population reproduces at a...

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230 Chapter 9 Applications of Integration For the non-constant solutions we use separation of variables. Note that ˙ y dt = dy dt dt = dy . Now ˙ y = 2(25 - y ) Z ˙ y 2(25 - y ) dt = Z 1 dt 1 2 Z 1 25 - y dy = Z 1 dt 1 2 ln | 25 - y | ( - 1) = t + C 0 ln | 25 - y | = - 2 t + ( - 2) C 0 = - 2 t + C | 25 - y | = e - 2 t e C y - 25 = ± e - 2 t e C y = 25 ± e - 2 t e C = 25 + Ae - 2 t . Finally, using the initial value, 40 = y (0) = 25 + Ae 0 15 = A, and so y = 25 + 15 e + - 2 t . Remember that y = 25 is also a solution. In the derivation, A = ± e C = 0, but if we allow A = 0 in the final solution, y = 25 + Ae - 2 t represents all solutions to the differential equation. EXAMPLE 9.47 Consider the differential equation ˙ y = ky . When k > 0, this describes certain simple cases of population growth: it says that the change in the population y is proportional to the population. The underlying assumption is that each organism in the
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Unformatted text preview: current population reproduces at a fixed rate, so the larger the population the more new organisms are produced. While this is too simple to model most real populations, it is useful in some cases over a limited time. When k < 0, the differential equation describes a quantity that decreases in proportion to the current value; this can be used to model radioactive decay. The constant solution is y ( t ) = 0; of course this will not be the solution to any interesting initial value problem. For the non-constant solutions, we proceed much as...
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