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Math Help - Range-Kutta

  1. #1
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    Range-Kutta

    Equation:
    I \theta '' + c \theta ' = Tm
    I = 0.9
    c = 0.15
    Tm = 0.3

    My solution:
    Reduced Equation 1 order:
    y_2' = \frac{T_m}{I} - \frac{Cy_2}{I}

    I used three steps to RK2 and found:
    y_1 = 0.01665
    y_2 = 0.03306
    y_3 = 0.04249

    Is that correct?
    How do I find the analytical solution?
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  2. #2
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    I don't know about the Range-Kutta part but as far as an analytical solution goes you've already taken the first step yourself.

    y' = \frac{T_m}{I} - \frac{C}{I}y

    This is a Linear First Order differential equation. You can, if you've ever covered linear first order ODE's, solve this for y = \theta' and then go from there.
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  3. #3
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    It seems to me the OP is looking for advice on how to solve the DE using Runge-Kutta not any other way.

    Follow this example Runge Kutta Methods
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  4. #4
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    Quote Originally Posted by Apprentice123 View Post
    Equation:
    I \theta '' + c \theta ' = Tm
    I = 0.9
    c = 0.15
    Tm = 0.3

    My solution:
    Reduced Equation 1 order:
    y_2' = \frac{T_m}{I} - \frac{Cy_2}{I}

    I used three steps to RK2 and found:
    y_1 = 0.01665
    y_2 = 0.03306
    y_3 = 0.04249

    Is that correct?
    What are the initial conditions and what the step size? Also are you not supposed to be finding $$ \theta ?

    CB
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Apprentice123 View Post
    Equation:
    I \theta '' + c \theta ' = Tm
    I = 0.9
    c = 0.15
    Tm = 0.3


    How do I find the analytical solution?
    This is an inhomogeneous linear constant coefficient ODE. You should have covered how to solve this.

    The general solution is the sum of the general solution of the homogeneous equation

    I \theta '' + c \theta ' = 0

    and a particular integral of the original equation, and a PI for this is T_m t.

    To get the actual solution you now apply the initial conditions to the general solution.



    CB
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  6. #6
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    The initial conditions are
    y_0 = 0 and x_0 = 0

    How to find the local and global errors?
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Apprentice123 View Post
    The initial conditions are
    y_0 = 0 and x_0 = 0

    How to find the local and global errors?
    Are you not supposed to be finding \theta(x) ? If so the initial conditions will specify \theta '(0) and \theta(0)

    CB
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by Apprentice123 View Post
    Equation:
    I \theta '' + c \theta ' = Tm
    I = 0.9
    c = 0.15
    Tm = 0.3

    My solution:
    Reduced Equation 1 order:
    y_2' = \frac{T_m}{I} - \frac{Cy_2}{I}
    You reduce the second order ODE to a first order system as shown in the second post in this thread: http://www.mathhelpforum.com/math-he...ta-162591.html

    CB
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