What am I doing wrong in the derivation of Bass diffusion model? I've been deriving Bass diffusion model and keep consistently finding a different result than Bass' original answer. To make things worse, every single link in the Google results page just copies the Bass original solution, while I am finding the different one. You don't have to know the model, just let me show you the math part, so you can check my solution.

The premise:

We have two formulations of the hazard rate --- as Bayesian conditional probability and as a linear function:

\begin{equation}
 \frac{f(t)}{1-F(t)} = p + q F(t)
\end{equation}

where $f(t) = \frac{dF(t)}{dt}$. This gives the following differential equation:

\begin{equation}
 \frac{dF(t)}{dt} = (1 - F(t)) (p + qF(t))
\end{equation}

My solution

Using Chain Rule, we rewrite this as:

\begin{equation}
 \int \frac{dF(t)}{(1-F(t))(p+qF(t))} = \int dt = t
 \label{eq:diff}
\end{equation}

Notice that:

\begin{equation}
 \frac{1}{(1-F(t))(p+qF(t))} = \left( \frac{1}{p+q} \right) \left( \frac{q}{p+qF(t)} + \frac{1}{1-F(t)} \right),
\end{equation}

substituting this to the above equation, implies:

\begin{equation}
 \int \frac{q}{p+qF(t)} dF(t) - \int \frac{-1}{1-F(t)} dF(t) = (p + q) t
\end{equation}

integrating yields:

\begin{equation}
 \log (p + qF(t)) - \log (1 - F(t)) = (p + q) t
\end{equation}

using log properties:

\begin{equation}
 \log \left( \frac{p + qF(t)}{1 - F(t)} \right) = (p + q) t
\end{equation}

or:

\begin{equation}
 \frac{p + qF(t)}{1 - F(t)} = e^{(p+q)t}
\end{equation}

cross producting the fraction yields:

\begin{equation}
 p + qF(t) = e^{(p+q)t} - e^{(p+q)t} F(t)
\end{equation}

finally:

\begin{equation}
 (q + e^{(p+q)t}) F(t) = (e^{(p+q)t} - p) 
\end{equation}

or:

\begin{equation}
 F(t) = \frac{e^{(p+q)t} - p}{e^{(p+q)t} + q}
\end{equation}

The problem:

But Bass, somehow found the following:


cancelling out my answer does not yield Bass' answer. The unnecessary $q$ is ruining everything.

Can you please help me with this inconsistency?

Thank you.
You are missing an integration constant

$$
\log\left(\frac{p + qF(t)}{1 - F(t)}\right) = (p + q)t + \color{red}{\tilde{C}}
$$

This constant you can name it whatever you want, I'm going to name it as

$$
\color{red}{\tilde{C}} = \color{blue}{C}(p + q) + \ln q
$$

where $C$ is just another constant. So I basically changed one constant for  another one (completely allowed). Now the problem becomes

\begin{eqnarray}
\ln\left(\frac{p + qF(t)}{1 - F(t)}\right) &amp;=&amp; (p + q)t + \color{blue}{C}(p + q) + \ln q = (p + q)(t + C) + \ln q\\
p + qF(t) &amp;=&amp; e^{(p + q)(t + C) + \ln q}(1 - F(t)) = qe^{(p + q)(t + C)}(1 - F(t)) \\
[q + qe^{(p + q)(t + C)}]F(t)&amp;=&amp; qe^{(p + q)(t + C)} - p \\
F(t) &amp;=&amp; \frac{1}{q} \frac{qe^{(p + q)(t + C)} - p}{1 + e^{(p + q)(t + C)}}
\end{eqnarray}

Rearranging a bit the terms

$$\bbox[5px,border:2px solid blue]
{
F(t) = \frac{1}{q}\frac{q - pe^{-(p + q)(t + C)}}{1 + e^{-(p + q)(t + C)}}
}
$$