Definition Equation $M(x,y)+N(x,y)y'=0$ is exact if $M(x,y)=\frac{\partial \Psi(x,y)}{\partial x}$ $N(x,y)= \frac{\partial \Psi(x,y)}{\partial y}$
Theorem For functions with continuous derivatives, equation $M(x,y)+N(x,y)y'=0$
- has integrating factor $\mu(x)$ if $\frac{M_y-N_x}{N}$ is a function of $x$ only.
- has integrating factor $\mu(y)$ if $\frac{N_x-M_y}{M}$ is a function of $y$ only.
Which of the following statements is correct for equation $1+(1+x y)y' =0$?
- It has integrating factor $\mu(x)$
- It has integrating factor $\mu(y)$
- None of the above
Answer: B
Solution We have $M(x,y)=1 $ and $N(x,y)=1+xy$ so $\frac{M_y-N_x}{N}=\frac{y}{1+xy}$ and $\frac{N_x-M_y}{M}=y$.
Which of the following statements is correct for equation $x y^2+e^y +(x^2 y+ e^x) y'=0$?
- It has integrating factor $\mu(x)$
- It has integrating factor $\mu(y)$
- None of the above
Answer: C
Solution We have $M(x,y)=x y^2+e^y $ and $N(x,y)=x^2 y+ e^x$. So $\frac{M_y-N_x}{N}=\frac{ e^y-e^x}{x^2 y+ e^x}$ and $\frac{N_x-M_y}{M}=\frac{e^x-e^y}{x y^2+e^y}$.
Which of the following statements is correct for equation $x^2+xy+y^2 +(x+y)^2y'=0$
- It has integrating factor $\mu(x)$
- It has integrating factor $\mu(y)$
- None of the above
Answer: C
Solution We have $M(x,y)=x^2+xy+y^2$ and $N(x,y)=(x+y)^2$. So $\frac{M_y-N_x}{N}=\frac{-x}{(x+y)^2}$ and $\frac{N_x-M_y}{M}=\frac{x}{x^2+xy+y^2}$.
Which of the following statements is correct for equation $y^2/2+(2xy-e^{-2y}) y'=0$
- It has integrating factor $\mu(x)$
- It has integrating factor $\mu(y)$
- None of the above
Answer: B
Solution We have $M(x,y)=y^2/2$ and $N(x,y)=(2xy-e^{-2y})$ So $\frac{M_y-N_x}{N}=\frac{-y}{2xy-e^{-2y}}$ and $\frac{N_x-M_y}{M}=\frac{y}{y^2/2}=2/y$.
Which of the following statements is correct for equation $(x+2)\sin y +(x \cos y) y'=0$
- It has integrating factor $\mu(x)$
- It has integrating factor $\mu(y)$
- None of the above
Answer: A
Solution We have $M(x,y)=(x+2)\sin y $ and $N(x,y)=(x \cos y)$ So $\frac{M_y-N_x}{N}=\frac{(x+1)\cos y}{(x \cos y}=\frac{x+1}{x}$ and $\frac{N_x-M_y}{M}=\frac{-(x+1)\cos y}{(x+2)\sin y}$.
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 $