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Runge kutta method

  1. Nov 3, 2009 #1
    1. The problem statement, all variables and given/known data
    apply the runge kutta of order 4 to solve the ode on [-1,1] with n = 5 of
    (e^x+y)dx-dy=0


    2. Relevant equations



    3. The attempt at a solution
    the problems i have done so far gave me an initial condition to find the k values, then to plus them into the formula...i don't know how to start with no initial conditions. please help
     
  2. jcsd
  3. Nov 3, 2009 #2
    i have x_0 = -1

    and for y_0 im guessing i solve the diff eq and plug in -1? if i do that i get y_0= 1/2e
     
  4. Nov 3, 2009 #3

    Mark44

    Staff: Mentor

    Are you given that x0 = -1? You didn't include that fact in the initial post.

    It seems to me that you need the initial condition to solve the differential equation, so you can't just solve the diff. equation to find the initial condition.

    What is the exact statement of the problem?
     
  5. Nov 3, 2009 #4
    the exact problem is 1) apply runge kutta of 4th order to solve the ode on [-1,1] with n = 5
    (e^x+y)dx-dy=0

    were i got x_0 is -1 is from my notes that said x_0 = a.
    in my notes it says that y(x_0) = y_0

    im pretty sure my x_0 is right but i don't know how to get y_0
     
  6. Nov 3, 2009 #5

    Mark44

    Staff: Mentor

    Maybe you're supposed to do hand calculations with those two values: x_0 = a and y(a) = y_0. If your notes say x_0 = a, how do you get x_0 = -1 from that?
     
  7. Nov 3, 2009 #6
    he knows x_0 is -1 from the interval [-1,1]
     
  8. Nov 3, 2009 #7
    yes thats how i got -1.
     
  9. Nov 4, 2009 #8

    Mark44

    Staff: Mentor

    Your differential equation is dy/dx = f(x, y) = ex + y . To use Runge-Kutta of order 4 you need to either be given h, the stepsize, or just choose one, if it's not given in the problem.

    The algorithm is yn + 1 = yn + 1/6(k1 + 2k2 + 2k3 + k4), with
    k1 = h*f(xn, yn)
    k2 = h*f(xn + h/2, yn + k1/2)
    k3 = h*f(xn + h/2, yn + k2/2)
    k4 = h*f(xn + h, yn + k3)

    For your problem, x0 = -1, and y0 = y(x0) = y0, according to your earlier post. It would be better if you had an actual value for y0.
     
  10. Nov 4, 2009 #9
    it would be better, but none is given. do you know how i can figure out y_0?

    i have h = .4 from b-a/n 1--1/5 = .4

    once i have y_0 i can finish the problem, this is what i need help with.
     
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