This was very quick, concise, and told me exactly what I needed to know. My group in my numerical analysis class is doing a term project on the Runge-Kutta method. Keep teaching! We need good math teachers!
Thank you so much. I was struggling with RK4 until I watched your video. After watching the video, I was able to immediately implement RK4 into an excel spreadsheet. 👍
Kind-of a centrally-weighted-k'luck-k'luck-k'luck-k'luck method...but, why, should we think that'd bracket it: where, did this method overstep so as to average in-between-? If we're hunting for a midpoint at which its derivative equals the slope, halfway h/2 works for y=x². _Period!_ Howabout then we bracket with its inverse, dx/dy, (shades of Newton-Raphson)... (Also, When dy/dx or dx/dy gets large, tracking errors get large: Shouldn't h slow-down?)
at 5.16 there is a mistake T4 (x1,y1,h) you forget to write the h (( which is the step size )) otherwise the video is nice. thank you it helped me in studying.
If the problem states h=0.1, Xsub 0 = 0 and there is an interval 0≤X≤1, does it mean that you will stop the solution when Xn = 1? Xsub 0 = 0 Xn = Xsub 0 + n(h)? 1 = 0 + n(0.1) n = 10 Does it mean I have to perform 10 times using the formula?
Approximate the solution for the third-order initial-value problem y^'''-ty^''-4y^'+3y=2t, for 0≤t≤1 With y(0)=3,y^' (0)=2,y^'' (0)=7 using Runge-Kutta method of order four by considering the step size h=0.2 help please????
Hello I was wondering if anyone knows of a way to take a system of equations derived from a circuit and apply the Runge-Kutta method to it. I'm looking for advice or links if available thank you.
When you are using the Runge-Kutta method, you are moving up a small step size every time you calculate the T4 value. You need to use several steps to get to the value you want. The smaller the step size used, the more accurate the answer will be. For every small step taken, this calculation is done again.
l = 120; %cm s = 80; % cm% R = 2*l/s; %l=connecting rod length % s is stroke d= 6.7 %cm Vd = (pi/4)*(d^2)*s; % d is bore % s is stroke a_1 = (R^2 - (sind(theta))^2)^-0.5; dV= (Vd/2)*sind(theta)*(1+(cosd(theta)*a_1)) Sir, i want to solve above equation dV with respect to theta usuing RK4 method in matlab plz help me
This was very quick, concise, and told me exactly what I needed to know. My group in my numerical analysis class is doing a term project on the Runge-Kutta method.
Keep teaching! We need good math teachers!
Me too, Elizabeth! :-)
I've having problem with Runge-Kutta Method. Your video is solid, simple and straight to the point. Thank you sir so much.
Thanks for ths amazing post, I understand the concept so much clearly and well explained. I hope you get rewarded for your efforts. God Bless
Thanks very much for the clear and concise video, it helped out a lot with my presentation last semester! (I totally rocked that presentation btw)
Thank you so much. I was struggling with RK4 until I watched your video. After watching the video, I was able to immediately implement RK4 into an excel spreadsheet. 👍
Thanks! as a non-math major this is much simpler than I though, it's basically addition, multiplication and division
Very nice. Your explanation is explicit and straight to the point. Thank you.
THANK YOU!! Everything great but man you should really work on those 7's they look like a mutant hybrid between 2 and 1.
Johan O 😂
lol
I just've watched the video and hoped to find this sorta comment. Hopefully i did
This screencast has been reviewed by faculty from other academic institutions.
How do you do second order linear differential equations with this?
@@gabor6259 have you tried converting the higher order DE to a system of first order DEs- I think that's how I'd approach 2nd order DEs
@@gabor6259 The method shown here is 4th Order Runge-Kutta, look into 2nd Order. It has different equations
Sweeeeeeeeeeeet, literally no idea what my lecturer was going on about. This seems a world of difference.
Watching this during my exam right now thanks man 👍
Lmao! Do you have your exam at home, or are you cheating at the campus?
@@hannananan9427 It was an online exam lol
Thank you Sir. I felt like I was being taught by Stan Lee himself
Thank you, as an introduction, it couldn't be clearer or more concise!
Thank you for this great video. It was really helpful going step by step, made everything more clear.
شكرا على الشرح❤
very nicely explained ! thank you !
how did you get x1? that corresponds to y1?
Senor chino The x changes by the size of the step. In this case the step size is 0.1. x0 = 1, so x1 = 1.1.
+Senor chino x1 = x0 + h;
Thank you so much it helped me I'm from University of Béjaïa Algeria
Awesome video sir. Thank you! Very useful!
Kind-of a centrally-weighted-k'luck-k'luck-k'luck-k'luck method...but, why, should we think that'd bracket it: where, did this method overstep so as to average in-between-? If we're hunting for a midpoint at which its derivative equals the slope, halfway h/2 works for y=x². _Period!_ Howabout then we bracket with its inverse, dx/dy, (shades of Newton-Raphson)... (Also, When dy/dx or dx/dy gets large, tracking errors get large: Shouldn't h slow-down?)
this is runge-kutta 4th order.
Great explanation sir!
Very nice video. Informative, quick and to the point.
Wish you could replace my lecturer lol.
at 5.16 there is a mistake
T4 (x1,y1,h)
you forget to write the h (( which is the step size ))
otherwise the video is nice. thank you it helped me in studying.
Thank you 🙏
If the problem states h=0.1, Xsub 0 = 0 and there is an interval 0≤X≤1, does it mean that you will stop the solution when Xn = 1?
Xsub 0 = 0
Xn = Xsub 0 + n(h)?
1 = 0 + n(0.1)
n = 10
Does it mean I have to perform 10 times using the formula?
Yes, it does. This is why a computer program is nice for these calculations.
thank you
Thank you sir.
thanks lot. its has been very helpful
How is possible applicate this method to find displacement from a set of acceleration data?
thank you so much !!!! great explanation 😍
Actually you do it right. But my text multiplies the values of k individually by h.
thank you . it was very helpful
Dog bark at 1:27 hahahah great video man helped me a lot and my test is in two days! cheers
hey! it's been 6 years, how was your test? cheers
very nice!!!!!! thank you
Approximate the solution for the third-order initial-value problem
y^'''-ty^''-4y^'+3y=2t, for 0≤t≤1
With y(0)=3,y^' (0)=2,y^'' (0)=7 using Runge-Kutta method of order four by considering the step size h=0.2
help please????
Nicely explained! Thanks a lot!
THIS VIDEO VERY HELPUL, THANKS
Thank you, pretty clear
your differential equation has to be from the 4th order to use runge-kutta 4th order ?
no your eq has to be 1st order
Aren't y values y + k/2 instead of y + (h/2)k?
I thought so too, maybe this is another more in depth Runge Kutta method? I'm surprised I actually found something useful in the comments.
nice explain the RK4 method ........ tankzzzzzzz
I want a general derivation of this method.
it's 2am on a tuesday and i'm desperately trying to understand this for a final project. I'll update if i decide to quit STEM
Nice Video ! Helpful
thanks for the help!
Thanks a lot
OH yea, this is just what I needed.
very nice.
Could you explain me what the step size is?
The step size is the amount by which x is increased for each step. So if Xn=1 and our step size is 0.1, X(n+1) is X+h, or 1.1.
wait sir how the k2 i try put in calculator and got 7.872 not 2.602?
7.872 is not the correct answer, it is 7.607 the 7 just looks similar to a 2
why do you have Y+K1H/2 for K2,k3,k4 It should be Y+K1/2 @ 2:36
GOOD JOB BRO
cool i now understand !!!
How to find the value of "h" when not given in question?
assume it
Hello I was wondering if anyone knows of a way to take a system of equations derived from a circuit and apply the Runge-Kutta method to it. I'm looking for advice or links if available thank you.
have got any idea on how to solve system of equation mate, even i am looking for it. could you please help me.
Like everyone else, "Thank You"
te amo
Y? Should not continue after K4
When you are using the Runge-Kutta method, you are moving up a small step size every time you calculate the T4 value. You need to use several steps to get to the value you want. The smaller the step size used, the more accurate the answer will be. For every small step taken, this calculation is done again.
excellent :)
good
This is wrong. You have to multiply the k values by h.
l = 120; %cm
s = 80; % cm%
R = 2*l/s; %l=connecting rod length % s is stroke
d= 6.7 %cm
Vd = (pi/4)*(d^2)*s; % d is bore % s is stroke
a_1 = (R^2 - (sind(theta))^2)^-0.5;
dV= (Vd/2)*sind(theta)*(1+(cosd(theta)*a_1))
Sir, i want to solve above equation dV with respect to theta usuing RK4 method in matlab plz help me
when ur prof too lazy to do examples himself and just throw u the link to this vid
I only found this video to know how Runge-Kutta is pronounced.
Good job! But, Can someone teach this guy how to write 7 PLEASE?
y1=2,7846, it's hard to understand your "7" kkkkk.
resolution is terrible, content good
Xanthi
Thank you.