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Math Help - 2 point BVP

  1. #1
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    2 point BVP

    questions is:

    >> -d/dx(e^x y') + 2ye^x = (2-x)e^x

    i need to simplify the equation to show that the finite difference method is given by:

    >> -(2-h)yi-1 +4(1+h^2)yi -(2+h)yi+1 = 2h^2(2-xi)

    I no how to apple the finite difference method so i get the difference equation to get the above but what is:

    >> -d/dx(e^x y') + 2ye^x = (2-x)e^x

    simplified so it is in the form containing y'' and y'?
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  2. #2
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    Re: 2 point BVP

    Quote Originally Posted by NFS1 View Post
    questions is:

    >> -d/dx(e^x y') + 2ye^x = (2-x)e^x

    i need to simplify the equation to show that the finite difference method is given by:

    >> -(2-h)yi-1 +4(1+h^2)yi -(2+h)yi+1 = 2h^2(2-xi)

    I no how to apple the finite difference method so i get the difference equation to get the above but what is:

    >> -d/dx(e^x y') + 2ye^x = (2-x)e^x

    simplified so it is in the form containing y'' and y'?
    Using the product rule:

    \frac{d}{dx}(e^xy')=e^xy'+e^xy''

    .
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  3. #3
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    Re: 2 point BVP

    How do you apply the finite difference method, once you've simplified the equation?
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  4. #4
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    Re: 2 point BVP

    Quote Originally Posted by KayAB View Post
    How do you apply the finite difference method, once you've simplified the equation?
    put:

    y(x_i)=y_i,

    y''(x_i)\approx \frac{y_{i+1}-2y_i+y_{i-1}}{h^2} ,

    y'(x_i)\approx \frac{ y_{i+1}- y_{i-1} }{2h}

    into the ODE and simplify.

    .
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  5. #5
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    Re: 2 point BVP

    After using the finite difference method I have solved the system using back substitution in matrix form to get:

    -(2-h)yi-1 + 4(1+h^2)yi -(2+h)yi+1=2h^2(2-xi) as the difference equation. And using h=0.2 and working to 4dp i got the solutions as:

    y1=0.4564
    y2=0.1503
    y3=0.3862

    How do i use this information to show that the analytical solution is: y=-0.6925e^-2x -0.0575e^x +0.75
    Last edited by NFS1; August 27th 2012 at 04:12 PM.
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  6. #6
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    Re: 2 point BVP

    Quote Originally Posted by NFS1 View Post
    After using the finite difference method I have solved the system using back substitution in matrix form to get:

    -(2-h)yi-1 + 4(1+h^2)yi -(2+h)yi+1=2h^2(2-xi) as the difference equation. And using h=0.2 and working to 4dp i got the solutions as:

    y1=0.4564
    y2=0.1503
    y3=0.3862

    How do i use this information to show that the analytical solution is: y=-0.6925e^-2x -0.0575e^x +0.75
    Please post the question exactly how it was originally worded.

    CB
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  7. #7
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    Re: 2 point BVP

    Quote Originally Posted by zzephod View Post
    Please post the question exactly how it was originally worded.

    CB
    You are given the following two point boundary value problem:

    >> -d/dx(e^x y') + 2ye^x = (2-x)e^x where y(0)=y(1)=0

    Simplify the equation and show that the resulting difference equation is (below) when the finite difference method is applied:

    >> -(2-h)yi-1 + 4(1+h^2)yi -(2+h)yi+1=2h^2(2-xi)

    Take h=0.2 and determine the approximate solution?

    I got the approximate solution as:

    y1=0.4564
    y2=0.1503
    y3=0.3862

    Show that the analytical solution is:

    >> y=-0.6925e^-2x -0.0575e^x +0.75

    And hence show the errors in your numerical approximation.
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  8. #8
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    Re: 2 point BVP

    Quote Originally Posted by NFS1 View Post
    You are given the following two point boundary value problem:

    >> -d/dx(e^x y') + 2ye^x = (2-x)e^x where y(0)=y(1)=0

    Simplify the equation and show that the resulting difference equation is (below) when the finite difference method is applied:

    >> -(2-h)yi-1 + 4(1+h^2)yi -(2+h)yi+1=2h^2(2-xi)

    Take h=0.2 and determine the approximate solution?

    I got the approximate solution as:

    y1=0.4564
    y2=0.1503
    y3=0.3862
    I assume y1 is the approximation to y(0.2), y2 to y(0.4), y3 to y(0.6), you should also give the next value y(0.8) and for that matter include the boundary points as well.

    Show that the analytical solution is:

    >> y=-0.6925e^-2x -0.0575e^x +0.75

    And hence show the errors in your numerical approximation.
    You just have to show that this solution satisfies the original equation, or its simplified form:

    y''+y'-2y=x-2

    by differentiating twice and substituting into the equation. Then comparing the exact solution at x=0, 0.2, 0.4, 0.6, 0.6, 1.0 with your numerical solution.

    .
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