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

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Homework Help Overview

The discussion revolves around finding an integration factor for the differential equation (Cos(2y)-Sin(x)) dx - 2 Tan(x) Sin(2y) dy = 0, which is not exact as determined by the original poster's calculations of the partial derivatives.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to make the equation exact by calculating the partial derivatives M_y and N_x, leading to the conclusion that the equation is not exact. They explore the expression (M_y - N_x)/(-N) to find an integration factor but express uncertainty about the next steps.
  • Some participants question the original poster's reasoning and the significance of their calculations, prompting further exploration of what the derived expression indicates about the integration factor.
  • One participant provides a detailed breakdown of the calculations and suggests that the resulting expression is a function of x only, which could imply that the integration factor might also depend solely on x.

Discussion Status

The discussion is active, with participants providing feedback and prompting further exploration of the integration factor. There is no explicit consensus, but some productive directions are being explored regarding the nature of the integration factor.

Contextual Notes

The original poster has not specified all previous attempts or the results of their calculations, which may limit the clarity of the discussion. The nature of the problem suggests that assumptions about the functions involved are being questioned.

ISU20CpreE
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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.
 
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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?
 
HallsofIvy said:
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?

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.
 
Last edited:
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]<br /> and, in order that this be an exact equation we must have<br /> (-2sin(2y)f+ (cos(2y)- sin(x))f'= 2tan(x)sin(2y)f'+ -2tan(x)sin(2y)f<br /> That is an equation in x only for f.[/tex]
 

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