Difference between revisions of "Tutorial Week 3"

From UBCMATH WIKI
Jump to: navigation, search
(Created page with "===Worksheet Questions=== #Using Reduction of Order, solve the following differential equations to find a second solution, and then state the general solution. ##$y''+y = 0$ ...")
 
 
(3 intermediate revisions by one user not shown)
Line 1: Line 1:
 
===Worksheet Questions===
 
===Worksheet Questions===
−
#Using Reduction of Order, solve the following differential equations to find a second solution, and then state the general solution.  
+
# Beer is being poured from a tap into a glass at a rate of $200$ml/min. The beer leaving the tap contains $0.005$g/ml of $\mathrm{CO}_2$. It is known that $\mathrm{CO}_2$ escapes beer at a rate proportional to its concentration, and that at a concentration of $0.1$g/ml, the concentration decreases at a rate of $0.05$g/(ml$\cdot$min).
−
##$y''+y = 0$ where $y_1(t) = cos(t)$ is a solution.  
+
## Write a differential equation with initial conditions for the total amount $Q(t)$ of $\mathrm{CO}_2$ (in grams) in the glass at time $t$.
−
##$2t^2y''+ty'-3y = 0$ where $y_1(t) = \dfrac{1}{t}$ is a solution.
+
## What would the steady state be if the glass were infinitely tall?
−
#Use the Method of Undetermined Coefficients to answer the following question.
+
## Solve the initial value problem from part (a).
−
##Determine a particular solution to $y''-4y'+4y = e^{2t}$
+
## The $\mathrm{CO}_2$ leaving the beer forms a foam. Each gram of $\mathrm{CO}_2$ that escapes produces $40$mm of foam. Additionally, the bubbles burst at a rate of $10\%$ per minute. Write and solve an equation for the height of the foam $F(t)$ at time $t$. (Hint: use your solution from (a) to determine the amount of $\mathrm{CO}_2$ escaping.)
−
##Write the general solution to the above equation.  
+
## Does the height of the foam $F(t)$ reach a steady state? If so, what is the limiting value?
−
##What does the general solution become if $y(0) = 2$ and $y'(0) = 5$?
+
#Reduction of Order is a technique for finding a second solution to a homogeneous second order ODE when a first solution has already been found. Consider the equation $t^2y''+ 4ty' + 6y = 0$ where $y_1(t) = t^2$ is our first solution.
−
#Optional: You have a garden and are beginning to see weeds. Weeds spread at a per-capita rate of 2/day, and naturally die at a rate of 0.3/day.  
+
##Assume the second solution is of the form $y_2(t)=y_1(t)v(t)$ and calculate $4ty_2'(t)$ and $t^2y_2''(t)$.
−
##Write a differential equation for the total number of weeds at time t, $\dfrac{dW_1}{dt}$.  
+
##Plug these in to the ODE and simplify to find a second order equation in terms of $v''(t)$ and $v'(t)$ only.
−
##You decide to take action and start weeding your garden, removing 20 weeds every day. What is the new differential equation,$\dfrac{dW_2}{dt}$, for the situation?
+
##Rename $v'(t)=w(t)$. Simplify and solve the resulting equation for $w(t)$.
−
##Find the solution to both differential equations if the initial number of weeds was 10.
+
##Find $v(t)$ and substitute back to get $y_2(t)$.
−
##Sketch a graph showing these two solutions.
+
−
##Is there ever a time in either situation where your garden is weed-free? If so, find the time.  
+
  
−
===Solutions===
+
[[Tutorial Week 3 Solutions]]
−
#
+
−
##$y(t) = c_1sint+c_2cost$
+
−
##$y(t) = c_1t^{3/2}+c_2\dfrac{1}{t}$
+
−
#
+
−
##$y_p(t) = \dfrac{1}{2}t^2e^{2t}$
+
−
##$y(t) = c_1e^{2t}+c_2te^{2t}+\dfrac{1}{2}t^2e^{2t}$
+
−
##$y(t) = \dfrac{1}{2}e^{2 t} (t^2+2 t+4)$
+
−
#
+
−
##$\dfrac{dW_1(t)}{dt} = 2W_1-0.3$
+
−
##$\dfrac{dW_2(t)}{dt} = 2W_2-20.3$
+
−
##$W_1 = 0.15+9.85e^{2t}, W_2 = 10.15-0.15e^{2t}$
+
−
##[https://www.desmos.com/calculator/fmzyes1ply graph]
+
−
##In situation 1, the weeds grow exponentially and the garden is never weed-free. In situation 2, the weeds decrease exponentially and the garden becomes weed-free after t = 1.85 days.
+

Latest revision as of 23:00, 30 December 2020

Worksheet Questions

  1. Beer is being poured from a tap into a glass at a rate of $200$ml/min. The beer leaving the tap contains $0.005$g/ml of $\mathrm{CO}_2$. It is known that $\mathrm{CO}_2$ escapes beer at a rate proportional to its concentration, and that at a concentration of $0.1$g/ml, the concentration decreases at a rate of $0.05$g/(ml$\cdot$min).
    1. Write a differential equation with initial conditions for the total amount $Q(t)$ of $\mathrm{CO}_2$ (in grams) in the glass at time $t$.
    2. What would the steady state be if the glass were infinitely tall?
    3. Solve the initial value problem from part (a).
    4. The $\mathrm{CO}_2$ leaving the beer forms a foam. Each gram of $\mathrm{CO}_2$ that escapes produces $40$mm of foam. Additionally, the bubbles burst at a rate of $10\%$ per minute. Write and solve an equation for the height of the foam $F(t)$ at time $t$. (Hint: use your solution from (a) to determine the amount of $\mathrm{CO}_2$ escaping.)
    5. Does the height of the foam $F(t)$ reach a steady state? If so, what is the limiting value?
  2. Reduction of Order is a technique for finding a second solution to a homogeneous second order ODE when a first solution has already been found. Consider the equation $t^2y''+ 4ty' + 6y = 0$ where $y_1(t) = t^2$ is our first solution.
    1. Assume the second solution is of the form $y_2(t)=y_1(t)v(t)$ and calculate $4ty_2'(t)$ and $t^2y_2''(t)$.
    2. Plug these in to the ODE and simplify to find a second order equation in terms of $v''(t)$ and $v'(t)$ only.
    3. Rename $v'(t)=w(t)$. Simplify and solve the resulting equation for $w(t)$.
    4. Find $v(t)$ and substitute back to get $y_2(t)$.

Tutorial Week 3 Solutions

Retrieved from "https://wiki.math.ubc.ca/mathbook/M256/index.php?title=Tutorial_Week_3&oldid=1546"