#2,pg.76: Determine an interval in which the solution of the given initial value problem is certain to exist: $t(t-4)y'+y=0;y(2)=1$.
Solution: First we need to express this equation in standard form: $$y' + \dfrac{1}{t(t-4)}y'+y=0.$$ The function $p(t)=\dfrac{1}{t(t-4)}$ has discontinuities at the values $t=0,4$. Therefore the possible choices for intervals of validity are $(-\infty,0)$, $(0,4)$, and $(4,\infty)$. Our initial condition is $y(2)=1$, which means we know a solution exists on the interval $(0,4)$, since $2$ is in that interval.
#4,pg.88: Sketch the graph of $y'$ vs $y$, determine the equilibrium solutions, draw a phase line for these solutions, and classify the equilibria of $y'=e^y-1$.
Solution: First we find the equilibrium solution(s): set $e^y-1=0$ and so we see that $y=\log(1)=0$. Now we plot $y'$ as a function of $y$:
This analysis yields the phase line . From this it is clear that the equilibrium solution $0$ is unstable.