Runge Kutta 4th Order Method: Example Part 1 of 2
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So my lecturer goes to us today, "So you should all be aware of the Ronge-Kutta and Euler methods". The whole class (first years) just looked at eachother with baffled faces. Thank you for uploading this, you have saved my Mathematical Physics GPA!!
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The correction for k3 has been posted
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All Comments (41)
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Macho pose at 8:16 by the way your brilliant
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Wow Mr. Numericalmethodsguy! You are the best! I would have failed my final exam without u! Thank you from Sydney Australia.
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Legg inn respons på denne videoen See 1:39
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@greynight234 He multiply it in the end instead. Here :y(n+1)= y(n) +1/6(k1+2k2+2k3+k4) since h is a constand which is in all k1,k2,k3,k4 it can be multiplied here instead.
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Hi, do you always use i=0 first? what if your asked to estimate the y(0.4)..would you use i=0.4 after using the i=0 ? please if you could answer this as i have an exam today. Thanks
arqi25 1 week ago
@arqi25 The variable i is the subscript, not the value of the independent variable. So if the ODE is of the form dy/dx=f(x,y), y(0)=y0, then x_sub_i+1=x_sub_i+h, x_sub_0=0; y_sub_0=y0. So if you want to estimate y(0.4), then what the value of i us at x=0.4 depends on the step size h. If h=0.2, it would be y_sub_2 that you will be seeking.
numericalmethodsguy 1 week ago
The formula which You used for k1, k2 k 3 and k4 is change what is in wikipedia and ...i have seen this formula as k1=h f(xn,yn)
h is multiplying in all k values but in yours it is not.. Kindly tell me ASAP tomorrow is my paper.
greynight234 5 months ago
@greynight234 h is multiplied separately. For more info go to numericalmethods(dot)eng(dot)usf(dot)edu and click on Keyword. Go to Ruge-Kutta 4th order method and read the textbook chapter!
numericalmethodsguy 5 months ago
can't you use the linear ODE equation for this?
Can you solve any nonlinear ODE with this?
presidentevil 1 year ago
You can solve any linear or nonlinear ODE of the form dy/dx=f(x,y) with Runge-Kutta methods.
numericalmethodsguy 1 year ago