Question

dP/dt = p(2-P) , p(0)=1/2 .... How do I solve for P(t) for the initial value?

Answer 1

It looks like it's more-or-less separable. We get.
\[\Large \frac{dP}{P(2-P)}=dt\]

Answer 2

Can you integrate both sides?

Answer 3

int(1/p(2-p))dp = 1 dt
then int by parts a/p +b/2-p
=int1/2p + int1/2(2-p)
=1/2(lnp) - 1/2

Answer 4

1/2(lnp) - 1/2(ln(2-p) = t +C

Answer 5

then, do I solve for C?
from here I'm not sure what to do...

Answer 6

Something's amiss... are you sure it isn't
\[\Large \frac1{2}\ln(P) - \frac12\ln(P-2)\]
?

Answer 7

well, after integration by parts i get \[1/2 \int\limits(1/p) + 1/2 \int\limits(1/2-p))\]
then after integrate shouldnt that give \[1/2(\ln(p)) - 1/2(\ln(2-p))\]?

Answer 8

the difference I see is in the ln(p-2) vs ln(2-p)

Answer 9

Hang on...
\[\Large \frac12\int\frac1p \color{green}{dp} + \frac12\int \frac1{2-p}\color{green}{dp}\]
right?

Answer 10

yes

Answer 11

First one's no problem:
\[\Large \frac12\ln(p) + \frac12\int\frac1{2-p}\color{green}{dp}\]

Answer 12

Hang on...

Answer 13

Wait, I see the problem... we're both right... I just forgot that the actual integral of 1/x
is
\[\Large \int \frac1xdx = \ln|x|+C\]
So it doesn't matter really if it were
\[\Large \ln(P-2) \] or \[\Large \ln(2-P)\]

LOL moving on...

Answer 14

But just to be sure, let's have it as
\[\Large \frac12\ln|P|-\frac12\ln|P-2| = t+C \]

Answer 15

lol, okay so assuming that correct I think I solve doe C using the initial values p(0)=1/2 then splve for P and plug in the C value... Is that the right idea?

Answer 16

Yup. Replace all t with 0 and all p with 1/2

Answer 17

okay.. I get C = (ln(3/4))/2

Answer 18

I think the arithmetic was a bit off...

Answer 19

Okay I need to crunch away on this for a while, as long as I know the process is correct I can do it. Thanks so much for your help!