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ModelingFirstOrderEq

# ModelingFirstOrderEq - Modeling First Order O.D.Es 1 Rate...

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Modeling First Order O.D.E’s 1- Rate of Growth and Decay The I.V.P. (1) dx dt = kx x(t 0 ) = x 0 ! " # \$ # where k is a constant, occurs in many physical sciences models that involve growth or decay. For example, in biology, it is often observed that the rate at which certain bacteria grow is proportional to the number of bacteria present at any time. Over short intervals of time the population of small animals, such as rodents, can be predicted fairly accurately by the solution of the I.P.V. (1). The constant k can be determined from the solution of the differential equation by using a subsequent measurement of the population at time t > t 0 . In physics I.V.P.(1) provides a model for approximating the remaining amount of a substance which is disintegrating through radioactivity. Also I.V.P. (1) determines the temperature in a cooling body. In chemistry the amount of a substance remaining during certain reaction is also described by I.V.P. (1). Example : 1) A culture initially has N 0 number of bacteria. At time t = 1 hour the number of bacteria is measured to be 3/2 N 0 . If the rate of growth is proportional to the number of bacteria present at any time, determine the time necessary for the number of bacteria to triple. Solution : Let’s write the I.V.P. that models the problem: dN dt = kN N(0) = N 0 ! " # \$ # First solve the first order equation dN dt = kN it is separable, and can be expressed as dN N = kt integrating, we get ln N = kt + c, where N > 0 or N(t) = ce kt Using the initial condition: N(0) = N 0 = ce 0 = c then N(t) = N 0 e kt Since at time t = 1 hour the population is 3/2N 0 ,

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then N(1) = 3/2 N 0 = N 0 e k or 3/2 = e k and solving for k, k = ln(3/2) = 0.4055 Therefore, N(t) = N 0 e 0.4055t Now, we must find the time t when the number of bacteria triple the original number. 3N 0 = N 0 e 0.4055t 3 = e 0.4055t taking natural logarithm both side ln3 = 0.4055t t = ln3 0.4055 = 2.71 hours Remark : The function e kt increases as t increases if k > 0 and decreases as t increases if k < 0. Thus, problems describing growth, such as population, bacteria, or even capital
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