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! :)

Printable View

- Feb 2nd 2008, 05:35 PMyellowroseDifferential Equations
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! :) - Feb 2nd 2008, 05:42 PMJhevon
- Feb 2nd 2008, 06:01 PMyellowrose
- Feb 2nd 2008, 06:12 PMJhevon
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