Question

differential equation, i dont get book's answer.
question: dP(t)/dt = kP(t) ( 1 - P(t)/M) , where P is population at time t, M is carrying capacity.books solution: P(t) = MCe^(kt)/ ( M + Ce^kt).
ok here is what i got.
first 1 - P/M = (M- P ) / M, so dP/dt = kP( M-P)/M , which is seperable autonomous.
we have integral M / [ P (M-P)] dP = integral k dt. using partial fractions i get
integral 1 / P + 1 / ( M-P) = kt + C
ln P - ln M-P = kt + C
ln | P / (M-P) | = kt + C
P /( M-P) = e^(kt + c)
P = (M-P) Ce^kt , where C = e^c
P = M Ce^kt -P*Ce^kt
P + P Ce^kt = M Ce^kt
P ( 1 + Ce^kt) = MCe^kt
P

Answer 1

this problem is posted here http://www.sosmath.com/diffeq/first/application/population/population.html

Answer 2

so i got P = MCe^kt / ( 1 + Ce^kt) , but the website has
P = MCe^(kt)/ ( M + Ce^kt)
theres a glaring discrepancy, i am missing an M

Answer 3

ok ill try to work it out and see if i get anything different

Answer 4

the only difference in the website is that they kept 1 - P/M in the partial fractions

Answer 5

whatcha got?

Answer 6

the suspense is killing me

Answer 7

ok yeah im getting the same result you are.
i tried both factoring out the 1/m and leaving it in when integrating fractions

Answer 8

this is so freaking frustrating!!!

Answer 9

last step P = MCe^(kt)/ ( 1 + Ce^kt), website has an extra M in there http://www.sosmath.com/diffeq/first/application/population/population.html
both ways seem sound . im stumped

Answer 10

Unless I'm missing something, surely it doesn't matter which you got? The constant comes out different anyway, so it (probably) leads to the same final results.

Answer 11

it should matter

Answer 12

ahh that could be it, by us splitting up the fraction we may have changed the constant

Answer 13

we get two different answers

Answer 14

After you have solved for the constant? (You may well do, I haven't tried after that)

Answer 15

oh i didnt try that

Answer 16

you have a point though, suppose i multiplied top and bottom by M, the C could absorb one of the M's

Answer 17

for my solution

Answer 18

no you cant do that, then the C's would be different

Answer 19

there is something seriously wrong here and i cant find it

Answer 20

OK, I have carried on from your answer (got a different constant than in the website's solution), and it leads to the same final p(t) - unless I made a mistake. Try it and see what happens.

In general, something can integrate to two different things (with a constant in both). The fact they both are general solutions / differentiate to the same thing does not imply they are the same.
e.g.
\[\int 2 \sec^2x\ \tan x = \tan^2x + c = \sec^2x + k \]
but
\[\tan^2x\ \not= \sec^2x \]

Answer 21

yes but i did precise partial fraction integration, this isnt like a trig antiderivative which is differing by a constant

Answer 22

what did you get ?

Answer 23

can i see what you got

Answer 24

thanks for your help btw

Answer 25

I carried on from your value:

P = MCe^(kt)/(1+Ce^(kt)).

When t = 0, solve for C in terms of P_O and get C = P_0/(M-P_0)

Put this back in and it results in the same value for P(t) in the final answer.

Answer 26

\[P = \frac{MCe^{kt}}{1-Ce^{kt}} \frac{}{} \implies C = \frac{P_0}{M-P_0} \]

My LaTeX is rendering horribly today, but maybe that shows up OK for you?

Answer 27

Sorry, first is 1+ Ce ... in the fraction :/

Answer 28

so ok

Answer 29

i see the latex

Answer 30

here is an idea, but it seems silly, define a new constant

Answer 31

multiply top and bottom by M, then define a new constant C* = M*C

Answer 32

P = M* M Ce^kt / [M( 1 + Ce^kt)] = M (MC)e^kt / [ M + MC e^kt ] ,
let C* = MC, so you have
P = MC* e^kt / (M + C*e^kt)

Answer 33

oh , thats bad symbols, one sec

Answer 34

P = M M Ce^kt / [M( 1 + Ce^kt)] = M (MC)e^kt / [ M + MC e^kt ] , let C` = MC, so you have P = MC` e^kt / (M + C`e^kt)

Answer 35

Yes, I don't see why you can't do that (but it seems unnecessary).

Answer 36

If you just solve for C as it is, you get the value they did divided by M, so all you're doing is getting the same constant times M. Like I said, unnecessary, but if it helps you think about why they're the same then fair enough).

Answer 37

well it doesnt explain why i get the same answer using partial fractions

Answer 38

also you should have a plus there, not a minus

Answer 39

you wrote P = M Ce^kt / ( 1 - Ce^kt), in latex

Answer 40

Oh, I corrected it the post below

Answer 41

ok thanks :)
so the final answer comes out the same ,

Answer 42

Indeed :D