# Application of Linear differential equation in solving problems

• MHB
• Help_me1
In summary, after 1 day, 5 students will be aware of the rumour and 10 students will be aware of the rumour after 7 days. It will take 850 students 7 days to be aware of the rumour.f

#### Help_me1

A rumour spreads through a university with a population 1000 students at a rate proportional to the product of those who have heard the rumour and those who have not.If 5 student leaders initiated the rumours and 10 students are aware of the rumour after one day:-
i)How many students will be aware of the rumour after 7 days.
ii)How long will it take for 850 students to hear the rumour

let $r$ = number of students who have heard the rumor

$(1000-r)$ = number who have not heard the rumor

$t$ is in days

$\dfrac{dr}{dt} = k \cdot r(1000-r)$, where $k$ is the constant of proportionality

you are given $r(0)=5$ and $r(1) = 10$

see what you can do from here …

Thank you Since this has been here a while and I just cannot resist:
dr/dt= kr(1000- r)
dr/[r(1000-r)]= kdt

To integrate on the left, separate using "partial fractions"- find constant A and B such that 1/[r(1000- r)]= A/r+ B/(1000- r)
Multiplying on both sides by r(1000- r)
1= A(1000- r)+ Br
Let r= 0: 1= 1000A so A= 1/1000.
Let r= 1000: 1= 1000B so B= 1/1000

1000 dr/r+ 1000 dr/(1000- r)= kdt

The integral of 1000 dr/r is 1000 ln(|r|).
To integrate 1000 dr/(1000- r) let u= 1000- r so that du= -dr.
then dr/(1000- r)= -du/u.
The integral is -1000 ln(|u|)= -1000 ln(|1000- r|)

We have 1000 ln(|r|)-1000 ln(|1000- r|)= kt+ C.
$1000 ln\left(|\frac{r}{1000- r}|\right)= kt+ C$
$ln\left(\left(\frac{r}{1000- r}\right)^{1000}\right)= kt+ C$
(I have dropped the absolute value since this is to an even power.)

Taking the exponential of both sides
$\left(\frac{r}{1000- r}\right)^{1000}= C' e^{kt}$
where $C'= e^C$.

Now, we are given that r(0)= 5 and r(1)= 10.
Setting t= 0, r= 5
$\frac{5}{9995}= \frac{1}{1999}=C'$
Setting t= 1, r= 10
$\frac{10}{9990}= \frac{1}{999}= \frac{e^k}{1999}$
$e^k= \frac{1999}{999}$
so $k= ln\left(\frac{1999}{999}\right)$
and $e^{kt}= (e^k)^t= \left(\frac{1999}{999}\right)^t$

$r(t)= \frac{\left(\frac{1999}{999}\right)^t}{1999}$.

You are putting A and B values in a wrong way ...these values will come in fraction