Відео

Good mathemation

Thank you so very much!❤

I enjoyed listening to you talk about this book, Mykyta. The power of one well written book to effect change is amazing!

Thanks for the video❤ I'll definitely put this book on my Christmas wish list

what would we do if we have 2 diff functions ?

Seriously, thank you so freaking much. You did such an amazing job with this video. I don’t think I’ve ever seen an excel math video that was so easy to follow before

can you do improved euler in excel too? great video btw.

Damn

thank you ao much

I love you

For the comparison with the exact solution, we need to get the integration of the derivative, y' = dy/dx = 2xy We divide both sides by y, 1/y (dy/dx) = 2x and then integrate both sides, ∫ (1/y) dy = ∫ 2x dx ∫ (1/y) dy = 2 * ∫ x dx the solution on both sides is given by, ln y = 2 * ((1/2) x^2 + C) ln y = (x^2 + C) and using the exponential (the antilog) on both sides, y = e^(ln y) = e^(x^2 + C) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) Now, the question is: From where we get the confirmations that C = -1? For y =1, and x =1 y = e^(ln 1) = e^(1^2 + C) y = e^(ln 1) = e^(1 + C) ln 1 = 1 + C 0 = 1 + C -1 = C Now, we write equation (1) as follows, y = e^[(x^2) -1] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . (2) Now, for x =1, y is given by, y = e^[(1^2) -1] = e^(1 -1) = e^0 = 1 and for x =1.1, y is given by, y = e^[(1.1^2) -1] = e^(1.21 -1) = e^0.21 = 1.233678 and for a list of x values, y is given as follow, x (x^2) (x^2)-1 e^[(x*2)-1] 1.00 1.000000 0.000000 1.000000 1.10 1.210000 0.210000 1.233678 1.20 1.440000 0.440000 1.552707 1.30 1.690000 0.690000 1.993716 1.40 1.960000 0.960000 2.611696 1.50 2.250000 1.250000 3.490343

A foto de EULER e´ na verdade a de Gauss. Corrija

Amazing!

very useful! thank you so muchhh

FUCK! thx for ur video men! Many vids describe this method about 2 houra, you showed us this in 10 minutes. And now I can use ur material in engineering practice.

Thanks a lot..!!

Thank you so much Sir. Please make more videos . Very much helpful.

GODDAMNIT THIS WAS SO GOOD GAH

Thank you so much! This is so helpful. I hope you get a raise for this :)

What an excellent tutor Micky is! Hope he releases more videos :D