An isolated college has a student population of 750. One student returns to college unwittingly carrying a 'flu virus. The rate at which the virus spreads is proportional to both the number of students, x, who have so far caught the virus and the number of students still unaffected. The situation can be represented by the differential equation dx/dt = kx(750 - x) where k is the proportional constant.

1) Find the general solution to this differential equation?
2) If the initial conditions are that, when t = 0 then x = 1, find the corresponding integration constant?
3) Given after 4 days (t=4), 25 students are suffereing from the virus, find a value for k and calculate the number of studetns who are suffering by the end of the first week?

2. Originally Posted by Natasha1

[snip]

The situation can be represented by the differential equation dx/dt = kx(750 - x) where k is the proportional constant.

1) Find the general solution to this differential equation?
So we want the general solution of:

$
\frac{dx}{dt}=kx(750-x)
$
]

Which id a first order ODE of variables separable type, ie:

$
\frac{1}{x(750-x)}\frac{dx}{dt}=k
$
,

which we know has solution:

$
\int\ \frac{1}{x(750-x)} dx=\int\ k\ dt+\ \mbox{const}
$

so:

$
\int\ \frac{1}{x(750-x)} dx=k\ t+\ \mbox{const}
$
.

The integral on the LHS can be done using partial fractions.

(I am now being called away for dinner (evening meal) and will continue this
if need be after I have eaten, and stuffed the washing up in the dishwasher,
and cleared away my share of the children's mess )

RonL

3. Anytime Captain. Thanks ever so much :-)

so continuing from this I get

1/750 [ln(x) + ln (750 -x)] = t + Constant

4. Originally Posted by Natasha1
Anytime Captain. Thanks ever so much :-)

so continuing from this I get

1/750 [ln(x) + ln (750 -x)] = t + Constant
That's almost right

$
1/750 [\ln(x) - \ln(750 -x)] = kt + \mbox{Const}
$

rewriting:

$
\ln(x/(750-x)=750kt + \mbox{different const}
$
,

or:

$
\frac{x}{750-x}=Ae^{750kt}
$
.

Rearranging again:

$
x(t)=\frac{750}{Be^{-750kt}+1}
$
,

for some constant $B$, which will give a sigmoid curve rising from
a small value at time 0 (assuming that we have such an initial condition) rising
untill it levels off at 750 infecteds.

The attached plot shows this for the initial conditions given in part (2), and
$k=0.01$

RonL

5. RoL,

When I equate I get

0=B-A
1=750A

Which gives me A and B = 1/750

So when I replace

1/x(750-x) = A/x + B/(750-x)

why don't I get

1/x(750-x) = 1/750 [ln(x) + ln (750-x)]

1/x(750-x) = 1/750 [ln(x) - ln (750-x)]

6. Originally Posted by Natasha1
RoL,

When I equate I get

0=B-A
1=750A

Which gives me A and B = 1/750

So when I replace

1/x(750-x) = A/x + B/(750-x)

why don't I get

1/x(750-x) = 1/750 [ln(x) + ln (750-x)]

1/x(750-x) = 1/750 [ln(x) - ln (750-x)]
When you differentiate $-\ln(750-x)$ you get $\frac{1}{750-x}$, because of the $-$ sign in front of the $x$.

RonL

7. oops!

8. Originally Posted by Natasha1
oops!
It happens to everyone sometime

(which is why you need to check what the helper here write )

RonL

9. Originally Posted by CaptainBlack
It happens to everyone sometime

(which is why you need to check what the helper here write )

RonL
Now for part 3?

10. Originally Posted by Natasha1
Now for part 3?
OK, we have:

$
x(t)=\frac{750}{749e^{-750kt}+1}
$
.

And we know that at $t=4, \ x=25$, so:

$
25=\frac{750}{749e^{-3000k}+1}
$
,

rearrange:

$
{749e^{-3000k}+1}=\frac{750}{25}=30
$
,

so:

$
e^{-3000k}=\frac{29}{749}
$
,

or:

$
-3000k=\ln(29/749)
$

so:

$
k=-\frac{\ln(29/749)}{3000}\approx 0.00108
$

11. RoL,

I want to stick to the first solution you got which was

x/ (750-x) = Ae^(750kt)

at the end of question 3 it is asked to calculate the number of students who are suffering by the end of the first week?

At t=7 x=?

my answer so far as we know that A=1/749

x/ (750-x) = (1/749)e^(750 * 0.00108 * 7)

749x/ (750-x) = e^(5.67)

Where do I go from here :-(

12. Originally Posted by Natasha1
RoL,

I want to stick to the first solution you got which was

x/ (750-x) = Ae^(750kt)

at the end of question 3 it is asked to calculate the number of students who are suffering by the end of the first week?

At t=7 x=?

my answer so far as we know that A=1/749

x/ (750-x) = (1/749)e^(750 * 0.00108 * 7)

749x/ (750-x) = e^(5.67)

Where do I go from here :-(
Cross multiply:

$
749x=e^{5.67}(750-x)
$

so:

$
(749+e^{5.67})x=750 e^{5.67}
$

and you should be able to take it from there (you can also check what you
get against the graph in my earlier post).

RonL