Difference between revisions of "Tutorial Week 3"

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(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. ##$y''+y = 0$ ...")
 
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===Worksheet Questions===
 
===Worksheet Questions===
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#Using Reduction of Order, solve the following differential equations to find a second solution, and then state the general solution.  
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# As a beer is poured, foam is added at the top of the beer at a constant rate of 2 mm per second and the bubbles burst at a continuous rate of 10 percent per second.
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##$y''+y = 0$ where $y_1(t) = cos(t)$ is a solution.  
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##Write a differential equation with initial condition for the depth of foam at time t, $\dfrac{dF}{dt}$.
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##$2t^2y''+ty'-3y = 0$ where $y_1(t) = \dfrac{1}{t}$ is a solution.
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##What would the steady state foam depth be if the glass were infinitely tall?
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#Use the Method of Undetermined Coefficients to answer the following question.
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##Solve the initial value problem.
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##Determine a particular solution to $y''-4y'+4y = e^{2t}$
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##Assume the beer is poured into a glass 15cm tall so that the depth of beer increases at a rate of 10 mm per second. How high is the beer and foam at time $t$?
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##Write the general solution to the above equation.
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##Write down an equation which when solved would give the time at which the bartender must stop pouring to ensure that no beer nor foam spill out of the glass. Give a rough approximation of the value of t along with an estimate of how close you approximation is.
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##What does the general solution become if $y(0) = 2$ and $y'(0) = 5$?
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#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 $2t^2y''+ty'-3y = 0$ where $y_1(t) = \dfrac{1}{t}$ is our first solution.
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#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.  
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##Assume the second solution is of the form $y_2(t)=y_1(t)v(t)$ and calculate $ty_2'(t)$ and $2t^2y_2''(t)$.
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##Write a differential equation for the total number of weeds at time t, $\dfrac{dW_1}{dt}$.  
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##Plug these in to the ODE and simplify to find a second order equation in terms of $v''(t)$ and $v'(t)$ only.
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##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?
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##Rename $v'(t)=w(t)$. Solve the resulting equation for $w(t)$.
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##Find the solution to both differential equations if the initial number of weeds was 10.
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##Find $v(t)$ and substitute back to get $y_2(t)$.
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##Sketch a graph showing these two solutions.
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##Is there ever a time in either situation where your garden is weed-free? If so, find the time.  
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===Solutions===
 
===Solutions===
 
#
 
#
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##$y(t) = c_1sint+c_2cost$
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##$\dfrac{dF}{dt}=2-0.1F$  '''(3 points)''', $\ F(0)=0$ '''(1 point)'''
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##$y(t) = c_1t^{3/2}+c_2\dfrac{1}{t}$
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##$F(t)=20mm$ '''(2 points)'''
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##$F(t)=20 -20e^{-0.1t}$. Integrating factor '''(1 point)''', antiderivatives '''(2 points)''', find F(t) '''(1 point)''', find C '''(1 point)'''
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##$H(t)=F(t)+B(t)=20(1-e^{-0.1t})+10t$ '''(1 point)'''
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## $H(t)=20(1-e^{-0.1t})+10t=150$ '''(1 point)''', arguing that $t\in(13,15)$ (steady state time and 0 foam time) '''(1 point)''', conclusion that $t$ must be closer to 13 '''(1 point)'''
 
#
 
#
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##$y_p(t) = \dfrac{1}{2}t^2e^{2t}$
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##$ty_2'(t)=ty_1'v+ty_1v'=-\dfrac{v}{t}+v'$ '''(2 points)''',  $ \quad 2t^2y_2''=2t^2y_1''v+4t^2y_1'v'+2t^2y_1v'' = \dfrac{4v}{t}-4v'+2tv''$ '''(2 points)'''
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##$y(t) = c_1e^{2t}+c_2te^{2t}+\dfrac{1}{2}t^2e^{2t}$
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## $v''-\dfrac{3}{2t}v'=0$ '''(2 points)'''
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##$y(t) = \dfrac{1}{2}e^{2 t} (t^2+2 t+4)$
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## $w'-\dfrac{3}{2t} w=0$ (separable equation '''1 point'''),  $w(t)=Ct^{3/2}$ (take the antiderivative '''2 points''').
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#
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## $v(t)=Ct^{5/2}+D$ (D=0 is ok) '''(2 points)'''  $ \quad y_2(t) = t^{3/2}$ '''(1 point)'''
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##$\dfrac{dW_1(t)}{dt} = 2W_1-0.3$
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##$\dfrac{dW_2(t)}{dt} = 2W_2-20.3$
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##$W_1 = 0.15+9.85e^{2t}, W_2 = 10.15-0.15e^{2t}$
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##[https://www.desmos.com/calculator/fmzyes1ply graph]
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##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.
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Revision as of 00:10, 29 January 2016

Worksheet Questions

  1. As a beer is poured, foam is added at the top of the beer at a constant rate of 2 mm per second and the bubbles burst at a continuous rate of 10 percent per second.
    1. Write a differential equation with initial condition for the depth of foam at time t, $\dfrac{dF}{dt}$.
    2. What would the steady state foam depth be if the glass were infinitely tall?
    3. Solve the initial value problem.
    4. Assume the beer is poured into a glass 15cm tall so that the depth of beer increases at a rate of 10 mm per second. How high is the beer and foam at time $t$?
    5. Write down an equation which when solved would give the time at which the bartender must stop pouring to ensure that no beer nor foam spill out of the glass. Give a rough approximation of the value of t along with an estimate of how close you approximation is.
  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 $2t^2y''+ty'-3y = 0$ where $y_1(t) = \dfrac{1}{t}$ is our first solution.
    1. Assume the second solution is of the form $y_2(t)=y_1(t)v(t)$ and calculate $ty_2'(t)$ and $2t^2y_2''(t)$.
    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 $v'(t)=w(t)$. Solve the resulting equation for $w(t)$.
    4. Find $v(t)$ and substitute back to get $y_2(t)$.

Solutions

    1. $\dfrac{dF}{dt}=2-0.1F$ (3 points), $\ F(0)=0$ (1 point)
    2. $F(t)=20mm$ (2 points)
    3. $F(t)=20 -20e^{-0.1t}$. Integrating factor (1 point), antiderivatives (2 points), find F(t) (1 point), find C (1 point)
    4. $H(t)=F(t)+B(t)=20(1-e^{-0.1t})+10t$ (1 point)
    5. $H(t)=20(1-e^{-0.1t})+10t=150$ (1 point), arguing that $t\in(13,15)$ (steady state time and 0 foam time) (1 point), conclusion that $t$ must be closer to 13 (1 point)
    1. $ty_2'(t)=ty_1'v+ty_1v'=-\dfrac{v}{t}+v'$ (2 points), $ \quad 2t^2y_2''=2t^2y_1''v+4t^2y_1'v'+2t^2y_1v'' = \dfrac{4v}{t}-4v'+2tv''$ (2 points)
    2. $v''-\dfrac{3}{2t}v'=0$ (2 points)
    3. $w'-\dfrac{3}{2t} w=0$ (separable equation 1 point), $w(t)=Ct^{3/2}$ (take the antiderivative 2 points).
    4. $v(t)=Ct^{5/2}+D$ (D=0 is ok) (2 points) $ \quad y_2(t) = t^{3/2}$ (1 point)
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