Verify that $x(t)=x_0e^{at}\text{ where }x_0=x(0)$ is the solution to the following differential equation: $$\frac{dx}{dt}=ax$$
This follows directly through differentiation:
$$\frac{dx}{dt}=x_0ae^{at}=a\frac{dx}{dt}$$
Verify that $\rho(t) = \rho_0e^{(a-b)t}\text{ where }\rho_0=\rho(0)$ is the expression for $\rho(t)=\frac{x(t)}{y(t)}$ where: $$\frac{dx}{dt}=ax\qquad\frac{dy}{dt}=by$$
Similarly, follows from differentiation and substitution:
$$\frac{d\rho}{dt}=\frac{\frac{dx}{dt}y-\frac{dy}{dt}x}{y^2}=\frac{axy-bxy}{y^2}=(a - b)\frac{x}{y}=(a-b)\rho$$
Using question 1 gives the required result.
What is the mean fitness in a constant sized population with:
This question is just a substitution exercise asking for $\phi=ax+by$
Show that the following system of differential equations has 3 potential stable solutions (what are they?) for $x+y=1$: $$\frac{dx}{dt}=x(a - \phi)\qquad\frac{dy}{dt}=y(b - \phi)$$
$$\frac{dx}{dt}=x(a-ax-b(1-x))=x(1-x)(a-b)$$
The three stable solutions are solutions to the equation $\frac{dx}{dt}=0$: