# separate the variables....

• Jan 30th 2011, 12:52 AM
slapmaxwell1
separate the variables....
dp/dt = P - P^2

ok i am supposed to separate the variables. im having a rough time with this problem..i know to get the p on one side and the dt on the other, and then i used partial fractions to try to integrate and thats when i had problems..

ln(p) + ln(1-p) = dt

i dont know im not even sure if thats right and suppose it is right, then when i raise it to e the p's cancel out?? any help would be appreciated.
• Jan 30th 2011, 12:56 AM
Prove It
You need to integrate both sides...

$\displaystyle \frac{dp}{dt} = p - p^2$

$\displaystyle \frac{1}{p - p^2}\,\frac{dp}{dt} = 1$

$\displaystyle \int{\frac{1}{p - p^2}\,\frac{dp}{dt}\,dt} = \int{1\,dt}$

$\displaystyle \int{\frac{1}{p-p^2}\,dp} = t + C_1$

$\displaystyle \int{\frac{1}{p(1-p)}\,dp} = t+C_1$.

Now follow your instincts to use partial fractions... You can then simplify using logarithm rules.
• Jan 30th 2011, 01:01 AM
slapmaxwell1
ok so the integral (A/P + B/(1-P)) ? right then 1 = A-AP + BP ?
• Jan 30th 2011, 06:59 AM
harish21
Quote:

Originally Posted by slapmaxwell1
ok so the integral (A/P + B/(1-P)) ? right then 1 = A-AP + BP ?

$\dfrac{1}{p(1-p)}=\dfrac{A}{p}+\dfrac{B}{1-p}$

$1\;=\;A(1-p)+B(p)$

find A and B and then integrate the right side of the above equation
• Jan 30th 2011, 09:32 AM
wonderboy1953
Quote:

Originally Posted by slapmaxwell1
ok so the integral (A/P + B/(1-P)) ? right then 1 = A-AP + BP ?

Partial fractions method works, but an easier method to break down the fraction (which I refer to as the A-S method or the addition-subtraction method) is to add and subtract p in the numerator of the fraction.