# Differential equation with two special functions

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: $$f(p)=-\frac{1}{\lambda}\left(\frac{p}{1-p}\right)^\tfrac{1}{\lambda}\int_1^p(1-s)^\tfrac{1-\lambda}{\lambda}s^{-\tfrac{\lambda+1}{\lambda}}\skew3\tilde{G}(s)ds+K\left(\frac{p}{1-p}\right)^\tfrac{1}{\lambda}$$