Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 1 - First-Order Differential Equations - 1.12 Chapter Review - Additional Problems - Page 111: 54

Answer

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Work Step by Step

Given: $P(t_0)=P_0=200,000\\P(t_1)=P_1=230,000\\P(t_2)=P_2=250,000$ The population growth can be expressed as: $\frac{dP}{dt}=(B_0-D_0P)P\\P_0=P(0)$ The logistic model of population is given as: $P(t)=\frac{CP_0}{P_0+(C-P_0)e^{-rt}}$ To find r, we form: $r=\frac{1}{t_1}\ln[\frac{P_2(P_1-P_0)}{P_0(P_2-P_1)}]\\ =\frac{1}{3}\ln[\frac{250,000(230,000-200,000)}{200,000(250,000-230,000)}]\\=\frac{1}{3}\ln[\frac{15}{8} \approx 0.21$ For $C$: $C=\frac{P_1[(P_0+P_2)-2P_0P_2]}{P_1^2-P_0P_2}\\=\frac{230,000[230,000(200,000+250,000)-2.200,000.250,000]}{230,000^2-200,000.250,000}\\=277,586$ Substitute back, we have $P(t)=\frac{277,586.200,000}{200,000+(277,586-200,000)e^{-0.21t}}$ To predict the population in 2018: $P(t)=\frac{277,586.200,000}{200,000+(277,586-200,000)e^{-0.21\times10}}\approx264,997$ The population in 2028 may be: $P(t)=\frac{277,586.200,000}{200,000+(277,586-200,000)e^{-0.21 \times 20}}\approx275,981$
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