Differential Equation Logistic Curve NOT A DUPLICATE - see comments below
I have to find P1 where the other question does not. Also the A = some function equation is different from mine.




I get this far and realize if I substitute nothing will really cancel except an X and P0. I think if I solve for X and substitute that in, it'll just make add a really messy polynomial on one side. I'll try that now, but I think I'm missing something on the set up.
I'll post the next attempt when it feels futile. I've already spent an hour on this problem and we didn't cover this section in class and I really want to see where I'm going wrong.
 A: I know this question was asked about 8 years ago, but it is never too late if the outcome is good!
Following the hint, we plug into (21) $\chi = \exp(-Ap_1t_a)$ and $\chi^2 = \exp(-Ap_1t_b)$, where
$$
\frac{p_ap_1\chi}{p_1-p_a(1-\chi)} = \frac{p_bp_1\chi^2}{p_1-p_b(1-\chi^2)} \Rightarrow \chi = \frac{p_a(p_1-p_b)}{p_b(p_1-p_a)}.
$$
Inserting this value into (21),
$$
\begin{aligned}
p_0 &= \frac{p_a^2p_1(p_1-p_b)}{p_1p_b(p_1-p_a) - p_a\left[p_b(p_1-p_a) - p_a(p_1-p_b)\right]} \\
&\Rightarrow (p_a^2-p_0p_b)p_1^2 = (p_ap_b - 2p_0p_b + p_0p_a)p_ap_1 \\
&\Rightarrow \boxed{p_1 = \frac{p_ap_b - 2p_0p_b + p_0p_a}{p_a^2-p_0p_b}p_a.}
\end{aligned}
$$
Notice that
$$
\begin{aligned}
p_1-p_b &= \frac{p_a^2p_b - 2p_0p_ap_b + p_0p_a^2 - p_a^2p_b + p_0p_b^2}{p_a^2-p_0p_b} = \frac{p_0(p_b - p_a)^2}{p_a^2 - p_0p_b}, \\
p_1-p_a &= \frac{p_a^2p_b - 2p_0p_ap_b + p_0p_a^2 - p_a^3 + p_0p_ap_b}{p_a^2-p_0p_b} = \frac{p_a(p_a-p_0)(p_b-p_a)}{p_a^2 - p_0p_b}.
\end{aligned}
$$
Therefore,
$$
\chi = \frac{p_a(p_1-p_b)}{p_b(p_1-p_a)} = \frac{p_0(p_b-p_a)}{p_b(p_a-p_0)},
$$
so
$$
\chi = \exp(-Ap_1t_a) \Rightarrow \boxed{A = \frac{1}{p_1t_a}\log\frac{p_b(p_a-p_0)}{p_0(p_b-p_a)}.}
$$
