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Linear-Shooting Method

  1. Mar 4, 2008 #1
    1. The problem statement, all variables and given/known data
    y''=2y'-y+x*exp(x)
    0<=x<=2 y(0)=0 , y(2)=-4, h = 0.2
    Approximate the solution using Euler's method

    2. Relevant equations



    3. The attempt at a solution

    So we'll have two guesses
    y1''=2*y1'-y1'+x*exp(x)-x y1(0)=0,
    y2''=2*y2'-y2'+x*exp(x)-x y2(0)=0,
    y=a1*y1+a2*y2
    So i need to find the coefficients

    What boundary conditions do I require? y1'(0)=? y2'(0)= ?
     
  2. jcsd
  3. Mar 4, 2008 #2
    Pick two random values for y1'(0) and y2'(0). This will give you two solutions to the ODE.
    Then y = a*y1 + b*y2 is also a solution.
    You know that y1(0)=0 and y(2) =0, so ay1(0)+by2(0)=0. So fix a = 1, and pick b such that y1(2) + b*y2(2) = y(2).
    So b = (y(2) - y1(2)) / y2(2), where y1(2) and y2(2) are the result of shooting with arbitrary guesses for y1'(0) and y2'(0).
     
  4. Mar 5, 2008 #3
    Okay so say if I choose y1'(0)=1 and y2'(0)=2

    y'=ay1'+by2' =>
    y'(0)=a*(1)+b*(2)=0
    => a=-2*b

    y(2)=-(2*b)*y1(2)+b*y2(2)=-4
    -2y1(2)+y2(2)=-4/b

    y2(2)=-4/b +2y1(2)

    b=(-4-y1(2))/(-4/b+2y1(2))

    b(-4/b+2y1(2)=-4-y1(2)

    => 3by1(2)=0
    b or y1(2) is 0.

    This correct so far?
     
    Last edited: Mar 5, 2008
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