Constant coefficient homogeneous equations: $ay''+by'+cy=0$ are solved by seeking solutions of the form $y=e^{rt}$.Characteristic equation $ar^2+br+c=0$ might have two real roots (Sect 3.1), two complex roots (Sect 3.3) or a double root (Sect 3.4).
For complex roots $r=\lambda \pm i\mu$ , instead of $C_1e^{t(\lambda+i\mu)}+C_2 e^{t(\lambda-i\mu)}$ the general solution is $$y(t)=C_1 e^{\lambda t}\cos(\mu t)+C_2 e^{\lambda t}\sin(\mu t)$$
Solve the initial value problem $y''+2 y'+5 y=0,\; y(0)=2,y'(0)=0$
Answer: $y=e^{-t}\sin (2t)+2e^{-t}\cos(2t) $
Solution
- Method 1: WolframAlpha
- Method 2: ODEsolver
- Method 3: by hand (read the description provided by ODESolver!)
Characteristic equaation $r^2+2r+5=0$ has roots $r_1=-1-2i$ and $r_2=-1+2i$. The general solution is $y=C_1 e^{-t}\cos(2t)+C_2e^{-t}\sin(2t)$. The system of equations is $y(0)=C_1=2$. Next we compute $y'(t)=-C_1 e^{-t} \cos (2t)-2C_1e^{-t}\sin(2t)-C_2e^{-t}\sin(2t)+2C_2e^{-t}\cos(2t)$. So $y'(0)=-C_1+2C_2$ and $C_2=1$.
Solve the initial value problem $y''-2 y'+5 y=0,\; y(0)=0,y'(0)=2$
Answer: $e^{t}\sin(2t)$
SolutionMethod 1: WolframAlpha Method 2: ODEsolver Method 3: by hand: read the description provided by ODESolver!
Solve the initial value problem $y''-2 y'+5 y=0,\; y(0)=5,y'(0)=5$
Answer: $y=5 e^t \cos(2t)$
SolutionMethod 1 (by hand): substituting $y=e^{rt}$ we get the characteristic equation $r^2-2r+5=0$. The roots are $r_1=1-2i$ and $r_2=1+2i$. The general solution is $y=C_1 e^{t}\cos(2t)+C_2e^{t}\sin(2t)$
The initial value problem requires solving system of equations $$ C_1=5$$ $$C_1+2 C_2=5$$ which is solved by $C_1=5, C_2=0$.Method 2: WolframAlpha Method 3: ODEsolver
$y=\cos(2t)-\sin(2t)$ is the solution of which initial value problem?
- $y''+4y=0$, $y(0)=0$, $y'(0)=1$
- $y''+4y=0$, $y(0)=1$, $y'(0)=0$
- None of the above
Answer: C
Solution We may first verify that this is a solution of the equation, and then check the initial conditions. But lets check the initial conditions: $y(0)=\cos 0=1$, which excludes (A). $y'(t)=-2 \sin (2t)+2\cos(2t)$ so $y'(0)=2$ which excludes (B)
$y=\cos (2t)+\sin(2t)$ is the solution of which initial value problem?
- $y''+4y=0$, $y(0)=1$ $y'(0)=2$
- $y''+4y=0$ , $y(0)=1$ $y'(0)=1$
- None of the above
Answer: A
Solution The general solution of the equation is $y=C_1\cos(2t)+C_2\sin(2t)$ so $y=\cos (2t)+\sin(2t)$ is one of the solutions. For this solution, we have $y(0)=1$ and $y'(t)=-2\sin(2t)+2\cos(2t)$, so $y'(0)=2$. This initial condition matches (A).
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 $