Question

A disease is infecting a heard of 100 cows. Let P(t) be the number of sick cows t days after outbreak. Suppose 15 cows initially had disease and suppose it is spreading at a rate proportional to the number of cows who do not have the disease.

Give the initial value problem for P.

Answer 1

Hmm these word problems are always a bit tricky.
\(P(t)\) represents the sick population.
\(P(0)=15\), this is the initial amount of sick cows.

The rate the disease is spreading \(\dfrac{dP}{dt}\), is proportional to,
the number of cows who are not sick.

So I'm trying to think of a good way to interpret 'not sick'.

I guess it would be.... \(100-P\) right?
P represents the sick population at time \(t\).
100 represents the total number of cows.

So total cows minus sick cows, equals healthy cows at time t.

Then we also nee a constant of proportionality.
So this might be what we're looking for,\[\large \frac{dP}{dt}=k(100-P)\]

Answer 2

I'm not certain, but you can at least see my thought process there.

Answer 3

why is there not a t though?

Answer 4

\[\large \frac{dP}{dt}=k\left(100-P(t)\right)\]P is a function of t.
This is a differential equation, it involves P and P's derivative.
t comes about by solving for P (separating variables and integrating).

Answer 5

so if i was to find the general solution to this equation would i have to find the integrating factor?

Answer 6

Using that method? Bah I forget what that's called. Method of Undetermined Coefficients or something..? yes you could.
This is also separable, see how there are no t's?

So you could approach it using either method.

Answer 7

\[\large \frac{dP}{dt}=k\left(100-P(t)\right) \qquad \rightarrow \qquad \frac{dP}{100-P}=k \;dt\]

Answer 8

That's probably the method I would use to solve this, but integrating factor seems fine also :)

Answer 9

well in the notes they take the differential equation

dy/dt = ky(M-y)

and they integrate and get

y(t) = MR / (R + (M - R)e^-Mkt

Answer 10

but the next part asks to find the general equation, but we dont know k or t?

Answer 11

So their equation started out as,\[\large \frac{dP}{dt}=kP(M-P)\](I don't see why they switched from P's to y's...)
Hmm so I didn't have the right setup? dang :( hmmmm

Answer 12

So if there is suppose to be a P on the outside like that, then we'll definitely want to approach this by separating variables.\[\large \frac{dy}{dt}=ky(M-y) \qquad \rightarrow \qquad \frac{dy}{y(M-y)}=k\;dt\]Then we would have to apply `Partial Fraction Decomp` on the y's.
Lot of weird stuff going on here, with t, k and m.

They gave us initial conditions though, see the initial population of sickness?
We can probably use that to solve for something that's missing.

Answer 13

the previous problem, which i have the answer to is very similar to this one, the only difference is the spreading rate.

Answer 14

oh that an example? lol my bad.

Answer 15

i just dont understand how they got it.

Answer 16

What was the rate proportional to in the other example you have?

Answer 17

Healthy times sick?

Answer 18

It says population is 1000. Initial with disease is 50. And the rate is proportional to the product of the time elapsed and the number who do not have disease.

The IVP for this one is

dP/dt = kt(1000-R)

Answer 19

which makes me think the IVP that you found was correct.

Answer 20

Yah me too :o

Answer 21

so if that is the case, then what do you think is the best method to find the general solution... seperable or using integrating factor

Answer 22

Bahhh I don't have a preference really :D
If you expand out the brackets on the right, and then move the P term to the left, I think we get this equation,\[\large P'+kP=100k\]
Which gives you a very nice easy integrating factor.

Answer 23

so we would multiply be e^(k^2/2)?

Answer 24

Woops! k is just a constant.
Our integrating factor would be with respect to ummmmmmmm, t.

Answer 25

\[\large \mu=e^{\int\limits k dt}\]

Answer 26

so just e^kt

Answer 27

yah there ya go c:

Answer 28

so i get (e^kt)p = integral (e^kt)100k

and we integrate again with respect to t... and the 100k is just a constant

Answer 29

yah looks good so far.

Answer 30

okay (e^kt)p = 100k(e^kt) + C

Im not sure when we take the integral of that do we bring down the constant again or is it just e^kt still.

Answer 31

We lose a k when we integrate.

If you take the derivative of your result, you'll see that a k comes down due to the chain rule.
So our answer should have a k divided out.

Answer 32

\[\large \int\limits e^{at}dt \quad = \quad \frac{1}{a}e^{at}\]Try to get comfortable with this idea because integrating exponentials will come up quite a bit in Diff EQ.

Answer 33

okay so then we are left with

P(t) = 100 + Ce^-kt

Answer 34

Yah looks good.

Answer 35

sweet, and the last part which says find the particular solution that satisfies the initial condition . all we are doing is finding what the C value has to be right?

Answer 36

I'm not quite sure how we find the particular solution.
It looks like we don't have enough information.
We've been given \(P(0)=15\).

But that's not enough to be able to solve for \(C\) and \(k\), our unknown constants.
We need another piece of information, I think... :\ hmm

Answer 37

in the second example, it looks like they just subtracted the 1000-50 and put in 950 for C in the general solution.... would it make sense to 100-15 and put 85 in for C

Answer 38

Oh I'm being stupid :) We can solve for C of course, since t=0 eliminates our k value as well.

Answer 39

\[\large P(0)=15 \quad \rightarrow \quad 15=100+Ce^0\]

Answer 40

I think we actually end up with -85.

Answer 41

fantastic... do you by chance know what the R is supposed to mean... we used the P instead, but in the book they use R.

Answer 42

Hmm i dunno :3

Answer 43

ohhh, i atleast understand what we did. thank you very much.

Answer 44

np \c:/