I am trying to return the following equation back to a differential equation form but I am rather stuck, and I am not sure how it can be done. I have read posts on here but to no avail.
I want to turn
$$ p(t) = \frac{\int_{t}^{\infty}\mu e^{\rho(\tau)} d\tau}{ 1 + \int_{t}^{\infty}\mu e^{\rho(\tau)} d\tau }, \quad \text{where} \quad \rho(t) = \int_t^\infty (\lambda(t') - \mu )dt'. $$
So far I've done the following steps: $$ \frac{p}{1 - p} = \int_{t}^{\infty}\mu e^{\rho(\tau)} d\tau $$ $$ \frac{p}{1-p} = q(t) = - \int_{\infty}^{t}\mu e^{\rho(\tau)} d\tau $$ $$ \frac{dq}{dt} = - \mu e^{\rho(t)} $$ $$ \frac{dp}{dt} = -(1-p)^2\mu e^{\rho(t)} $$ However I am almost certain that the solution should have no exponetial term on the RHS and should be a quadratic with $\lambda(t) \text{ and } \mu$ as coefficients.
I would appreciate any and all help, thanks!