Let $\alpha>0$, $p>1$, $T>0$, $c\ge0$ and $y_0>0$. Are we able to find an explicit form for the solution $y:[0,T]\to\mathbb R$ of $$y(t)=y_0-ct+\alpha\int_0^t(t-s)y(s)^p\:{\rm d}s$$ for all $t\in[0,T]$?
It seems like this is some kind of Bernoulli equation, but I'm not sure ...
Differentiating the integral equation twice we obtain $$ y'(t)=-c+\alpha\int_0^t y(s)^p\,ds,\qquad y''(t)=\alpha y(t)^p. \tag{1} $$ The first integral of $y''=\alpha y^p$ is $$ \frac{1}{2}(y')^2=\frac{\alpha}{p+1}\left(y^{p+1}-y_0^{p+1}\right)+\frac{1}{2}c^2, \tag{2} $$ where we used the initial conditions $y(0)=y_0$ and $y'(0)=-c$. Integrating again we obtain $$ t=-\int_{y_0}^{y}\frac{dz}{\sqrt{c^2+\frac{2\alpha}{p+1}\left(z^{p+1}-y_0^{p+1}\right)}}. \tag{3} $$ I suspect this is the farther one can go without specifying the value of $p$.