For the Lotka-Volterra system below $$\frac{dF}{dt}=-aF+\alpha FR$$ $$\frac{dR}{dt}=bR-\beta FR$$ show that $V=R^aF^b e^{-\alpha R-\beta F}$ is a first integral, that is, $V(t)$ is constant along any trajectory. What can you conclude about the behaviour of the solutions?
My attempt:
$$\frac{dF}{dt}=-aF+\alpha FR \iff \frac{dF}{-aF+\alpha FR}=dt \ \ \ \ \ (1)$$ $$\frac{dR}{dt}=bR-\beta FR\iff \frac{dR}{bR-\beta FR}=dt \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ Equating $(1)$ and $(2)$ \begin{align} \frac{dF}{-aF+\alpha FR}&=\frac{dR}{bR-\beta FR} \\ \int \frac{b}{F}-\beta \ dF&=\int -\frac{a}{R}+\alpha \ dR \\ \ln|F^b|+\ln|R^a|&=\alpha R+\beta F+C \\ F^bR^a&=e^{\alpha R+\beta F}e^C \\ V&=R^aF^be^{-\alpha R-\beta F} \ \ \ \ \ (V=e^C\in\mathbb{R}) \end{align} Is this a correct method? I do not know what this tells us about the behaiour of the solutions.
The question didn't ask you to find a first integral, you just need to verify the one suggested to you. Suppose that we have a solution $(F(t),R(t))$, and consider $$\frac{\mathrm{d}}{\mathrm{d} t} V(F(t),R(t)) =\frac{\partial V}{\partial F} \frac{\mathrm{d} F}{ \mathrm{d} t}+\frac{\partial V}{\partial R} \frac{\mathrm{d} R}{\mathrm{d} t}=\left(b R^a F^{b-1}\mathrm{e}^{-\alpha R-\beta F}- \beta R^a F^b \mathrm{e}^{-\alpha R-\beta F} \right)\left(-a F+\alpha F R \right)+\left(a R^{a-1} F^b \mathrm{e}^{-\alpha R-\beta F}-\alpha R^a F^b \mathrm{e}^{-\alpha R-\beta F} \right) \left( bR+\beta FR \right). $$ This is zero if you actually meant to use $-\beta$ (or $-\alpha$) in your original equations.