Thank you. Please subscribe and ask your friends to subscribe - our goal is to get to 100,000 subscribers by the end of 2021. To get even more help, subscribe to the numericalmethodsguy channel ua-cam.com/users/numericalmethodsguy, and go to MathForCollege.com/nm and MathForCollege.com/ma for more resources. Follow the numerical methods blog at AutarKaw.org. You can also take a free massive open online course (MOOC) at canvas.instructure.com/enroll/KYGTJR Please share these links with your friends and fellow students through social media and email. Support the channel if you able to do so at ua-cam.com/users/numericalmethodsguy/store
Very educating! Please can you refer me to the electronic circuit question of this solution? I mean with regards to its capacitor, diode, and resistor values. I need to write some codes as regards Runge-Kutta method on matlab with all the error estimates. Thanks.. I will be so glad if you reply to this. Appreciate :)
Dear Sir, Thank you very much for this amazing video but at the end why did you change the sign of the function (3e^-x -0.4*y) in the calculation of K4 you changed it to +0.4y.???!!! Again appreciating your explanation.....
The procedure is not any different. Remember v is a function of t, so if dv/dt is a function of v, it eventually is a function of t also. Step size does effect as step size is not t but deltat.
On reflection I think this example from numericalmethodsguy is confusing as this method is better described with a time variable t instead of x, and then use x as the variable which changes over time. So then the equation is dx/dt = f(x). Although this gets confusing in two or more dimensions where x is a vector e.g. in 3 dimensions write x (underlined) and use variables x, y and z, then write dx/dt=f1(x, y, z) dy/dt=f2(x, y, z) etc.
I think you are talking about 2nd order ODEs whch are set up as two simultaneous 1st order ODEs. See examples of this by 2nd order (not 4th order) method by going to numericalmethods(dot)eng(dot)usf(dot)edu and click on Keyword. Click on Higher Order ODEs. You will see more resources.
Answer to Quân Nguyễn Minh: I solved the differential equation of this video using a general formula for this type of equation. The differential equation of this video was: y’+0.4y= 3exp(-x) y(0)=5 this is a first order linear differential equation which has this general form: y’+f(x)y=r(x) The formula to solve it is the following: y=exp(-h)( integral(exp(h)r(x)dx) + c ) where h is: h= integral (f(x)dx) In the equation of the video we have that: f(x)=0.4 h=integral(f(x)dx)=integral(0.4dx)=0.4x r(x)= 3exp(-x) I substituted in the general formula to solve the equation: y=exp(-0.4x)( integral(exp(0.4x)3exp(-x)dx) + c ) y=exp(-0.4x)( integral(3exp(0.4x-x)dx) + c ) y=exp(-0.4x)( integral(3exp(-0.6x)dx) + c ) y=exp(-0.4x)( 3exp(-0.6x)/(-0.6) + c ) y=exp(-0.4x)( 3exp(-0.6x)/(-3/5) + c ) y=exp(-0.4x)( -(5/3)3exp(-0.6x) + c ) y=exp(-0.4x)( -5exp(-0.6x) + c ) y=-5exp(-0.4x)exp(-0.6x) + c exp(-0.4x) y=-5exp(-x) + c exp(-0.4x) I calculated c using the initial condition y(0)=5 y(0)=-5exp(-0) + c exp(-0.4(0))=5 -5exp(0) + c exp(0)=5 -5+c=5 c=5+5=10 Therefore the solution is: y=-5exp(-x)+10exp(-0.4x) So the exact solution for y(3) shown in this video is: y(3)=-5exp(-3)+10exp(-0.4(3)) y(3)=-5exp(-3)+10exp(-1.2) y(3)=-0.248935+3.011942 y(3)=2.763007≈2.763 I hope this information was useful for you. I am a graduate student of specialization in electric power systems at Central University of Venezuela in Caracas. Best regards from Venezuela.
h is something you will choose. To get an accurate value choose different step sizes such as 0.1, 0.05, 0.025 and so on. Find the value of y(1.1) and see what the absolute relative approximate error is for each step size. When it is below the prespecified tolerance, you have your answer to the accuracy you have chosen.
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I'm in IITDelhi and I'm studying this one night before my major exam.
very nicely explained
his channel is goldmine for numerical analysis
It's always great to see videos like this get the good ratings and view count they deserve. Keep up the good work!
seriously man, thank you very much !
CLEAR AND EASY TO UNDERSTAND
thax so much,examples makes us to understand it much better.
much thanks for the videos. you are much better at explaining this then my professor
@ingcarcruz k1, k2, k3 and k4 are all slopes of dy/dx. The weighted average gets multiplied later by h.
this is pretty awesome stuff, I am finding this really useful for numerical methods and its applications in programming, thank you for the upload!
Fantastic explanation; thank you!
Thank you. Please subscribe and ask your friends to subscribe - our goal is to get to 100,000 subscribers by the end of 2021.
To get even more help, subscribe to the numericalmethodsguy channel ua-cam.com/users/numericalmethodsguy, and go to MathForCollege.com/nm and MathForCollege.com/ma for more resources.
Follow the numerical methods blog at AutarKaw.org. You can also take a free massive open online course (MOOC) at canvas.instructure.com/enroll/KYGTJR
Please share these links with your friends and fellow students through social media and email.
Support the channel if you able to do so at ua-cam.com/users/numericalmethodsguy/store
Very educating! Please can you refer me to the electronic circuit question of this solution? I mean with regards to its capacitor, diode, and resistor values. I need to write some codes as regards Runge-Kutta method on matlab with all the error estimates. Thanks.. I will be so glad if you reply to this. Appreciate :)
Thank u. To the point explanation
very good explanations,makes my project much easier to do, thanks!
very fast clean and clear!!!!!!!!!
You 're the best!!!!!!!!!!!!!!!!
Thank you Sir, it was very helpful
lmao...uz ah bosss!!! made something difficult become so simple. thanks man
Thanks a lot... It made it so easy to understand the complicated runge kutta method... Thanks heaps
Dear Sir,
Thank you very much for this amazing video but at the end why did you change the sign of the function (3e^-x -0.4*y) in the calculation of K4 you changed it to +0.4y.???!!!
Again appreciating your explanation.....
loved it... very helpful ty😊
Thank you very much for this. It helped me a lot.
Thanks for your video, can you explain in detail about how to modelling of Runge - Kutta 4th order method.
The procedure is not any different. Remember v is a function of t, so if dv/dt is a function of v, it eventually is a function of t also. Step size does effect as step size is not t but deltat.
thank you so much for this lecture , sir ! :) really helpful for me..
Go to MathForCollege.com/nm for more resources. Follow my numerical methods blog at AutarKaw.org
Thank you very much for this
Thank you! Such a helpful video!
Good example, although as ingcarcruz pointed out below the correct equation for k1 is f(x(t), y(t))h.
On reflection I think this example from numericalmethodsguy is confusing as this method is better described with a time variable t instead of x, and then use x as the variable which changes over time. So then the equation is dx/dt = f(x). Although this gets confusing in two or more dimensions where x is a vector e.g. in 3 dimensions write x (underlined) and use variables x, y and z, then write dx/dt=f1(x, y, z) dy/dt=f2(x, y, z) etc.
he multiplied the 'h' in the y(i) equation later
thanks a lot for this great job
Great job. Thanks heaps.
I think you are talking about 2nd order ODEs whch are set up as two simultaneous 1st order ODEs. See examples of this by 2nd order (not 4th order) method by going to numericalmethods(dot)eng(dot)usf(dot)edu and click on Keyword. Click on Higher Order ODEs. You will see more resources.
hello sir, is it ok if i send you some of my work for review?
Thank you so much!
hi sir, may i know how to get the exact value??
Please go here. autarkaw.wordpress.com/2011/01/21/classical-solution-technique-to-solve-a-first-order-ode/
Damn! This actually makes sense!!
how I solve the second order ordinary DE using Runge Kutta Method ??
nm.mathforcollege.com/topics/higherorder_ode.html. See resources at this site.
Srsly,tnks a lot
Hi sir I have a question for you: how to find exact y(3)? I don't understand 2.763.
See solution at autarkaw.org/2011/01/21/classical-solution-technique-to-solve-a-first-order-ode/
thank you, sir
sir, could you make a clip of curve fitting ?
Curve fitting has two topics - regression and interpolation. Go here to learn about interpolation and regression. nm.mathforcollege.com
Answer to Quân Nguyễn Minh:
I solved the differential equation of this video using a general formula for this type of equation.
The differential equation of this video was:
y’+0.4y= 3exp(-x) y(0)=5
this is a first order linear differential equation which has this general form:
y’+f(x)y=r(x)
The formula to solve it is the following:
y=exp(-h)( integral(exp(h)r(x)dx) + c )
where h is:
h= integral (f(x)dx)
In the equation of the video we have that:
f(x)=0.4
h=integral(f(x)dx)=integral(0.4dx)=0.4x
r(x)= 3exp(-x)
I substituted in the general formula to solve the equation:
y=exp(-0.4x)( integral(exp(0.4x)3exp(-x)dx) + c )
y=exp(-0.4x)( integral(3exp(0.4x-x)dx) + c )
y=exp(-0.4x)( integral(3exp(-0.6x)dx) + c )
y=exp(-0.4x)( 3exp(-0.6x)/(-0.6) + c )
y=exp(-0.4x)( 3exp(-0.6x)/(-3/5) + c )
y=exp(-0.4x)( -(5/3)3exp(-0.6x) + c )
y=exp(-0.4x)( -5exp(-0.6x) + c )
y=-5exp(-0.4x)exp(-0.6x) + c exp(-0.4x)
y=-5exp(-x) + c exp(-0.4x)
I calculated c using the initial condition y(0)=5
y(0)=-5exp(-0) + c exp(-0.4(0))=5
-5exp(0) + c exp(0)=5
-5+c=5
c=5+5=10
Therefore the solution is:
y=-5exp(-x)+10exp(-0.4x)
So the exact solution for y(3) shown in this video is:
y(3)=-5exp(-3)+10exp(-0.4(3))
y(3)=-5exp(-3)+10exp(-1.2)
y(3)=-0.248935+3.011942
y(3)=2.763007≈2.763
I hope this information was useful for you. I am a graduate student of specialization in electric power systems at Central University of Venezuela in Caracas. Best regards from Venezuela.
By using RK4 method compute y(1.1) for the equation y'=3x+y^2 ,y(1)=1.2?
What is the value of h?
h is something you will choose. To get an accurate value choose different step sizes such as 0.1, 0.05, 0.025 and so on. Find the value of y(1.1) and see what the absolute relative approximate error is for each step size. When it is below the prespecified tolerance, you have your answer to the accuracy you have chosen.
Putting an example is very helpfull . I am surprised by the accuracy of this method . 0,15% of error if a step of 1,5 . Are you kidding ?
sir how x2=3
h=1.5; x0=0; x1=x0+h=1.5; x2=x1+h=3
Who the hell disliked this video? Come back and apologize now! >:(
exam in 2 hours, haven't even opened the book yet..
1 person prefers direct methods...