The following equation has come up in my research and I am lost at where to start. I have tried guessing forms of the solution and Mathematica is not helpful. Any help pointing me in the right direction or useful resources would be greatly appreciated.
$$p''(x) + \bigg(1-K\Big(N-\int_0^\infty tp(t)dt\Big)\bigg)p'(x)=0$$
The equation is defined in real space, all the parameters are positive, and $\Big(N-\int_0^\infty tp(t)dt\Big)>0$.
On
Starting from @Masd's solution $$ p(x) = -\frac{\dot p(0)}{(1−K(N−\int_0^{\infty} tp(t)dt))}e^{-Ax}+C = -\frac{\dot{p}(0)}{A}e^{-Ax} + C, $$ we can immediately see that $C=0$ is necessary for the improper integral to be convergent. Now define $M = \int_0^\infty tp(t)~\mathrm{d}t$ so that $A = 1-K(N-M)$. Now multiply $p$ by $x$ and integrate on $(0,\infty)$ to obtain $$ M = -\frac{\dot{p}(0)}{A^3} = -\frac{\dot{p}(0)}{(1-K(N-M))^3}, $$ which yields a nonlinear equation for $M$ in terms of $\dot{p}(0)$: $$ M(1-K(N-M)^3) +\dot{p}(0)=0. $$ This polynomial has even order in $M$ and hence there are some values of $\dot{p}$ which will yield no real solutions. From this we can also recover the trivial solution corresponding to $\dot{p}(0) = M = 0$. Assuming nontrivial solutions, we can look for solutions to $$ 1-K(N-M)^3 = -\frac{\dot{p}(0)}{M}. $$ A pair of roots $M_- <0 < M_+$ exists for every $\dot{p}(0)<0$ by the intermediate value theorem by noting that for $M>0$, the LHS is bigger for sufficiently large $M$ and smaller for $M$ sufficiently close to zero, and vice versa for $M<0$.
Note that once a root $M^*$ is found, the solution is completely determined to be $$ p(x) = -\frac{\dot{p}(0)}{1−K(N−M^*)}e^{-(1−K(N−M^*))x}. $$ Also note that $M=N$ cannot be a root for $K,N\neq 0$
Edit: $M<0$ for $\dot{p}(0)<0$ seem to be incompatible with the convergence of the integral, so perhaps more constraints on the roots are necessary.
If $$(1−K(N−\int_0^{\infty} tp(t)dt))$$ is not dependent on the variable $x$, we can just say $$A = (1−K(N−\int_0^{\infty} tp(t)dt))$$ and we get $$ \ddot p(x) + A \dot p(x) = 0$$ We can also say $$R(x) = \dot p(x)$$ $$ \mathcal{L}(\dot R(x)) = sr(s)-R(0)$$ $$ \mathcal{L}(-R(x)A) = -Ar(s)$$ $$ sr(s)-R(0) = -Ar(s)$$ $$ r(s)(s+A) = R(0)$$ $$ r(s) = \frac{R(0)}{s+A}$$ $$ \mathcal{L}^{-1}(\frac{R(0)}{s+A}) = R(0)e^{-Ax}$$ So $$p(x)=\dot p(0)\int {e^{-Ax}}dx$$ $$p(x) = \frac{-\dot p(0)}{A}e^{-Ax}+C$$ $$p(x) = -\frac{\dot p(0)}{(1−K(N−\int_0^{\infty} tp(t)dt))}e^{-(1−K(N−\int_0^{\infty} tp(t)dt))x}+C$$ !Do note that actually $$\int_0^{\infty} tp(t)dt$$ is a constant. this is obvious because this integral is definite. Perhaps the easiest way to evaluate it is using known data at points. E.g $p(0) = a$ and $p(1)=b$ and so on on till we could evaluate all the unknown constants. It is unlikely that is possible to understand the value of the double integral of this function at $\infty$ and $0$, and so with the first integral.