Solve by separation of variables:
dP/dt = P-P^2
This is what I have so far:
P/(P-1)=ce^t
but I need to get that to:
P=(ce^t)/(1+ce^t)
Any help would be greatly appreciated.
Thanks!
oh wait. i made an error. i wrote P(P - 1) when it should have been P(1 - P). that changes things a bit.
$\displaystyle \frac {dP}{dt} = P - P^2 = P(1 - P)$
$\displaystyle \Rightarrow \frac {dP}{P(1 - P)} = dt$
$\displaystyle \Rightarrow \left( \frac 1P + \frac 1{1 - P} \right)dP = dt$ ......through algerba, or partial fractions if you want to work hard
$\displaystyle \Rightarrow \ln |P| - \ln |1 - P| = t + C$
$\displaystyle \Rightarrow \ln \left| \frac P{1 - P} \right| = t + C$
$\displaystyle \Rightarrow \frac P{1 - P} = Ae^t$
$\displaystyle \Rightarrow P - P^2 = Ae^t$
$\displaystyle \Rightarrow P^2 - P + Ae^t = 0$
this is a quadratic in P, use the quadratic formula to solve for P, so:
$\displaystyle P = \frac {1 \pm \sqrt{1 - 4Ae^t}}{2}$
now simplify ... er, if you can, or want to