Question

(1 point) Suppose that news spreads through a city of fixed size of 800000 people at a time rate proportional to the number of people who have not heard the news.

(a.) Formulate a differential equation and initial condition for ?(?), the number of people who have heard the news ?t days after it has happened.

No one has heard the news at first, so ?(0)=0. The "time rate of increase in the number of people who have heard the news is proportional to the number of people who have not heard the news" translates into the differential equation

????=?()

where ? is the proportionality constant.

(1 point) Suppose that news spreads through a city of fixed size of 800000 people at a time rate proportional to the number of people who have not heard the news. (a.) Formulate a differential equation and initial condition for y(t), the number of people who have heard the news t days after it has happened. No one has heard the news at first, so y(0) = 0. The "time rate of increase in the number of people who have heard the news is proportional to the number of people who have not heard the news" translates into the differential equation dy = kl 800000-y where k is the proportionality constant. (b.) 6 days after a scandal in City Hall was reported, a poll showed that 400000 people have heard the news. Using this information and the differential equation, solve for the number of people who have heard the news after t days. yt) = |

(b.) 6 days after a scandal in City Hall was reported, a poll showed that 400000 people have heard the news. Using this information and the differential equation, solve for the number of people who have heard the news after ?t days.

?(?)=y(t)=

Answer 1

Let y people heard news at t days after it is has happened.

So there are 800,000 -y people unheard this news.

a>

Rate of increase of the number of people who have heard the news is proportional to (800,000-y)$\Rightarrow \frac{dy}{dt}=k(800,000-y)$answer

b> Integrating the above differential equation:

$\\\int_{800,000}^{y} \frac{dy}{800,000-y}=\int_{0}^{t}kdt\\\\\\\\-\ln\left [ 800,000-y \right ]^y_{800,000}=kt$ {integration of 1/(a+by) is (ln(a+by))/b, here a = 800,000 and b = -1}

$ln\left ( \frac{800000}{800000-y} \right )=kt$

for t = 6 days , y = 4000000

$\\ln\left ( \frac{800000}{800000-4000000} \right )=k6\\\\ln2=6k\\\\ \Rightarrow k = ln(2)/6\approx 0.1155$

Thus,

$ln\left ( \frac{800000}{800000-y} \right )=0.1155t$

applying antilog both side-

$\\\left ( \frac{800000}{800000-y} \right )=e^{0.1155t}\\\\y(t)=800,000(1-e^{-0.1155t})$answer