§7.4 Separable differential equations (continued)

Solving separable differential equations

For $\frac{dy}{dt} = \frac{t}{y^2}$, separation followed by integration gives

The general solution of $\frac{dy}{dt} = 3y$ is

A town's population grows at a continuous annual rate of $3\%$. If $P(t)$ is the population in year $t$, the differential equation is

For a population satisfying $\frac{dP}{dt} = kP$ with $k \gt 0$, the doubling time equals

Coffee at $T(0) = 105^\circ$F in a $75^\circ$ room obeys $\frac{dT}{dt} = -k(T-75)$. If the coffee cools at exactly one degree per minute at that moment, then

Solving $\frac{dT}{dt} = -\frac{1}{30}(T-75)$ by separation yields the family

The equation $\frac{dy}{dt}-(2-t)y = 2-t$ has general solution

The equation $\frac{1}{t}\frac{dy}{dt} = e^{t^2-2y}$ has general solution

For the town with $\frac{dP}{dt} = 0.03P$, the doubling time is $\frac{\ln 2}{0.03}$. Report the value in years, rounded to three decimals.

The town's population obeys $P = 10000e^{0.03t}$. Evaluate $P$ at the doubling time $t = \frac{\ln 2}{0.03}$, to the nearest whole person.

With $k = \frac{1}{30}$ and initial temperature $105^\circ$F, the solution is $T = 75+30e^{-t/30}$. Compute $T(20)$, rounded to one decimal.

The coffee model $T = 75+30e^{-t/30}$ reaches $80^\circ$F when $t = -30\ln\left(\frac{5}{30}\right)$. Report the time in minutes, rounded to one decimal.

The initial value problem $y' = 2y+2$, $y(0) = 2$ has solution $y = -1+3e^{2t}$. Compute $y(1)$, rounded to three decimals.

The initial value problem $y' = 2y^2$, $y(-1) = 2$ has solution $y = -\frac{2}{4t+3}$. Compute $y(0)$, rounded to three decimals.

For the coffee model $T = 75+Ce^{-t/30}$, every solution satisfies $T \to 75$ as $t \to \infty$.