§7.2 Qualitative behavior of solutions to differential equations (continued)

Summary

For the differential equation $\frac{dy}{dt} = t-y$, along the line $y = t$ every segment of the slope field has slope

For $\frac{dy}{dt} = t-y$, the initial value $y(0) = -1$ picks out which solution?

In the mouse population model with predation term $M(y) = \frac{y}{2+y}$, the predation factor behaves how as $y$ grows large?

With $C = 60$ cats, the mouse model $\frac{dy}{dt} = 20y-\frac{60y}{2+y}$ has equilibria $y = 0$ and $y = 1$ (in thousands). Their stability is

For the fish model $\frac{dP}{dt} = P(6-P)-h$ with harvest $h$ thousand fish per month, a positive stable equilibrium exists precisely when

For the modified fish model $\frac{dP}{dt} = P(6-P)-1$, the smaller equilibrium solves $P^2-6P+1 = 0$. Compute it as a decimal, rounded to three places.

In the mouse model $\frac{dy}{dt} = 20y-C\frac{y}{2+y}$, a positive equilibrium exists only when $C \gt 40$. What is the smallest whole number of cats that caps the population?

For $\frac{dy}{dt} = t-y$ with $y(0) = -1$, take one Euler step with $\Delta t = 0.5$. Compute $y_1$.

For $\frac{dy}{dt} = t-y$, along the line $y = t-1$ every slope field segment has slope $1$.

For solutions of $\frac{dy}{dt} = t-y$, a local minimum can occur only where $t = y$, since the rate vanishes exactly there.

In the threshold model with equilibria $0$, $1$, $3$ and positive rate only on $(1,3)$, a population starting with $P(0) \gt 3$ decreases toward $3$ and stays above it.