Homework Help: Diff Equations

1. Sep 7, 2006

ISU20CpreE

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

$$(Cos(2y)-Sin (x)) dx-2 Tan (x) Sin (2y) dy = 0$$

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

So, my

$$M_{y}$$ is equal to $$-2 Sin (2y)-0$$ and,

my $$N_{x}$$ is equal to $$-2(Sec^{2}(x)) Sin (2y)$$

Which makes it not exact. So, then I tried using
$$\frac{M_{y}-N_{x}}{-N}$$ 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.

2. Sep 8, 2006

HallsofIvy

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

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

3. Sep 8, 2006

ISU20CpreE

Sorry for not specifying I got:

$$\frac {2Sin(2y)}{Cos^2 x} -Sin (2y)$$

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

Last edited: Sep 8, 2006
4. Sep 8, 2006

HallsofIvy

Well, that can't be right. I thought that the whole reason you mentioned (Mx- 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.