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anonymous
 5 years ago
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( MP)/M , which is seperable autonomous.
we have integral M / [ P (MP)] dP = integral k dt. using partial fractions i get
integral 1 / P + 1 / ( MP) = kt + C
ln P  ln MP = kt + C
ln  P / (MP)  = kt + C
P /( MP) = e^(kt + c)
P = (MP) 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
 5 years ago
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( MP)/M , which is seperable autonomous. we have integral M / [ P (MP)] dP = integral k dt. using partial fractions i get integral 1 / P + 1 / ( MP) = kt + C ln P  ln MP = kt + C ln  P / (MP)  = kt + C P /( MP) = e^(kt + c) P = (MP) 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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this problem is posted here http://www.sosmath.com/diffeq/first/application/population/population.html

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok ill try to work it out and see if i get anything different

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the only difference in the website is that they kept 1  P/M in the partial fractions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the suspense is killing me

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0ok yeah im getting the same result you are. i tried both factoring out the 1/m and leaving it in when integrating fractions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is so freaking frustrating!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0last 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
 5 years ago
Best ResponseYou've already chosen the best response.0Unless 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.

dumbcow
 5 years ago
Best ResponseYou've already chosen the best response.0ahh that could be it, by us splitting up the fraction we may have changed the constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we get two different answers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0After you have solved for the constant? (You may well do, I haven't tried after that)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have a point though, suppose i multiplied top and bottom by M, the C could absorb one of the M's

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no you cant do that, then the C's would be different

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there is something seriously wrong here and i cant find it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OK, 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
 5 years ago
Best ResponseYou've already chosen the best response.0yes but i did precise partial fraction integration, this isnt like a trig antiderivative which is differing by a constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can i see what you got

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks for your help btw

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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/(MP_0) Put this back in and it results in the same value for P(t) in the final answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[P = \frac{MCe^{kt}}{1Ce^{kt}} \frac{}{} \implies C = \frac{P_0}{MP_0} \] My LaTeX is rendering horribly today, but maybe that shows up OK for you?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, first is 1+ Ce ... in the fraction :/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here is an idea, but it seems silly, define a new constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0multiply top and bottom by M, then define a new constant C* = M*C

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0P = 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
 5 years ago
Best ResponseYou've already chosen the best response.0oh , thats bad symbols, one sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0P = 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, I don't see why you can't do that (but it seems unnecessary).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0well it doesnt explain why i get the same answer using partial fractions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also you should have a plus there, not a minus

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you wrote P = M Ce^kt / ( 1  Ce^kt), in latex

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, I corrected it the post below

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok thanks :) so the final answer comes out the same ,
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