I've been trying to solve this one for a while, but I still can't make it. Here's the problem.
I have a $f_1(t;\rho,\nu)$ that for $t\to\infty$ and for $\rho>\nu$ goes as $f_1\sim t^{\nu/\rho}.$
Saying that, the problem is with a second function $f_2(t, \rho\, \nu)$ which obeys the equation:
$\frac{df_2(t)}{dt}=\frac{\nu f_2(t) + (\nu+1)f_1(t)}{\rho t + (\nu+1)[f_1(t)+f_2(t)]}$
with $\rho,\nu\in\mathbb{N}^+$ and $f_2(t)>0$ (and increasing).
I am interested in the long term behavior as for $f_1$. I know (from simulations) that for $t\to\infty$ it behaves similarly to $f_1$, but with a different exponent. I've tried to plug in the $f_1$ written above (assuming $\rho>\nu)$ and also to force $f_2(t)\sim ct^\alpha$ as a solution, ultimately looking for an expression for $\alpha$ as a function of the other parameters, but I only arrived to contradict myself.
I've also tried on Mathematica with initial conditions $f_2(0)=0$ but without succeeding.
Any tips?
Thanks in advance!!
It seems to me that it might have the same exponent, but logarithmic terms. Rather than try to solve the differential equation for $f_2(t)$, I tried plugging in a form for $f_2(t)$ and solving for $f_1(t)$. I found that $$ f_2(t) = \frac{\nu + 1}{\rho} \log(t)\; t^{\nu/\rho} $$ is a solution with $$ f_1(t) = {\frac {1}{\rho} \left( {t}^{{\frac {\nu+ \rho}{\rho}}}{\rho}^{3}+\ln \left( t \right) {t}^{2\,{\frac {\nu}{ \rho}}} \left( \nu+1 \right) ^{2} \left( \ln \left( t \right) \nu+ \rho \right) \right) \left( - \left( \nu+1 \right) \left( \ln \left( t \right) \nu+\rho \right) {t}^{{\frac {\nu}{\rho}}}+{\rho}^{2 }t \right) ^{-1}} \sim t^{\nu/\rho} $$