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

- anonymous

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- anonymous

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

- anonymous

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

- dumbcow

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

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## More answers

- anonymous

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

- anonymous

whatcha got?

- anonymous

the suspense is killing me

- dumbcow

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

- anonymous

this is so freaking frustrating!!!

- anonymous

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

- anonymous

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.

- anonymous

it should matter

- dumbcow

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

- anonymous

we get two different answers

- anonymous

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

- anonymous

oh i didnt try that

- anonymous

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

- anonymous

for my solution

- anonymous

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

- anonymous

there is something seriously wrong here and i cant find it

- anonymous

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
\[\tan^2x\ \not= \sec^2x \]

- anonymous

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

- anonymous

what did you get ?

- anonymous

can i see what you got

- anonymous

thanks for your help btw

- anonymous

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.

- anonymous

\[P = \frac{MCe^{kt}}{1-Ce^{kt}} \frac{}{} \implies C = \frac{P_0}{M-P_0} \]
My LaTeX is rendering horribly today, but maybe that shows up OK for you?

- anonymous

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

- anonymous

so ok

- anonymous

i see the latex

- anonymous

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

- anonymous

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

- anonymous

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)

- anonymous

oh , thats bad symbols, one sec

- anonymous

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)

- anonymous

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

- anonymous

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).

- anonymous

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

- anonymous

also you should have a plus there, not a minus

- anonymous

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

- anonymous

Oh, I corrected it the post below

- anonymous

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

- anonymous

Indeed :D

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