Is There a Function G Such That G_x Equals -F_y and G_y Equals F_x in R?

[Medicol]

Suppose all second partial derivatives of F = F (x, y) are continuous and F_{xx} + F_{yy} = 0 on an open rectangle R.
Show that F_ydx - F_xdy = 0 is exact on R, and therefore there’s a function G such that
G_x = −F_y and Gy = F_x in R.
≈≈≈≈≈≈≈≈

To prove that F_ydx + F_xdy = 0 is exact on R,
I have F_{xx} + F_{yy} = 0
which is F_{xx}=-F_{yy}
Integrating both sides and cancel out the constants I obtain
F_x=-F_y
This proves the function F_ydx - F_xdy = 0 is exact on R​

Could you help me prove the existence of G ? Thank you...

 

[WWGD]

Suppose all second partial derivatives of F = F (x, y) are continuous and F_{xx} + F_{yy} = 0 on an open rectangle R.
Show that F_ydx - F_xdy = 0 is exact on R, and therefore there’s a function G such that
G_x = −F_y and Gy = F_x in R.
≈≈≈≈≈≈≈≈

To prove that F_ydx + F_xdy = 0 is exact on R,
I have F_{xx} + F_{yy} = 0
which is F_{xx}=-F_{yy}
Integrating both sides and cancel out the constants I obtain
F_x=-F_y
This proves the function F_ydx - F_xdy = 0 is exact on R​

Could you help me prove the existence of G ? Thank you...

If your rectangle lives in the plane, then just show the form F_ydx - F_xdy = 0 is closed, since in a rectangle, every closed form is exact. If you haven't seen this, what results can you use?

 

[Borek]

F_{xx}=-F_{yy}
Integrating both sides and cancel out the constants

How do you know constants are identical?

 

[Medicol]

I answered so, I meant to choose 2 exact constants to give both a go.
I'm thinking because I already answered the first which is also the main part of the problem, I may continue to put "the given function has become exact on R, so there must be a function G that satisfies both of the given conditions in R"

Is this a correct solution ?