# Math Help - Differential Equations

1. ## Differential 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!

2. Originally Posted by yellowrose
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!
you have a problem here. i get (P - 1)/P = Ae^t

this yields: P = 1/(1 + Ae^t)

3. Originally Posted by Jhevon
you have a problem here. i get (P - 1)/P = Ae^t

this yields: P = 1/(1 + Ae^t)
Would you be able to tell me how you got (P-1)/P=Ae^t?

4. Originally Posted by yellowrose
Would you be able to tell me how you got (P-1)/P=Ae^t?
oh wait. i made an error. i wrote P(P - 1) when it should have been P(1 - P). that changes things a bit.

$\frac {dP}{dt} = P - P^2 = P(1 - P)$

$\Rightarrow \frac {dP}{P(1 - P)} = dt$

$\Rightarrow \left( \frac 1P + \frac 1{1 - P} \right)dP = dt$ ......through algerba, or partial fractions if you want to work hard

$\Rightarrow \ln |P| - \ln |1 - P| = t + C$

$\Rightarrow \ln \left| \frac P{1 - P} \right| = t + C$

$\Rightarrow \frac P{1 - P} = Ae^t$

$\Rightarrow P - P^2 = Ae^t$

$\Rightarrow P^2 - P + Ae^t = 0$

this is a quadratic in P, use the quadratic formula to solve for P, so:

$P = \frac {1 \pm \sqrt{1 - 4Ae^t}}{2}$

now simplify ... er, if you can, or want to