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

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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

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this problem is posted here http://www.sosmath.com/diffeq/first/application/population/population.html
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
ok ill try to work it out and see if i get anything different

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the only difference in the website is that they kept 1 - P/M in the partial fractions
whatcha got?
the suspense is killing me
ok yeah im getting the same result you are. i tried both factoring out the 1/m and leaving it in when integrating fractions
this is so freaking frustrating!!!
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
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.
it should matter
ahh that could be it, by us splitting up the fraction we may have changed the constant
we get two different answers
After you have solved for the constant? (You may well do, I haven't tried after that)
oh i didnt try that
you have a point though, suppose i multiplied top and bottom by M, the C could absorb one of the M's
for my solution
no you cant do that, then the C's would be different
there is something seriously wrong here and i cant find it
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 \[\tan^2x\ \not= \sec^2x \]
yes but i did precise partial fraction integration, this isnt like a trig antiderivative which is differing by a constant
what did you get ?
can i see what you got
thanks for your help btw
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.
\[P = \frac{MCe^{kt}}{1-Ce^{kt}} \frac{}{} \implies C = \frac{P_0}{M-P_0} \] My LaTeX is rendering horribly today, but maybe that shows up OK for you?
Sorry, first is 1+ Ce ... in the fraction :/
so ok
i see the latex
here is an idea, but it seems silly, define a new constant
multiply top and bottom by M, then define a new constant C* = M*C
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)
oh , thats bad symbols, one sec
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)
Yes, I don't see why you can't do that (but it seems unnecessary).
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).
well it doesnt explain why i get the same answer using partial fractions
also you should have a plus there, not a minus
you wrote P = M Ce^kt / ( 1 - Ce^kt), in latex
Oh, I corrected it the post below
ok thanks :) so the final answer comes out the same ,
Indeed :D

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