A Flipped Differential Equation

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[@skwbusaidi]

You can also solve by multiplying by 1-cosz
We get
Cosz(1-cosz)/sin^2 z
= (cosz-cos^2 z)/ sin^2 z
= cscz cotz - cot^2 z
= cscz coz +1-csc^2 z
Now integerate
-cscz + z + cotz = x+c
Final answer
y + cot(x+y) - csc(x+y) = c

[@ozgurtankut7404]

Is it impossible that we just rewrite the question as "dy/dx=cos(y+x)", solve it for y and then name all x's as y and all y's as x? I mean, cant we just substitue y with x and x with y and at the end turn them again?

[@ldanielmule8]

I've got with that as well and though the inverse derivative theorem could simplify it a lot, but i'm not sure

[@Chrisoikmath_]

Can't we fine exactly y in terms of x and c only?

[@bkkboy-cm3eb]

dx/dy=cos(x+y)
x+y=u dx/dy+1=du/dy
cos(u)+1=du/dy
dy = du/(cos(u)+1)
= (1-cos(u))/sin²(u)·du
y = -1/tan(u) + 1/sin(u) + C
y +1/tan(x+y) - 1/sin(x+y) = C

[@NowInAus]

Lovely and don’t apologise for going over 10 minutes!

[@SyberMath]

🥰

[@subversively6680]

PLZ, answer this
Why flipping dx/dy
Can not be solved as x is function of y?????

[@kevinmadden1645]

I believe you are correct .

[@Chrisoikmath_]

Sure! That's why the final equation is symmetric with respect to x and y.

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@dominicellis1867]

Does an explicit expression exist for this implicit solution?

[@tixanthrope]

flipped does not mean much as the RHS is symmetrical in x and y.

[@84com83]

Solving for "y", why not Forex?

[@SyberMath]

😜

[@DonEnsley]

Problem
dx/dy = cos(x+y)
Invert.
dy/dx = 1 / cos(x+y)
Let z = x + y
y = z-x
dy/dx = dz/dx - 1
dz/dx = 1 / cos z + 1
= ( 1 + cos z )/ cos z
∫[ cos z / (1+cos z) ] dz = x + k
∫[ (cos z +1-1)/ (1+cos z) ] dz = x + k
∫dz - ∫ [ 1 / (1+cos z) ] dz = x + k
z- ∫ 1 / [1+cos 2(z/2) ] dz = x + k
z- ∫ 1 / [1+2cos²(z/2) -1] dz = x + k
z- ∫ 1 / [2 cos² (z/2)] dz = x + k
u = z/2
dz = 2 du
z- ∫ 1 / [cos² (u)] du = x + k
z- ∫ sec² (u) du = x + k
z - tan u = x + k
z - tan (z/2) = x + k
x+y - tan((x+y)/2) = x + k
y - tan((x+y)/2) = k
x + y = 2 tan⁻¹(y - k)
x = 2 tan⁻¹(y - k) - y
x = ± 2 tan⁻¹(y - k) - y
This gives you
dx/dy = [1-(y-k)²]/[1+(y-k)²]
tan [(x+y)/2] = y - k
So
dx/dy = {1-tan²[(x+y)/2]}/{1+tan²[(x+y)/2]}
Multiplying top and bottom by cos² [(x+y)/2]
dx/dy = cos² ((x+y)/2) - sin² ((x+y)/2)
Or
dx/dy = cos(x+y)
Solution:
x = ± 2 tan⁻¹(y - k) - y
Now y in terms of x I could not find.