I was wondering if there is a special technics on this problem. I have two differential equation with respect to time $t$.
$\dot{p}=p(1-p) \lambda $ where $\lambda>0$
$\dot{\pi} = \pi (1-\pi) \alpha$ where $\lambda>\alpha>0$
And finally $p_{0}=\pi_{0}=c$ stating that the initial points are the same.
Once solved the differential equations, I know that $p=\frac{e^{\lambda t}}{c+e^{\lambda t}}$ and $\pi=\frac{e^{\alpha t}}{c+e^{\alpha t}}$
However, my problem is that I have another differential equation like $f(p)= g(p,\pi, \frac{d \pi}{dp})+f’(p)p(1-p))\lambda $
I guess there would be no closed-form solution, but at least I would like to know what kind of problem I am trying to solve.
The equation $f(p)$ per se is not expressed with time variable $t$ but with $p$ and $\pi$. However, both are functions of $t$.
Thank you very much!
Since $\pi$ can be expressed in terms of $p$: $$ \pi(p)=\frac{\left(\frac{cp}{1-p}\right)^\tfrac{\alpha}{\lambda}}{c+\left(\frac{cp}{1-p}\right)^\tfrac{\alpha}{\lambda}} $$ The problem boils down to: $$ f'(p)=f(p)\frac{1}{\lambda p(1-p)}+\skew3\tilde{G}(p) $$ This is a linear first order ODE and can be solved explicitly: $$ f(p)=-\frac{1}{\lambda}\left(\frac{p}{1-p}\right)^\tfrac{1}{\lambda}\int_1^p(1-s)^\tfrac{1-\lambda}{\lambda}s^{-\tfrac{\lambda+1}{\lambda}}\skew3\tilde{G}(s)ds+K\left(\frac{p}{1-p}\right)^\tfrac{1}{\lambda} $$