## 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( 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

1. anonymous

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

2. 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

3. anonymous

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

4. anonymous

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

5. anonymous

whatcha got?

6. anonymous

the suspense is killing me

7. anonymous

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

8. anonymous

this is so freaking frustrating!!!

9. 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

10. 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.

11. anonymous

it should matter

12. anonymous

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

13. anonymous

14. anonymous

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

15. anonymous

oh i didnt try that

16. anonymous

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

17. anonymous

for my solution

18. anonymous

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

19. anonymous

there is something seriously wrong here and i cant find it

20. 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 $\tan^2x\ \not= \sec^2x$

21. anonymous

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

22. anonymous

what did you get ?

23. anonymous

can i see what you got

24. anonymous

25. 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.

26. 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?

27. anonymous

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

28. anonymous

so ok

29. anonymous

i see the latex

30. anonymous

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

31. anonymous

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

32. 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)

33. anonymous

oh , thats bad symbols, one sec

34. 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)

35. anonymous

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

36. 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).

37. anonymous

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

38. anonymous

also you should have a plus there, not a minus

39. anonymous

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

40. anonymous

Oh, I corrected it the post below

41. anonymous

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

42. anonymous

Indeed :D