How to solve this IVP? Could you please help me solve this IVP? 
A certain population grows according to the diﬀerential equation: 
$$\frac{\mathrm{d}P}{\mathrm{d}t} = \frac{P}{20}\left(1 − \frac{P}{4000}\right) $$
and the initial condition $P(0) = 1000$. What is the size of the population at time $t = 10$? 
The answer is
$$P(10)=\frac{4000}{(1 + 3e^{1/2})}$$
but I can't seem to get it. I have tried to bring all the right hand side terms to integrate with respect to $P$. But I end up getting stuck with this equation $$20\ln{(P)} + \frac{80000}{P} = 90 + 20\ln{(1000)}$$ and don't know how to isolate $P$...
Thanks so much! :)
 A: I'm not entirely sure where your equation came from, but this solves the problem.
$$\frac{\mathrm{d}P}{\mathrm{d}t}=\frac{P}{20}\left(\frac{4000-P}{4000}\right)=\frac{P(4000-P)}{80000}$$
Then we want to separate variables and get:
$$\frac{\mathrm{d}P}{P(4000-P)}=\frac{\mathrm{d}t}{80000}$$
Where you can employ the method of partial fractions and integrate. This yields
$$\frac{1}{4000}\int\left[\frac{1}{P}+\frac{1}{4000-P}\right]\;\mathrm{d}P=\frac{1}{80000}\int\mathrm{d}t$$
$$\ln{(P)}-\ln{(4000-P)}=\frac{t}{20}+c$$
where $c$ is the combined constant of integration from both sides.
$$\ln{\left(\frac{P}{4000-P}\right)}=\frac{t}{20}+c$$
$$\frac{P}{4000-P}=e^{t/20+c}$$
$$\frac{P}{4000-P}=ke^{t/20}$$
where $k=e^c$
$$P=(4000-P)ke^{t/20}$$
$$P=4000ke^{t/20}-Pke^{t/20}$$
$$P(1+ke^{t/20})=4000ke^{t/20}$$
$$P(t)=\frac{4000ke^{t/20}}{1+ke^{t/20}}$$
Now we need to find $k$. When $t=0$ we have
$$P(0)=\frac{4000k}{1+k}=1000$$
which has a solution of $k=1/3$
$$P(t)=\frac{\frac{4000}{3}e^{t/20}}{1+\frac{1}{3}e^{t/20}}$$
Divide by $\frac{1}{3}e^{t/20}$ on top and bottom.
$$P(t)=\frac{4000}{1+3e^{-t/20}}$$
Now just plug in $t=10$ and we get
$$P(10)=\frac{4000}{1+3e^{-1/2}}$$
This is off by a negative from the answer you provided, but extensive checking (plugging it back into the starting formula and verifying with WolframAlpha) confirms my answer. Could you share the steps you took to get to your logarithmic equation? 
A: The solution of your logistic differential equation, $$P'(t) = \alpha P(t) (1-\beta P(t)),$$ with $\alpha = 1/20$, $\beta = 1/4000$ is given by:
$$\color{blue}{P(t) = \frac{1}{\beta + k\, e^{-\alpha  t}}} \tag{1}, $$ with $k$ a constant of integration. Now set the initial condition to have:
$$P(0) = P_0 =1000 = \frac{1}{\beta + k} \Rightarrow k = \frac{1}{P_0}-\beta,$$ substituting back all this for $t=10$ you have:
$$ P(t=10) = \frac{1}{1/4000 + (1/1000 - 1/4000) e^{-1/2}} = \frac{4000}{1 + 3e^{-1/2}}, $$ as @BeaumontTaz has shown previously.
Note that I just want to highlight that this kind of equations finds applications in a range of fields, including artificial neural networks, biology, biomathematics, chemistry, demography, economics, geoscience, mathematical psychology, probability, sociology, political science, and statistics [Source].
Cheers!
Edit: I will add the derivation of the solution just for completeness. 
Follow the steps I'm going to describe below:
\begin{align}
\frac{dP/dt}{\alpha P ( 1- \beta P)} & = 1 \\
\frac{dP}{\alpha P(1-\beta P)} & =  dt \\
\int^P_{P_0} \frac{dP}{\alpha P(1-\beta P)} & = \int^t_0 dt = t. \\
\end{align} Let's focus on the LHS integral:
\begin{align}
\int^P_{P_0} \frac{dP}{P(1-\beta P)} & = \int^P_{P_0}  \left( \frac{1}{P} + \frac{\beta}{1-\beta P}\right) \, dP\\
& = \left.[\log{|P|} - \log{|1-\beta P|}]\right|^P_{P_0} \\
& = \log{ \left|\frac{P (1-\beta P_0)}{P_0 (1-\beta P)} \right|}. 
\end{align} Therefore, we have:
$$\log{ \frac{P (1-\beta P_0)}{P_0 (1-\beta P)} } = \alpha t, $$ take exponentials on both side to have:
$$ \frac{P (1-\beta P_0)}{P_0 (1-\beta P)}  = e^{\alpha t} \Rightarrow \color{blue}{P(t) = \frac{P_0 e^{\alpha t}}{1- \beta P_0 + \beta P_0 e^{\alpha t}}} \tag{2}, $$ which you can check matches the previous result after further simplification.
A: We have that $\frac{dP}{dt}=\frac{P(4000-P)}{80,000}$, so substituting $z=\frac{1}{P}=P^{-1}$ gives $\frac{dz}{dt}=-P^{-2}\frac{dP}{dt}$.  
Then multiplying by $-P^{-2}$ gives $-P^{-2}\frac{dP}{dt}=\frac{-P^{-1}(4000-P)}{80,000}$ or $\frac{dz}{dt}=\frac{1}{80,000}-\frac{1}{20}z$.
Then $\frac{dz}{dt}+\frac{1}{20}z=\frac{1}{80,000}$, so multiplying both sides by
$e^{\frac{1}{20}t}$ yields
$\displaystyle e^{\frac{1}{20}t}\left(\frac{dz}{dt}+\frac{1}{20}z\right)=\frac{1}{80,000}e^{\frac{1}{20}t}$ and therefore $\displaystyle\left(e^{\frac{1}{20}t}z\right)^{\prime}=\frac{1}{80,000}e^{\frac{1}{20}t}.$
Integrating gives $\displaystyle e^{\frac{1}{20}t}z=\frac{1}{4000}e^{\frac{1}{20}t}+C$, 
so $\displaystyle z=\frac{1}{4000}+Ce^{-\frac{1}{20}t}=\frac{1+De^{-\frac{1}{20}t}}{4000}$ (where $D=4000C$).
Taking reciprocals gives $\displaystyle P=\frac{4000}{1+De^{-\frac{1}{20}t}},\;\;$ and $P(0)=1000\implies \frac{4000}{1+D}=1000\implies D=3$.
Therefore $\displaystyle P(10)=\frac{4000}{1+3e^{-\frac{1}{2}}}.$
A: $\textbf{Given}$
$\dfrac{dP}{dt} = \dfrac{P}{20}\left(1-\dfrac{P}{4000}\right)$, $P(0) = 1000$. 
$\textbf{Find}$ 
$P(10)$
$\textbf{Analysis}$
$\dfrac{dP}{dt} = \dfrac{P}{20}\left(1-\dfrac{P}{4000}\right)$.
First divide out $P\left(1-\dfrac{P}{4000}\right)$ from the RHS and multiply $dt$ over from the LHS: 
$$
\dfrac{dP}{P(1-P/4000)} = \dfrac{dt}{20}.  \hspace{3in} (1)
$$
Next use partial fraction decomposition on the LHS: 
\begin{align*}
\dfrac{dP}{P(1-P/4000)} & = \dfrac{A\ dP}{P} + \dfrac{B\ dP}{(1-P/4000)} \\
                        & \Rightarrow 1 = A\left(1-\dfrac{P}{4000}\right) + BP\\ 
                        & \Rightarrow \left\{\begin{array}{lr}
                                              A = 1\\
                                              B =  1/4000
                                      \end{array}\right.
\end{align*}
The solutions for $A,B$ comes from equating coefficients. 
Hence, we may rewrite equation (1) as 
$$
\dfrac{dP}{P} + \dfrac{dP}{4000-P} = \dfrac{dt}{20}.
$$ 
Integrate both sides 
$$
\int \left(\dfrac{1}{P} + \dfrac{1}{4000-P}\right)dP = \dfrac{1}{20}\int dt,
$$
$$
\ln\left(\dfrac{P}{4000-P}\right) = \dfrac{t}{20} + C.
$$
Exponentiate both sides 
$$
\dfrac{P}{4000-P} = Ke^{t/20} \hspace{1cm} \Rightarrow  \hspace{1cm} P = (4000-P)Ke^{t/20}
$$
so that
$$
P = 4000\dfrac{Ke^{t/20}}{1 + Ke^{t/20}}.
$$
Now, the initial condition is $P(0) = 1000$, thus 
$$
P(0) \equiv 1000 = 4000\dfrac{K}{1 + K} \hspace{1cm} \Rightarrow \hspace{1in} K=\dfrac{1}{3}.
$$
So $P(t)$ is given explicitly by 
$$
P(t) = \dfrac{4000}{1+3e^{t/20}},
$$
and $P(10) = \dfrac{4000}{1+3e^{-1/2}}$.
