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Homework Help: Diff Equations

  1. Sep 7, 2006 #1
    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.
  2. jcsd
  3. Sep 8, 2006 #2


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    Science Advisor

    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?
  4. Sep 8, 2006 #3
    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: Sep 8, 2006
  5. Sep 8, 2006 #4


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    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.
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