Answer:
1.[tex]\frac{dy}{dt}=ky[/tex]
2.543.6
Step-by-step explanation:
We are given that
y(0)=200
Let 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 constant
2.[tex]\frac{dy}{y}=kdt[/tex],y'(0)=100
Integrating on both sides then, we get
[tex]lny=kt+C[/tex]
We have y(0)=200
Substitute 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=2
Then, we get [tex]y=200e^{\frac{2}{2}}=200e[/tex]
[tex]y=200(2.718)=543.6=543.6[/tex]
e=2.718
Hence, the number of bacteria after 2 hours=543.6
