Finding an Integration Factor for (Cos(2y)-Sin (x)) dx-2 Tan (x) Sin (2y) dy = 0

[ISU20CpreE]

Hi there i am trying to make this equation look exact.

[tex](Cos(2y)-Sin (x)) dx-2 Tan (x) Sin (2y) dy = 0 [/tex]

What I've done so far is take the partial with respect to x and y.

So, my

[tex]M_{y}[/tex] is equal to [tex]-2 Sin (2y)-0[/tex] and,

my [tex]N_{x}[/tex] is equal to [tex]-2(Sec^{2}(x)) Sin (2y)[/tex]

Which makes it not exact. So, then I tried using
[tex]\frac{M_{y}-N_{x}}{-N}[/tex] and,

here is where I have tried so many times to find out a way to find an Integration Factor (I.F.)

Any suggestions will help, thanks for your time.

 

[HallsofIvy]

But, other than saying that you calculated (M_y- N_x)/(-N) (and you don't say what you got for that), you don't say what you have tried!

Why did you calculate (M_y- N_x)/N (which was in fact a very good thing to do!) ? What did it tell you?

 

[ISU20CpreE]

But, other than saying that you calculated (M_y- N_x)/(-N) (and you don't say what you got for that), you don't say what you have tried!

Why did you calculate (M_y- N_x)/N (which was in fact a very good thing to do!) ? What did it tell you?

Sorry for not specifying I got:

[tex] \frac {2Sin(2y)}{Cos^2 x} -Sin (2y) [/tex]

and my integration factor is the one i need help on.

 

[HallsofIvy]

Well, that can't be right. I thought that the whole reason you mentioned (M<sub>x- Ny)/N was because it gave you something worthwhile!

M= cos(2y)- sin(x) so My= -2sin(2y). N= 2tan(x)sin(2y) so Nx= 2sec²(x)sin(2y). My- Nx= -2sin(2y)- 2tan(x)sin(2y)= -2sin(2y)(1- tan(x)). (My- Nx)/N= -2sin(2y)(1- tan(x))/2tan(x)sin(2y)= -2(1- tan(x))/2tan(x) which is a function of x only.

I thought the reason you mentioned (My- Nx)/ was the fact that you recognized that that was a function of x only and therefore that an integrating factor would be a function of x only.

If we multiply the equation by some f(x) we get
[tex](cos(2y)-sin(x))f(x)dy- 2tan(x)sin(2y)f(x)dx= 0[/itex]
and, in order that this be an exact equation we must have
(-2sin(2y)f+ (cos(2y)- sin(x))f'= 2tan(x)sin(2y)f'+ -2tan(x)sin(2y)f
That is an equation in x only for f.</sub>