# Math Help - I need help solving a simple differential equation?

1. ## I need help solving a simple differential equation?

hello!

I have never learned how to solve differential equations using partial fractions (i think that is what it is called?). I was hoping someone could post the solution to the following differential equation - I do not need to know the derivation, I just need the correct solution:

$dP/dt=K*P(1-P/L)$

I suppose it is not simple, but I think it is quite a well known equation for population growth! I appreciate any solutions!

2. Originally Posted by Machete
hello!

I have never learned how to solve differential equations using partial fractions (i think that is what it is called?). I was hoping someone could post the solution to the following differential equation - I do not need to know the derivation, I just need the correct solution:

$dP/dt=K*P(1-P/L)$

I suppose it is not simple, but I think it is quite a well known equation for population growth! I appreciate any solutions!
It is the Logistic Equation see >>here<<

CB

3. sweet thanks

4. Originally Posted by Machete
hello!

I have never learned how to solve differential equations using partial fractions (i think that is what it is called?). I was hoping someone could post the solution to the following differential equation - I do not need to know the derivation, I just need the correct solution:

$dP/dt=K*P(1-P/L)$

I suppose it is not simple, but I think it is quite a well known equation for population growth! I appreciate any solutions!
$\frac{dP}{dt} =K\cdot P\cdot \left(1-\frac{P}{L}\right)$
$\displaystyle\frac{dP}{P\left(1-\frac{P}{L}\right)} =K dt$
$\displaystyle\frac{dP}{P\cdot \frac{1}{L}\left(L-P\right)} =K dt$
$L\cdot\displaystyle\frac{dP}{P\left(L-P\right)} =K dt$
$\displaystyle\frac{dP}{P\left(L-P\right)} =\frac{K}{L} dt$
$\displaystyle\frac{1}{P\left(L-P\right)} =\frac{1}{LP}+\frac{1}{L(L-P)}$
$\displaystyle\int\frac{dP}{P\left(L-P\right)} =\int\left(\frac{1}{LP}+\frac{1}{L(L-P)}\right) dP=\frac{1}{L}\ln P-\frac{1}{L}\ln(L-P)=$ $\displaystyle=\frac{1}{L}\left(\ln P -\ln (L-P)\right)=\frac{1}{L}\ln\frac{P}{L-P}$
$\displaystyle \int \frac{dP}{P(L-P)}=\int\frac{K}{L}dt$
$\displaystyle \frac{1}{L}\ln\frac{P}{L-P}=\frac{K}{L} t + C$