Logistic model differential equation I need to find the solution of the IVP $$\frac{dp}{dt}=ap-bp^2, p(0)=p_0$$
where $a,b$ are constants.
I have found $$p(t)=\frac{ap_0}{bp_0+(a-bp_0)e^{-at}}$$
Now if $p(t_1)=p_1$ and $p(t_2)=p_2$ with $t_2=2t_1$ I need to find
$$a=\frac{1}{t_1}\ln \frac{p_2(p_1-p_0)}{p_0(p_2-p_1)}\,,\,b=\frac{a}{p_1}\frac{p_1^2-p_0p_2}{p_1p_2-2p_0p_2+p_0p_1)}$$
The parentheses in the denominator of $b$ is not a typo on my part, it's in the notes. I chose to ignore it and tried to prove that $b=\dfrac{a}{p_1}\dfrac{p_1^2-p_0p_2}{p_1p_2-2p_0p_2+p_0p_1}$, but no luck.
Questions:


*

*What is $b$ exactly? Is removing the parentheses right?

*Assuming $b$ doesn't have the parentheses, can you please show how to prove that $a$ and $b$ are what they are supposed to be? I've tried it so many times with no success.

 A: We have
$$
\frac{p(t)}{a-b p(t)}=\frac{p_0}{a-b p_0}e^{at}
$$
Then
\begin{equation}
\frac{p_1}{a-b p_1}=\frac{p_0}{a-b p_0}e^{at_1}
\end{equation}
$$
\frac{p_2}{a-b p_2}=\frac{p_0}{a-b p_0}e^{at_2}=\frac{p_0}{a-b p_0}e^{2at_1}
$$
From here we get
$$
\frac{p_2(a-bp_1)}{p_1(a-b p_2)}=e^{at_1}
$$
$$
\frac{p_1(a-b p_0)}{(a-b p_1)p_0}=e^{at_1}
$$
Then
$$
p_2(a-bp_1)(a-b p_1)p_0=p_1(a-b p_2)p_1(a-b p_0)
$$
Simplifying:
$$
a (p_0 p_2-p_1^2)+b(p_1^2(p_0+p_2)-p_0p_2(2p_1))=0
$$
and we obtain the second relation:
$$
b=\frac{a (-p_0 p_2+p_1^2)}{p_1^2(p_0+p_2)-p_0p_2(2p_1)}=\frac{a}{p_1}\frac{p_1^2-p_0p_2}{p_0p_1+p_2p_1-2p_0p_2}
$$
The first relation can be obteined from here:
$$
a-bp_1=a\left(1-\frac{ p_1^2-p_0p_2}{p_0p_1+p_2p_1-2p_0p_2}\right)
$$
$$
a-bp_0=a\left(1-\frac{p_0}{p_1}\frac{ p_1^2-p_0p_2}{p_0p_1+p_2p_1-2p_0p_2}\right)
$$
Then
$$
e^{at_1}=\frac{p_1}{p_0}\frac{1-\frac{p_0}{p_1}\frac{ p_1^2-p_0p_2}{p_0p_1+p_2p_1-2p_0p_2}}{1-\frac{ p_1^2-p_0p_2}{p_0p_1+p_2p_1-2p_0p_2}}=\frac{1}{p_0}\frac{p_1(p_0p_1+p_2p_1-2p_0p_2)-p_0(p_1^2-p_0p_2)}{p_0p_1+p_2p_1-p_0p_2-p_1^2}\\
=\frac{p_2}{p_0}\frac{p_1^2-2p_0p_1+p_0^2}{(p_2-p_1)(p_1-p_0)}=\frac{p_2(p_1-p_0)}{p_0(p_2-p_1)}
$$
