# Differential Equations

• Feb 2nd 2008, 05:35 PM
yellowrose
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! :)
• Feb 2nd 2008, 05:42 PM
Jhevon
Quote:

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)
• Feb 2nd 2008, 06:01 PM
yellowrose
Quote:

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?
• Feb 2nd 2008, 06:12 PM
Jhevon
Quote:

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.

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

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

$\displaystyle \Rightarrow \left( \frac 1P + \frac 1{1 - P} \right)dP = dt$ ......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