A model for the population P(t) in a suburb of a large city is given by the initial-value problem dPdt=P(10−1−10−7P), P(0)=5000 where t is measured in months. What is the limiting value of the population? At what time will the population be equal to one-half of this limiting value?

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A model for the population P(t) in a suburb of a large city is given by the initial-value problem  dPdt=P(10−1−10−7P), P(0)=5000  where t is measured in months. What is the limiting value of the population? At what time will the population be equal to one-half of this limiting value?

George Woody · Accepted Answer

∴dP&amp;nbsp;dt&amp;nbsp;=P(10−1−10−7&amp;nbsp;P)→dPP(10−1−10−7&amp;nbsp;P}=&amp;nbsp;dt&amp;nbsp;&amp;nbsp;∴∫dPN(10−1−10−7&amp;nbsp;P}=∫&amp;nbsp;dt&amp;nbsp;&amp;nbsp;We shall use partial fractions to solve the L.H.S integral.As a result of employing the cover-up rule, we obtain:1P(10−1−10−7&amp;nbsp;P}=10P+10−610−1−10−7&amp;nbsp;P&amp;nbsp;Thus, the integral becomes&amp;nbsp;∫(10P+10−61−1−10−7&amp;nbsp;P})dP=∫&amp;nbsp;dt&amp;nbsp;&amp;nbsp;∴10ln|P|−10ln|10−1−10−7&amp;nbsp;P|=t+c1&amp;nbsp;P10−1−10−7&amp;nbsp;P=Cet10&amp;nbsp;Now, applying theinitial condition&amp;nbsp;P(0)=5000&amp;nbsp;to get&amp;nbsp;50000.1−0.0005=C→C=50000.0995&amp;nbsp;Now, in the solution, substitute by this result to get&amp;nbsp;P(t)=500et100.0995+0.0005et10→P(t)=5000.005+0.995e−t10

A model for the population P(t) in a suburb of

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