Please can someone solve this ordinary differential equation

Question

The equation is similar to the generalised bass model

$y'(t) = g(t)\left(p + \frac{q}{n} y(t)\right) \left(n - y(t)\right) $

where the initial conditions g(0) = 0, y(0) = 1, the constant c = 1.

This is part of a market penetration model I am working one. Please show workings as I am not really good with ODE.

Thank you

Answer 1

First a substitutions $q/n=a$

$\frac{y'(t)}{(p+ay)(n-y)}=g(t)$

$\int\frac{dy(t)}{(p+ay)(n-y)}=\int g(t) dt$

$\int (\frac{1}{an+p}(\frac{a}{(p+ay)}+\frac {1}{(n-y)})dy=\int g(t)dt $

$ ln(p+ay)-ln(n-y)=(an+p)\int g(t)dt$

$ ln\frac{p+ay}{n-y}+y(0)=(an+p)\int g(t)dt +C $

Maybe this helps... If g(t) is a linear function then there would be no problems in integrating...

$ln\frac{p+ay}{n-y}=(an+p)\int g(t)dt +C-y(0)$

$\frac{p+ay}{n-y}=e^{(an+p)\int g(t)dt}$

and now you can easily get y(t)...

$K=e^{(an+p)\int g(t)dt}$

$\frac{p+ay}{n-y}=K$

$p+ay=Kn-Ky$

$y(a+K)=Kn-p$

$y=\frac{n e^{(an+p)\int g(t) dt}-p}{a+ e^{(an+p)\int g(t)dt} }$

Answer 2

This is a separable equation hence the usual methods apply. To wit, $y$ solves this if and only if $$ \frac{ny'(t)}{(np + q y(t))(n - y(t))}=g(t). $$ Note that $$ \frac{n(q+p)}{(np + q y(t))(n - y(t))}=\frac{q}{np + q y(t)}+\frac1{n - y(t)}, $$ hence $$ \frac{np+qy(t)}{n-y(t)}=c\,\mathrm e^{(q+p)G(t)}, $$ for some constant $c$, where $$ G(t)=\int_0^tg(s)\mathrm ds. $$ that is, $$ y(t)=n\,\frac{c\,\mathrm e^{(q+p)G(t)}-p}{c\,\mathrm e^{(q+p)G(t)}+q}. $$ If $y(0)=1$, then $$ c=\frac{np+q}{n-1}, $$ hence $$ y(t)=n\,\frac{(np+q)\,\mathrm e^{(q+p)G(t)}-(n-1)p}{(np+q)\,\mathrm e^{(q+p)G(t)}+(n-1)q}. $$