anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
this problem is posted here http://www.sosmath.com/diffeq/first/application/population/population.html
anonymous
  • 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
  • dumbcow
ok ill try to work it out and see if i get anything different

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

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