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
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.
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
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
What an excellent tutor Micky is! Hope he releases more videos :D
Thank you so much! This is so helpful. I hope you get a raise for this :)
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.
Thank you so much Sir.
Please make more videos .
Very much helpful.
GODDAMNIT THIS WAS SO GOOD GAH
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
I also use this way
very useful! thank you so muchhh
Good mathemation
Thank you so very much!❤
Amazing!
Thanks a lot..!!
thank you ao much
what would we do if we have 2 diff functions ?
A foto de EULER e´ na verdade a de Gauss. Corrija
Damn
I love you