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Thread: 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
    Senior Member zzephod's Avatar
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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:

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

    .
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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
    Senior Member zzephod's Avatar
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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:

    $\displaystyle y(x_i)=y_i$,

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

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

    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; Aug 27th 2012 at 04:12 PM.
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  6. #6
    Senior Member zzephod's Avatar
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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
    Senior Member zzephod's Avatar
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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 $\displaystyle y(0.2)$, y2 to $\displaystyle y(0.4)$, y3 to $\displaystyle y(0.6)$, you should also give the next value $\displaystyle y(0.8)$ 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:

    $\displaystyle y''+y'-2y=x-2$

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

    .
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