A Very Nice Differential Equation | Surprising Subs
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- Опубліковано 6 жов 2024
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Method 3 multiply top and buttom by 1- cosz and continue
But would yiu agree most ppl will not think to replace x plus y with z maybe either..
@@leif1075 this is famous subsitution for who is take differential equations becase z' and y' differ only by a constant
4:00 I was hoping you'll get to that half-angle identity. It's a huge shortcut. 1/1 + cosx should be 1/cos^2(x/2)
@3:40 - according to wikipedia, it's misattributed to Weierstrass, but it was first used by Euler: en.wikipedia.org/wiki/Tangent_half-angle_substitution
The complete solution is y = 2*arctan(x+c)-x+2n*pi
Thank you
Huh. The second answer from Wolfram Alpha is equivalent to your answer. (1/2)(c + 2x) = c/2 + x, and c/2 is just another constant.
I suspect the other answer has something to do with how you define the range of the arctan function. Because arctan(tan(x)) is not always x. But I don't know why it would just be negative, and not involve something like adding 2nπ to the answer somewhere.
Half angle formula is best 😎...
Absolutely!
@SyberMath would you agree most nobody would think.of that sub. And maybe sub z equals cosine y..that's what I thought of or z equals cos(×+y) or z equals sine (×+y)..did you think of those?
@@leif1075it matters on what kind of problems you have experience with...
Half angle solution was pretty trivial to me but it might be different for others...
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Pythagoras Is very powerful.
those two answers are the same as inverse tangent is odd
What about y=pi-x, can it be a solution?
Cool
Very easy
let u = x + y
u' = y' + 1
y' = u' - 1
u' = 1 + cos u
∫du/(1 + cos u) = ∫dx
∫du/(2 cos^2 (u/2)) = ∫dx
1/2 ∫ sec^2 (u/2) du = ∫dx
tan(u/2) = x + c
u = 2 arctan(x + c)
x + y = 2 arctan(x + c)
y = 2 arctan(x + c) - x
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Nice. Surprising sub😂😂😂
a+bi dude be teaching me calc
a + bi dude does complex numbers only! 😍
dizzy, dizzy, dizzy.....