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Differential equations, euler's method and bisection method

  1. Nov 26, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi guys, I have the following problem and I dont know how to start.

    I am given that W = 0.5, X(0) = 0, Iab = 20m and ha = 5m [tex]\frac{dy}{dx}=0[/tex]


    \frac{dy^2}{d^2x}=\frac{W}{T}\sqrt{1 + (\frac{dy}{dx})^2}

    I am told to convert the 2nd order ODE to two 1st order ODEs at first. Then I am asked to choose an arbitrary value of T and integrate the system of equations using Euler's method from 0-Iab, where Iab=20, with 3 different step sizes. Then, I have to select the step size that gives an accurate value of y(Iab) (i.e. approximate relative error less than 0.1% if comparing with ha = 5).. Then I am asked to chose two values of T such that one gives y(Iab) smaller than ha and the other one gives y(Iab) bigger than ha and use those two values with the bisection method

    2. Relevant equations

    Euler's method: [tex]yi+1 = yi + f(xi,yi)*h[/tex] where [tex]f(x,y) = \frac{dy}{dx}[/tex]

    3. The attempt at a solution

    At first I broke it to two 1st order ODES:

    [tex]\frac{dy}{dx}=g[/tex] and [tex]\frac{dg}{dx}=\frac{W}{T}\sqrt{1 + g^2}[/tex]

    Then, for Euler's method: [tex]yi+1 = yi + gi*h[/tex] and [tex]gi+1 = gi + (\frac{W}{T}\sqrt{1+gi^2})*h[/tex]

    My problem is, I dont know where to start in order to find y(Iab) (because the problem is asking for an accurate value of y(Iab) )... and what step size (h) to use. .. I also dont know which equation shall be used with the bisection method..

    any help would be really appreciated. thank you
  2. jcsd
  3. Nov 26, 2009 #2


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    I would suggest you go back and reread the problem. You say that X(0)= 0 but your differential equation has y as a function of x. What does X(0) mean?
  4. Nov 26, 2009 #3
    Sorry, typo mistake. It says that the slope, at x=0, is dy/dx=0
  5. Nov 26, 2009 #4
    and from the Picture I have, the graph, when x=0 -> y=0.

    So I suppose for euler's method, suppose I choose a step size of ie, h=5, then it would be

    [tex]y(5) = y(0) + g(0)*h[/tex] and [tex]g(5) = g(0) + (\frac{W}{T}\sqrt{1+g(0)^2})*h[/tex]

    Is that correct ? And since I have to integrate from 0-20, I suppose I'll have to do y(10), y(15) and y(20) aswell, right ? (Same for g(10),g(15),g(20) ) ?
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