Question

1.The yield y(t) (in bushes)per acre of a corn crop satisfies the equation dy/dt+y=100+e^-t

if y(0)=0 find y at any time t

2.The population N(t) of a species of micro-organism in a laboratory setting at any time t is established to vary under the influence of a certain chemical at a rate given by dN/dt=t-2e^t show that N(t)=t-2e^t+c.hence if N(0)=400 determine the population when t=5

3.A single individual starts a rumour in a community of 200 people. the rumour spread at a rate proportional to the number of people who have not yet heard the rumour. After 2days 10 people have heard the rumour

1. how many people will have heard the rumour after 5days
2. how long will it take for the rumour to spread to 100 people

Answer 1

Answer:-

1.

$$y'+y=100+e^{-t}$$

Integrating factor: $\mu=e^t$

$$(e^ty)'=100e^t+1$$ $$e^ty=\int(100e^t+1)dt$$ $$e^ty=100e^t+t+c$$ $$y(t)=100+te^{-t}+ce^{-t}$$

$y(0)=0$

$$0=100+0+ce^{-0}=>c=-100$$ $$y(t)=100+te^{-t}-100e^{-t}$$

2.

$\frac{dN}{dt}=1-2e^t.$

$N=\int(1-2e^t)dt=t-2e^t+C.$

$N(0)=400\to -2+C=400\to C=402.$

$N(5)=5-2e^5+402=110.$

3.

If $N$ denotes the number of people who have heard the rumor, then $200-N$ represents the number of people who haven’t heard the rumor.

Then

$$\dfrac{dN}{dt}=k(200-N)$$ $$\dfrac{dN}{200-N}=kdt$$

Integrate

$$\ln|200-N|=-kt+\ln C$$ $$N=200-Ce^{-kt}$$ $$N(0)=200-C=0=>C=200$$ $$N=200-200e^{-kt}$$ $$N(2)=200-200e^{-k(2)}=10$$ $$e^{-k(2)}=0.95$$ $$2k=-\ln0.95$$ $$k=-\dfrac{1}{2}\ln0.95$$ $$N(t)=200-200e^{(\ln0.95)t/2}$$ $$N(t)=200-200(0.95)^{t/2}$$

1.

$$N(5)=200-200(0.95)^{5/2}=24$$

2.

$$N(t)=200-200e^{(\ln0.95)t/2}=100$$ $$e^{(\ln0.95)t/2}=0.5$$ $$(\ln0.95)t=2\ln0.5$$ $$t=\dfrac{2\ln0.5}{\ln0.95}$$ $$t=27\ days$$