Which is which? Draw a directional field! Or make a phase diagram like in the book.
If $y'=(10-y)(20-y)$ and $y(0)=$ , without solving the equation determine $\lim_{t\to\infty}y(t)$.
Solution $y=10$ is a stable equilibrium and $y=20$ is an unstable equilibrium. So all initial conditions above 20 will give $\lim_{t\to\infty}y(t)=\infty$ and all initial conditions below 20 will give $\lim_{t\to\infty}y(t)=10$. This is easier to see from the equation than from the solution WolfA Answer: $10$
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 $t$. Without solving the equation determine the amount of salt in the tank after long time.
Solution The "rate in - rate out" formula gives $A'=2\times \frac12-2\times \frac{A}{100}=1-A/50$. So the equation is $A'=1-A/50$. The equilibrium $A=50$ is a stable equilibrium, so $\lim_{t\to \infty}A(t)=50$.
Note that this answer is "intuitively obvious": eventually we will have 100 gallons of water with concentration 1/2 Lb/gallon
If $y'=y(10-y)(y-20)$ and $y(0)= $ , without solving the equation determine $\lim_{t\to\infty}y(t)$.
Solution There are three equilibria at $y=0,10,20$, $y=0$ and $y=20$ are both stable equilibria and $y=10$ is an unstable equilibrium. So all initial conditions above 10 will give $\lim_{t\to\infty}y(t)=20$ and all initial conditions below 10 will give $\lim_{t\to\infty}y(t)=0$. Answer: $20$
If $y'=y(y-10)(20-y)$ and $y(0)=$ , without solving the equation determine $\lim_{t\to\infty}y(t)$.
Solution There are three equilibria at $y=0,10,20$, $y=0$ and $y=20$ are both stable equilibria and $y=10$ is an unstable equilibrium. So all initial conditions above 10 will give $\lim_{t\to\infty}y(t)=20$ and all initial conditions below 10 will give $\lim_{t\to\infty}y(t)=0$. Answer: $0$
If $y'=\cos y$ and $y(0)=1$, without solving the equation determine $\lim_{t\to\infty}y(t)$.
Solution There are infinitely many equilibria $\dots, -\frac{\pi}{2},\frac{\pi}{2},\frac{3}{2}\pi, \frac{5}{2}\pi,\dots$. The two closest to the initial value 1 are $y=-\pi/2$, which is unstable, and $y=\pi/2$ which is stable. So $\lim_{t\to\infty}y(t)=\frac{\pi}{2}$.
If $y'=\sin y$ and $y(0)=1$, without solving the equation determine $\lim_{t\to\infty}y(t)$.
Solution The equlibria closest to $y_0=1$ are $y=0$, which is unstable, and $y=\pi$ which is stable. So $\lim_{t\to\infty}y(t)=\pi$
Cooling Type B: moderate
Roasting instruction on a frozen turkey wrapping is as follows.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 $0.2$
- 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.
Solution Let $T=T(t)$ denote the temperature of the turkey as in increases during roasting. 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 $V(t)=625 + 1375 e^{-0.8 t}$ years. According to this model, the infection will never clear on its own. But it will subside to a 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 she needs 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 overespil?l
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.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!)
- 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.
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 $T_o=31/14\times 180=398.6\approx 400 F$.
Population Type C:
Solution
Money Type D:
A college graduate borrows 50,000 to buy a Tesla S at an interest rate of 6%. Anticipating steady salary increases, he expects to steady increase his monthly payments, with $m(t)=500+10 t$ after $t$ months.Assuming that this payment schedule can be maintained, what will be the amount of loan in 5 years? Will the loan be fully paid in 6 years?
SolutionLets use months as units of $t$. The loan amount changes at the rate $B'= \frac{.06}{12} B - (500+ 10 t)$ with the initial condition $B(0)=50000$. This is a linear but non-separable equation. Solution from WolframAlpha is $B(t)=500000. - 450000. e^{0.005 t} + 2000. t$. Answer: $B(60)=12563.5$ (From the graph of $B(t)$, the loan will be paid of in about 71.17 months.)
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 overespil?l
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 $