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Homework Help: ODE Solution Query

  1. Jan 12, 2009 #1
    1. The problem statement, all variables and given/known data

    This is a solution detailed in a past paper,

    http://stuff.bsodmike.com/sensors_pastpaper.png [Broken]


    2. Relevant equations
    Utilise an integrating factor to solve as detailed by myself https://www.physicsforums.com/showthread.php?t=283610".


    3. The attempt at a solution

    It can be said that,

    [tex]\dfrac{d\theta}{dt}+\dfrac{\theta}{\beta}=\dfrac{\theta_m}{\beta}[/tex]

    Hence, the solution by employing an integrating factor would yield,

    [tex]e^{t/\beta}\theta=\dfrac{\theta_m}{\beta}\int{e^{t/\beta}\,dt}[/tex]

    [tex]\theta=\dfrac{\theta_m}{\beta}+Ce^{-t/\beta}[/tex]

    The problem is that I'm having the extra 'beta' term as shown above. Any ideas?

    Thanks!
    Mike
     
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 12, 2009 #2
    A more direct approach, solving the homogenous part:

    [tex]\int{\dfrac{1}{\theta}\,d\theta}=\int{-\dfrac{1}{\beta}\,dt}[/tex]

    [tex]ln(\theta)=-\dfrac{t}{\beta}+C[/tex]

    [tex]\theta=e^{-t/\beta+C}[/tex]

    Now, my memory here is fuzzy but [tex]e^{-t/\beta+C} = e^{-t/\beta}+e^C[/tex] ??
     
  4. Jan 12, 2009 #3
    Ah, another small blunder, of course!

    Now, [tex]\int{e^{t/\beta}\,dt}=\beta e^{t/\beta}+C[/tex]. thus, the 'beta's cancel,

    [tex]e^{t/\beta}\theta=\theta_m e^{t/\beta}+C[/tex]

    Dividing by [tex]e^{t/\beta}[/tex], yields,

    [tex]\theta=Ce^{-t/\beta}+\theta_m[/tex]
     
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