Question Number 63424 by Rio Michael last updated on 04/Jul/19
$${A}\:{colony}\:{of}\:{bacteria}\:{if}\:{left}\:{undisturbed}\:{will}\:{grow}\:{at}\:{a}\:{rate} \\ $$$${proportional}\:{to}\:{the}\:{number}\:{of}\:{bacteria},\:{P}\:{present}\:{at}\:{time},{t}. \\ $$$${However},{a}\:{toxic}\:{substance}\:{is}\:{being}\:{added}\:{slowly}\:{such}\:{that} \\ $$$${at}\:{time}\:{t},\:{the}\:{bacteria}\:{also}\:{die}\:{at}\:{the}\:{rate}\:\mu{Pt}\:{where}\:\mu\:{is} \\ $$$${a}\:{positive}\:{constant}. \\ $$$$\left({a}\right)\:\:{Show}\:{that}\:{at}\:{time}\:{t}\:{the}\:{rate}\:{of}\:{growth}\:{of}\:{the}\:{bacteria}\:{in} \\ $$$${the}\:{colony}\:{is}\:{governed}\:{by}\:{the}\:{differential}\:{equation} \\ $$$$\:\frac{{dP}}{{dt}}=\:\left({k}−\mu{t}\right){p}\:{where}\:{k}\:{is}\:{apositive}\:{constant}. \\ $$$${when}\:{t}=\mathrm{0},\:\frac{{dP}}{{dt}}=\mathrm{2}{P}\:{and}\:{when}\:{t}=\mathrm{1},\:\frac{{dP}}{{dt}}=\frac{\mathrm{19}}{\mathrm{10}}{P} \\ $$$$\left({b}\right)\:{show}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{dP}}{{dt}}=\:\frac{\mathrm{1}}{\mathrm{10}}\left(\mathrm{20}−{t}\right){P}. \\ $$$$\:{Sir}\:{Forkum}\:{Michael}. \\ $$