§7.1 An introduction to differential equations (continued)

Solving a differential equation

Summary

For the differential equation $\frac{dv}{dt} = 1.5-0.5v$, the function $v(t) = 3+2e^{-0.5t}$ is

The function $v(t) = 3$ is a solution to $\frac{dv}{dt} = 1.5-0.5v$ because

The family $v(t) = 3+Ce^{-0.5t}$ with arbitrary constant $C$ consists of

Testing $v(t) = 1.5t-0.25t^2$ in the equation $\frac{dv}{dt} = 1.5-0.5v$ shows that

Coffee obeying $\frac{dT}{dt} = -\frac{1}{15}T+5$ with $T(0) = 105$ sits in a room whose temperature is

Coffee at $T(0) = 105^\circ$F cools with $\frac{dT}{dt}\big|_{T=105} = -2$ degrees per minute. Use the linear approximation to estimate $T(1)$.

Bananas on a scale oscillate with $h(t) = 4\sin(3t)$ satisfying $\frac{d^2h}{dt^2} = -kh$. Find the spring constant $k$.

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

The function $y(t) = t+1+2e^t$ solves $\frac{dy}{dt} = y-t$. Compute its value $y(0)$.

For $\frac{dy}{dt} = y-t$, which of the listed functions fails to be a solution?

For an autonomous population model $\frac{dP}{dt} = f(P)$ whose graph crosses zero at $P = 1$, $3$, and $4$ with $f \gt 0$ on $(1,3)$, a population starting between $1$ and $3$ increases toward $3$.

In the threshold growth model with $f \lt 0$ on $(0,1)$ and $f \gt 0$ on $(1,3)$, an initial population with $0 \lt P(0) \lt 1$ declines toward extinction at $P = 0$.