All of the story problems below and most, but not all, in Section 2.3 rely on a differential equation of the form $y'=a y+b$ with constant $a,b$.
Types of questions
Find the amount of substance, dollar amount, population, temperature, etc., after a given time
Population
Find the time after which amount of substance, dollar amount, population, temperature, etc., after a given time.
Determine values of the parameters that yield the desired outcome.
Mixing Population
Solve a practical problem using information about the model provided in the question.
Money
Cooling Population
Problem to solve
Money Type A: easy
A college freshmen would like to purchase of a used Tesla S as a graduation present for himself in 5 years. He decides to save 500 a month for 5 years, which he invests in CDs that bring annual interest rate of 5%. Determine the price of a car that he will be able
to purchase.
Solution His yearly savings are $500\times 12=6000$, so we want to solve
$$B'=.05 B+6000$$ with the initial condition $B(0)=0$, and determine the value of $B(5)$.
Solution from WolframAlpha
is $B(t)=-120000 + 120000 e^{0.05 t}$.
Answer: $B(5)=120000 e^{0.25}-120000\approx 34083.1$
Mixing Type A: easy
A tank originally contains 100 gal of fresh water. Then water
containing ${\frac 12}$ lb of salt per gallon is poured into the tank at a
rate of 2 gal/min, and the mixture is allowed to leave at the same rate.
Setup the differential equation for the amount $A(t)$ of salt in the
tank at time $t$.
Find the amount of salt in the tank after 25 minutes.
Solution
The "rate in - rate out" formula gives $A'=2\times \frac12-2\times \frac{A}{100}$.
Solving the equation $A'(t)=1-A/50$ by hand, we get:
$$
\frac{dA}{dt}=-\frac{A-50}{50}
$$
So variables separate,
$$
\frac{dA}{A-50}=-\frac{dt}{50}
$$
and we can integrate
$$
\int \frac{dA}{A-50}=-\int\frac{ dt}{50}
$$
Integrating, we get the implicit solution $\ln |A-50|=-t/50+c$ which yields the explicit solution $A(t)=50+Ce^{-t/50}$. Since $A(0)=0$ (fresh water), we get
$C=-50$ so
$$
A(t)=50-50e^{-t/50}
$$
The answer is: $A(25)=50-50/\sqrt{e}\approx 19.67$
Cooling Type A: easy Roasting instruction on a frozen turkey
wrapping is as follows.
Thaw the turkey to 40F
Preheat oven to 350F
Put the turkey into the oven for 3 hours.
What is the temperature of the turkey after the roast?
Use Newton's law of cooling: the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Use proportionality constant of $1/5$.
Solution Let $T=T(t)$ denote the temperature of the turkey as in increases during baking.
The Newton law of cooling says that $T'=k(350-T), T(0)=40$ and the statement gives the value of $k=1/5$.
This is a linear and separable equation, Solving the equation as separable, by hand:
$$
\frac{dT}{dt}=-\frac15(T-350)
$$
$$
\frac{dT}{T-350}=-\frac15dt
$$
$$
\int \frac{dT}{T-350}=-\int \frac15dt
$$
$$
\ln |T-350|=-t/5+C
$$
$$
T -350=\pm e^{-t/5+C}=\pm e^C \times e^{-t/5}=C_1 e^{-t/5}
$$
So $T=350+=C_1 e^{-t/5}$ and since $T(0)=40$, we get $T(t)=350-310 e^{-t/5}$. After $3$ hours, the temperature is $T(3)=350-310 e^{-3/5}=179.86839281085\approx 179.9$
Money Type B: moderate, but on the easy side A college graduate would like to purchase of a Tesla S for 50,000 as a belated graduation present for himself. He decides to save 500 a month, which he invests in CDs that bring annual interest rate of 5%.
How long does he have to wait?
Solution His yearly savings are $500\times 12=6000$, so we want to solve
$$B'=.05 B+6000$$ with the initial condition $B(0)=0$, and determine the value of $B(5)$.
Solution from WolframAlpha
is $B(t)=-120000 + 120000 e^{0.05 t}$. Using again WolframAlpha, he solves the equation $-120000 + 120000 e^{0.05 t}=50000$ for $t$ and gets his answer: $t=20\ln \frac{17}{12}\approx 6.97$ years.
Mixing Type B: moderate, but on the easy side
A tank originally contains 100 gal of fresh water. Then water
containing ${\frac 12}$ lb of salt per gallon is poured into the tank at a
rate of 2 gal/min, and the mixture is allowed to leave at the same rate.
Setup the differential equation for the amount $A(t)$ of salt in the
tank at time $t$.
At what moment of time the amount of salt in the tank exceeds 25 lb?
SolutionSolution
The "rate in - rate out" formula gives
$A'(t)=1-A/50$ Solving the equation by hand, we get:
$$
\frac{dA}{dt}=-\frac{A-50}{50}
$$
So variables separate,
$$
\frac{dA}{A-50}=-dt/50
$$
and we can integrate
$$
\int \frac{dA}{A-50}=-\int dt/50
$$
Integrating, we get $\ln |A-50|=-t/50+c$ or $A(t)=50+Ce^{-t/50}$. Since $A(0)=0$ (fresh water), we get
$$
A(t)=50-50e^{-t/50}
$$
$A(t)=25$ when $e^{-t/50}=1/2$, or $t=50\ln 2\approx 34.7$ min.
Cooling Type B: moderate Roasting instruction on a frozen turkey
wrapping is as follows.
Thaw the turkey to 32F
Preheat oven to 350F
Put the turkey into the oven.
Your turkey is ready to eat when its temperature reaches 180F.
How long should the turkey be on the oven?
Use Newton's law of cooling: the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Use proportionality constant of $0.2$
Solution Let $T=T(t)$ denote the temperature of the turkey as in increases during baking.
The Newton law of cooling says that $T'=k(350-T), T(0)=40$ and the statement gives the value of $k=0.2=1/5$.
WolframAlpha says that
$T(t)=350-318 e^{-t/5}$. Solving the equation $350-310 e^{-t/5}=180$ we get $t=5 \ln\frac{159}{51}\approx 3.13$ hours.
Money Type B: moderate, but on a harder side A person got infected with 2000 viruses. The immune system clears 80% of viruses daily (daily rate of .8) but additional viruses arrive at a rate of 500 per day.
When will the virus load fall below a dangerous level of 1000? When will the infection clear?
Solution Daily load of viruses $V=V(t)$ changes according to differential equation $V'=-.8 V+ 500$
We solve it with the initial condition $V(0)=2000$, and determine the value of $t$ when $V(t)=1000$.
Solution from WolframAlpha
is $V(t)=625 + 1375 e^{-0.8 t}$. The infection will never clear on its own. It will subside to manageable viral load of 1000 in about
$t=\frac45 \ln \frac{11}{3} \approx 1.62$ days.
Money Type C: harder
A college freshmen would like to purchase of a Tesla S for 50,000 as a graduation present for herself. She will be graduating in 4 years, and she found a high yield CD (Certificate of Deposit) with annual interest rate of 5%.
How much does she need to be saving per month?
Solution Her yearly savings are $12 m$, where $m$ is the number to be determined. So she sets up and solves differential equation
$$B'=.05 B+12 m$$ with the initial condition $B(0)=0$,
Solution from WolframAlpha is
is $B(t)=\left(-240 + 240 e^{0.05 t}\right) m$. The equation for the unknown value $m$ is $B(4)=50000$ i.e. $ (-240 + 240 e^{0.05 \times 4}) m=50000$. Using her scientific calculator, she determines the required monthly rate of saving to be $m=940.97$.
Answer: $m=940.97 $
Type D: hard
A tank originally contains 100 gal of fresh water and has capacity of 200 gal. Then water
containing ${\frac 12}$ lb of salt per gallon is poured into the tank at a
rate of 3 gal/min, and the mixture is allowed to leave at a rate of 2 gal/min.
Setup the differential equation for the amount $A(t)$ of salt in the
tank at time $t$.
What is the amount of salt in the tank at the time of overspill?
Solution Water level increases at the rate of 1 gal/min, so the volume of water at time $t$ is $V(t)=100+t$. The water with overspill at $t=100$ min.
The "rate in - rate out" formula gives
$A'(t)=3\times \frac12 -2\times \frac{A}{100+t}$ and $A(0)=0$ (fresh water initialy). Solving the equation by
WolframAlpha
we get $A(t)=\frac{t (30000 + 300 t + t^2)}{2 (100 + t)^2}$ and $A(100)=\frac{175}{2}=87.5 $
Type C: harder
Roasting instruction on a frozen turkey
wrapping is as follows.
Thaw the turkey to 40F
Preheat oven to 350F
Put the turkey into the oven for 3 hours.
Your turkey is ready to eat when its temperature reaches 180F.
I don't have time to thaw my turkey.
I need to roast my frozen turkey (0F) in
the same amount of time (3 hours).
What
temperature should I set on the oven to preheat? Use Newton's law of cooling: the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings.
(proportionality constant is not given!)
Solution
Let $T=T(t)$ denote the temperature of the turkey as in increases during baking. Let $T_o$ denote the
unknown setting for the temperature
of the oven.
The Newton law of cooling says that $T'=k(350-T), T(0)=40$ in the first case, and
$T'=k(T_o-T), T(0)=0$ in the second case.
Thus $T(t)=350-310e^{-kt}$ for the first case and $T(t)=T_o(1-e^{-kt})$ for the second one.
Now compute $T_o$ from the condition that $T(3)=180$ in both cases. From the
first equation $e^{-3k}=17/31$ thus $1-e^{-3k}=\frac{14}{31}$ and $T_o=31/14\times 180=398.6\approx 400 F$.