Constant coefficient homogeneous equations: ay″ 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