@@SyberMath You are the best. Best wishes from an Indian from Dubai. I have watched all your videos. And this brings a lot of joy to me. Thank you so much.
y'dy'/dy = 2yy' y' = 0 or ∫dy' = 2∫ydy y = const or y' = y^2 + C1 y = C or ∫dy/(y^2 + C1) = ∫dx y = C or arctan(y/C2)/C2 + C3 = x y = C or y = C2 tan(C2(x - C3))
The solution has VERY different behavior when c0, you actually get the particle to reach infinity in finite time (since you have tan(kt+c) in there)
Will you please provide a PDF of difficult maths problems ?
I am 42 years old. And i loved maths when I was younger, and your videos make my love for maths to another level.
I'm glad to hear that! Thanks for watching 😍
@@SyberMath You are the best. Best wishes from an Indian from Dubai. I have watched all your videos. And this brings a lot of joy to me. Thank you so much.
I am 50 and feel the same way. As a matter of fact, I told him the same thing a couple years ago.
Nice!
Thanks!
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That is a known result!
y=kTan(kx+a)
if one can observe the soln is easy as on right side we derivative of Y^2 wrt x
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If c=0? That value is missed. If c=0, that integral is -1/y=x+k...
Also if c is negative
what if c
In the end, put c as you want it and that should answer your question
@@Roq-stone did you try to understand yourself what you have wrote? For example the solutions of z'=z^2+c for c=5 and c=-5 are totally different.
Please sir❤❤
Have you replaced y in terms of his solution?
Hello sir
You didn't consider the cases c=0 and c
trivial cases. y = 0
(y')'=(y^2)'...y'=y^2+c,..dy/(y^2+c)=dx...(1/√c)arctg(y/√c)=x+C1...y/√c=tg(√c(x+C1))
y'dy'/dy = 2yy'
y' = 0 or ∫dy' = 2∫ydy
y = const or y' = y^2 + C1
y = C or ∫dy/(y^2 + C1) = ∫dx
y = C or arctan(y/C2)/C2 + C3 = x
y = C or y = C2 tan(C2(x - C3))
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