A Very Nice Differential Equation | Surprising Subs

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[@skwbusaidi]

Method 3 multiply top and buttom by 1- cosz and continue

[@leif1075]

But would yiu agree most ppl will not think to replace x plus y with z maybe either..

[@skwbusaidi]

@@leif1075 this is famous subsitution for who is take differential equations becase z' and y' differ only by a constant

[@Sg190th]

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)

[@jimschneider799]

@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

[@hlee4248]

The complete solution is y = 2*arctan(x+c)-x+2n*pi

[@ZipplyZane]

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.

[@Aman-simple]

Half angle formula is best 😎...

[@SyberMath]

Absolutely!

[@leif1075]

​@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?

[@arnavdeep8396]

​@@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...

[@user-hl7lt8ge7o]

«have sence? Ku-ku!” - it was fun! Thank you:)))))))

[@agrimmittal]

Thank you

[@rob876]

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

[@sunildhuri8421]

Very easy

[@tommyjenga5976]

Cool

[@yoav613]

Nice. Surprising sub😂😂😂

[@andypandy6063]

Pythagoras Is very powerful.

[@ymj5161]

those two answers are the same as inverse tangent is odd

[@hafez591]

What about y=pi-x, can it be a solution?

[@barakathaider6333]

👍

[@maballshurt]

a+bi dude be teaching me calc

[@SyberMath]

a + bi dude does complex numbers only! 😍

[@師太滅絕]

dizzy, dizzy, dizzy.....