@numericalmethodsguy You cleared up my confusion. I just want to say that your teaching method is very effective. It makes a seemingly difficult subject a lot easier to understand.Thank you for the explanation and the videos.
@jmlew The initial condition is given as y(0)=5 which means y(x=0)=5. xsub0 is the value of x at which the initial condition is given, and ysub0 is the value of y at xsub0. So xsub0=0, ysub0=5. There is no assumption involved. Please reply if you have any issues with understanding my explanation.
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.
We do not. They are all 2nd order accurate. nm.mathforcollege.com/blog/html/compare_runge2.html math.stackexchange.com/questions/3337299/among-midpoint-method-heuns-method-and-ralston-method-which-method-of-solving math.stackexchange.com/questions/3302467/in-runge-kutta-methods-why-midpont-method-is-accurate-than-modified-euler-method
I think I figured it out, Ill write it here encase anyone has the same problem. for k1=f(y) k2=f(y+h*k1), and just exclude x. Seems simple now that I think back on it....
Andrew Walsh You should still treat it as function of x and y, f(x,y) and treat 3*exp(y)-0.04y+0*x. This allows you to proceed to think what the f(x,y) stand for at different values of the independent variable.
Jithin Kurien I know I’m Late but still if you still need help then here is the explaination: K1 and K2 are nothing but the slope of the function at different point. We take average of two slope. First is at the start and the other one is at the end( step size) so yeah it’s correct. hf(x,y) gives us the difference in function value between those two point which has nothing to do with slopes therefore, K1 is not equal to hf(x,y)
@numericalmethodsguy
You cleared up my confusion. I just want to say that your teaching method is very effective. It makes a seemingly difficult subject a lot easier to understand.Thank you for the explanation and the videos.
You have saved my life by your videos!! You are amazing!!
@jmlew The initial condition is given as y(0)=5 which means y(x=0)=5. xsub0 is the value of x at which the initial condition is given, and ysub0 is the value of y at xsub0. So xsub0=0, ysub0=5. There is no assumption involved. Please reply if you have any issues with understanding my explanation.
@nq316meb Heun's method may refer to the improved or modified Euler's method, just the predictor-corrector is not used.
Thank you so much. I’d like to send a shout out to the director(s) of your videos coz they are amazing and so is the camerawork.
@sakkarugzo What do you get? As given in the video, the exact answer is y(3)=2.763. Your procedure is correct. Please check again.
For i=0, are you assuming that x(0)="x not"=0 or did you somehow plug in y(0)=5 into the differential equation to solve for x(0)="x not"=0 ?
Great lecture series, nice quick movement, with clear explanation!
Where did you get the exact value to compare to?
Amazing lectures
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.
Hi Sir, there are 3 methods in runge Kutta 2nd order, how you know when to use the respective methods as the general formula is the same?
We do not. They are all 2nd order accurate.
nm.mathforcollege.com/blog/html/compare_runge2.html
math.stackexchange.com/questions/3337299/among-midpoint-method-heuns-method-and-ralston-method-which-method-of-solving math.stackexchange.com/questions/3302467/in-runge-kutta-methods-why-midpont-method-is-accurate-than-modified-euler-method
@@numericalmethodsguy thank you sir
Please verify with my calculations and reply.
k1 = 1
k2 = -1.93061
y1 = 4.30204
k1 = -1.05143
k2 = -0.9406
y2 = 2.80802
2023 and i am here now
thanks sir
Thank you! You're my hero!
what happens if f(x,y) is only a function of f(y) and independent of x? ex-> if your function was 3e^(-y)-.04y
I think I figured it out, Ill write it here encase anyone has the same problem. for k1=f(y) k2=f(y+h*k1), and just exclude x. Seems simple now that I think back on it....
Andrew Walsh You should still treat it as function of x and y, f(x,y) and treat 3*exp(y)-0.04y+0*x. This allows you to proceed to think what the f(x,y) stand for at different values of the independent variable.
Your approximation value might be wrong - after writing a code I get it to be 2.6503.
Anyway amazing lecture series!
1:46 i think it is wrong k1=h*f(x1,x2) please reply sir
Jithin Kurien I know I’m Late but still if you still need help then here is the explaination: K1 and K2 are nothing but the slope of the function at different point. We take average of two slope. First is at the start and the other one is at the end( step size) so yeah it’s correct. hf(x,y) gives us the difference in function value between those two point which has nothing to do with slopes therefore, K1 is not equal to hf(x,y)
How to find exact value?
Look for almost a similar problem for exact value. nm.mathforcollege.com/chapter-08-01-exact-solution-of-1st-order-ode/
hello please how did find the exact value 2.763 ???
autarkaw.org/2011/01/21/classical-solution-technique-to-solve-a-first-order-ode/
thank you very very much
Could you explain what happens if I have a vectorial function?
Like;
x1(dot)=2/3*x1-4/3*x1*x2
x2(dot)=x1*x2-x2
x1=x2=1
Thank you ?
You need to follow the resources given here: nm.mathforcollege.com/topics/higherorder_ode.html
Go to nm(dot)mathforcollege(dot)com and click on Keyword. Click on PRIMER ON ORDINARY DIFFERENTIAL EQUATIONS. You will see more resources.
Isn't he supposed to find "y(3)=?", he stopped at x=1.5... Someone explain!
At 8:29, you see the value of y(3).
Thank u so much sir 😊
just amazing! thanks a lot! :D!!!!!
Yes, apologies, found a small mistake! :)
is this not first order???
so y1=-1.04 yeah?