A solution initially contains 200 bacteria. 1. Assuming the number y increases at a rate proportional to the number present, write down a differential equation connecting y and the time t. 2. If the rate of increase of the number is initially 100 per hour, how many bacteria are there after 2 hours? Solution:

Accepted Solution

Answer:1.[tex]\frac{dy}{dt}=ky[/tex]2.543.6Step-by-step explanation:We are given that y(0)=200Let y be the number of bacteria at any time [tex]\frac{dy}{dt}[/tex]=Number of bacteria per unit time[tex]\frac{dy}{dt}\proportional y[/tex][tex]\frac{dy}{dt}=ky[/tex]Where k=Proportionality constant2.[tex]\frac{dy}{y}=kdt[/tex],y'(0)=100Integrating on both sides then, we get[tex]lny=kt+C[/tex]We have y(0)=200Substitute the values then , we get [tex]ln 200=k(0)+C[/tex][tex]C=ln 200[/tex]Substitute the value of C then we get [tex]ln y=kt+ln 200[/tex][tex]ln y-ln200=kt[/tex][tex]ln\frac{y}{200}=kt[/tex][tex]\frac{y}{200}=e^{kt}[/tex][tex]y=200e^{kt}[/tex]Differentiate w.r.t[tex]y'=200ke^{kt}[/tex]Substitute the given condition then, we get[tex]100=200ke^{0}=200 \;because \;e^0=1[/tex][tex]k=\frac{100}{200}=\frac{1}{2}[/tex][tex]y=200e^{\frac{t}{2}}[/tex]Substitute t=2Then, we get [tex]y=200e^{\frac{2}{2}}=200e[/tex][tex]y=200(2.718)=543.6=543.6[/tex]e=2.718Hence, the number of bacteria after 2 hours=543.6