Constant coefficient homogeneous equations: $ay''+by'+cy=0$ are solved by seeking solutions of the form $y=e^{rt}$.
Solve the initial value problem $y''+3 y'-10y=0,\; y(0)=2,y'(0)=-3$
Answer: $y=e^{2t}+e^{-5t}$
Solution
- Method 1 (by hand): substituting $y=e^{rt}$ we get the characteristic equation $r^2+3r-10=0$. The roots are $r_1= -5$ and $r_2= 2$. By the principle of superposition, the general solution is $y=C_1 e^{-5t}+C_2e^{2t}$.
We now solve the system of equations $y(0)=2$ and $y'(0)=-3$. We get $$ C_1+C_2=2$$ $$-5C_1+2C_2=-3$$ After some work, or by the method of determinants the solution is $$C_1=\frac{\left| \begin{matrix}2&1\\-3&2\end{matrix}\right|}{\left| \begin{matrix}1&1\\-5&2\end{matrix}\right|}=1, \; C_2=\frac{\left| \begin{matrix}1&2\\-5&-3\end{matrix}\right|}{\left| \begin{matrix}1&1\\-5&2\end{matrix}\right|}=1 $$- Method 2: WolframAlpha
- Method 3: ODEsolver
Solve the initial value problem $y''+5 y'+6 y =0, y(0)=0,y'(0)=1$
Answer: $e^{-2t}-e^{-3t}$
SolutionMethod 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 $y=C_1 e^{-2t}+C_2e^{-3t}$
The initial value problem requires solving system of equations $$ C_1+C_2=0$$ $$-2C_1-C_2=1$$ The initial value problem is solved by $C_1=1, C_2=-1$Method 2: WolframAlpha Method 3: ODEsolver
Solve the initial value problem $y''-5 y'+6 y =0$, $y(0)=0,y'(0)=1$
Answer: $y=e^{3t}-e^{2t}$
Solution SolutionMethod 1 (by hand): substituting $y=e^{rt}$ we get the characteristic equation $r^2-5r+6=0$. The roots are $r_1=2$ and $r_2=3$. The general solution is $y=C_1 e^{2t}+C_2e^{3t}$
The initial value problem requires solving system of equations $$ C_1+C_2=0$$ $$2C_1+3C_2=1$$ which is solved by $C_1=-1, C_2=1$.Method 2: WolframAlpha Method 3: ODEsolver
$y=e^{2t}-e^{-3t}$ is the solution of which initial value problem?
- $y''+y'-6y=0$, $y(0)=0$, $y'(0)=1$
- $y''+y'-6y=0$, $y(0)=0$, $y'(0)=5$
- None of the above
Answer: B
Solution We may first verify that this is a solution of the equation, and then check the initial conditions. We compute $y'(t)=2e^{2t}+3e^{-3t}$ and $y''(t)=4e^{2t}-9e^{-3t}$. Plugging these into the equation we get $$y''+y'-6y=(4e^{2t}-9e^{-3t})+(2e^{2t}+3e^{-3t}) -6(e^{2t}-e^{-3t})= (4+2-6)e^{2t}+(-9+3+6)e^{-3t}=0$$ For the initial condition, we have $y(0)=1-1=0$ and $y'(0)=2+3=5$.
$y=e^{t/2}+e^{-t}$ is the solution of which initial value problem?
- $2y''+y'-y=0$, $y(0)=2$, $y'(0)=2$
- $2y''+y'-y=0$ , $y(0)=2$, $y'(0)=0$
- None of the above
Answer: C
Solution For this solution, $y(0)=2$ and $y'(0)=-1/2$, so none of the initial conditions is a match.
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 $