Chapter 08.03: Lesson: Runge Kutta Second Order Method Derivation: Part 2 of 2
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- Опубліковано 19 жов 2024
- Learn how Runge-Kutta 2nd order method of solving ordinary differential equations is derived. For more videos and resources on this topic, please visit nm.mathforcolle...
I am so confused! I'm sure it was a good explanation but far too fast! Need to watch it again about 7 more times! Thanks for making the video available to the world :)
@0wannabegeek0 The issue to note is that x and y are not independent variables, y is dependent on x, hence the chain rule to find f'(x,y).
@0wannabegeek0 The motivation was to solve the ODE numerically to 2nd order accuracy (using three terms of Taylor series) without having to find the first derivative of f(x,y) (that is, y") as that would involve symbolic application of chain rule. That is the beauty of Runge-Kutta 2nd and 4th order methods.
Thank you for the derivation - the only thing I think needs clarification is where the function K2 actually comes from? It seems somewhat arbitrary.
@0wannabegeek0 If you have DE with more than one independent variable, you have partial differential equations, and the numerical methods for such equations are different.
Thank you for derivation. Much Appreciated.
Where did you get the expression of k2 from??
See the pdf file for proof. mathforcollege.com/nm/mws/gen/08ode/mws_gen_ode_txt_runge2nd.pdf
Thank You so much! It helped me a lot
Thank you. To get even more help, go to MathForCollege.com/nm for more resources and share the link with your friends. Follow my numerical methods blog at AutarKaw.org. You can also take a free online course at www.canvas.net/?query=numerical%20methods
Thank you for these videos, I'm studying by myself this quarantine. It's a great help :) I hope you can tell us about you reference book
Go here for all the references: nm.mathforcollege.com/ The textbook is here nm.mathforcollege.com/textbook-numerical-methods-with-applications/ Tell your friends.
any of you guys did the homework? I really need it. :( please?