Constant coefficient homogeneous equations: $ay''+by'+cy=0$ are solved by seeking solutions of the form $y=e^{rt}$.
Which of the following is the general solution of $y''+5y'+6y =0$?
- $y=C_1e^{2t}+C_2e^{3t}$
- $y=C_1e^{-3t}+C_2e^{-2t}$
- None of the above
Answer: B
Solution
- Method 1 (by hand): substituting $y=e^{rt}$ we get the characteristic equation $r^2+5r+6=0$. The roots are $$r_1=\frac{-5-\sqrt{5^2-4\times 6}}{2}=\frac{-5-1}{2}=-3$$ and $r_2=\frac{-5+1}{2}=-2$. By the principle of superposition, the general solution is (B)
- Method 2: WolframAlpha with warning!
- Method 3: ODEsolver
Which of the following is the general solution of $y''+3y'-2y =0$?
- $y=C_1e^{2t}+C_2e^{t}$
- $y=C_1e^{-t}+C_2e^{-2t}$
- None of the above
Answer: C
Solution
- Method 1 (by hand): substituting $y=e^{rt}$ we get the characteristic equation $r^2+3r-2=0$. The roots are $$r_1=\frac{-3-\sqrt{3^2-4\times (-2)}}{2}=\frac{-3-\sqrt{17}}{2}$$ and $r_2=\frac{-3+\sqrt{17}}{2}$. By the principle of superposition, the general solution is $C_1 e^{\frac{-3-\sqrt{17}}{2}t}+C_2 e^{\frac{-3+\sqrt{17}}{2}t}$ which is nether (A) nor (B)
- Method 2: WolframAlpha with warning!
- Method 3: ODEsolver
$y=C_1e^{5t}+C_2e^{-2t}$ is the general solution of which equation?
- $y''-3y'-10y=0$
- $y''+3y'-10y=0$
- None of the above
Answer: A
Solution The roots $r_1=5, r_2=-2$ are from equation $(r-5)(r+2)=r^2-3r-10=0$
$y=C_1e^{2t}+C_2e^{-3t}$ is the general solution of which equation?
- $y''+5y'+6 y=0$
- $y''+y'-6y=0$
- None of the above
Answer: B
Solution The roots $r_1=2, r_2=-3$ are from equation $(r-2)(r+3)=r^2+r-6=0$
$y=C_1e^{t/2}+C_2e^{-t}$ is the general solution of which equation?
- $2y''+y'-y=0$
- $y''+\tfrac12 y'+\tfrac12 y=0$
- None of the above
Answer: A
Solution The roots $r_1=1/2, r_2=-1$ are from equation $(r-1/2)(r+1)=r^2+r/2-1/2=0$ This is the same equation as $2r^2+r-1=0$
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 $