We are given a logistic growth model with constant harvesting as:
$\frac{dN}{dt} = rN(1-\frac{N}{K})-Y_0$
We are asked to show that the recovery time for harvesting a yield $Y_0$, $T_R(Y_0)$, satisfies
$\frac{T_R(Y_0)}{T_R(0)} = \frac{1}{(1-\frac{Y_0}{Y_M})^{1/2}}$
We define recovery time as the time taken for a pertubation to decrease by a factor of e.
I am aware of the fact that: $T_R(Y_0) = \frac{1}{r}$
However, apart from this I am mostly lost. How would one approach this problem? Would we find the stable points and use linearization from that point, by finding dn/dt? Or would we directly solve the integral and find the recovery time?
First of all, you have $$ T_R(0)=\frac{1}{r}\,, $$ not $T_R(Y_0)$.
To do the computations, you also need first to show that the maximal yield in this model is given by $$ Y_m=\frac{rK}{4}\,. $$
After this is just arithmetic. First, for $Y_0>0$ you have two equilibria, but only the larger one (stable) is of interest. This equilibrium has the coordinate $$ \hat N=\frac{-r+\sqrt{r^2-\frac{4rY_0}{K}}}{\frac{-2r}{K}}\,. $$ Since you need to linearize about this point, plug $N=u+\hat N$ in your equation and keep only the linear terms in $u$, you should find $$ \dot u=\left(r-\frac{2r \hat N}{K}\right)u+\ldots, $$ now $$ T_R(Y_0)=\frac{1}{r-\frac{2r \hat N}{K}}=\frac{1}{\sqrt{r^2-\frac{4rY_0}{K}}}. $$ Now you can calculate that $$ \frac{T_R(Y_0)}{T_R(0)} $$ is exactly what is being asked in your problem.