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Help with integrating factor

  1. Apr 6, 2009 #1
    I can see why y2 is an integrating factor, but I don't know how this answer can be derived. Do you use:

    [tex]
    \frac{N_{x}(x,y) - M_{y}(x,y)}{M(x,y)}
    [/tex]
    Where M (x,y) = 4(x3/y2)+(3/y)]dx, and N = 3(x/y2)+4y dy

    When I try and solve this, I get a very complex integrating factor. Does anyone have any suggestions on solving these types of problems?
     
  2. jcsd
  3. Apr 6, 2009 #2

    djeitnstine

    User Avatar
    Gold Member

    [tex]

    \frac{M_{y}(x,y) - N_{x}(x,y)}{N(x,y)}

    [/tex]

    is equally valid
     
  4. Apr 6, 2009 #3
    :confused: Isn't the integrating factor, a function of y (y2)? I know how to do other problems like this, but I just can't figure out this one. Would you be able to provide a better hint? Thanks.
     
  5. Apr 29, 2009 #4
    I still am having trouble with integrating factors....any ideas on the problem above?
     
  6. Apr 29, 2009 #5
    What did you get for [tex] \frac{N_{x}(x,y) - M_{y}(x,y)}{M(x,y)}[/tex]?

    If your integrating factor say, [tex] \mu [/tex] is a function of x only, then

    [tex] \frac{d \mu}{dx} = \frac{N_{x}(x,y) - M_{y}(x,y)}{M(x,y)} \mu [/tex]

    From here you can solve for the exact equation after multiply the integrating factor by the original equation.
     
  7. Apr 29, 2009 #6
    My=-8x3y-3-3y-2
    Nx=3y-2

    I think I see what I've been doing wrong now....I've always canceled out the two 3y-2, when I should have added. I can't believe I did that.....

    The reason I posted again, was partly due to another problem involving integrating factors.
    -xsin(y) dy + (2+x)cos(y) dy ---- Find the integrating factor to make this exact.

    Since I thought I didn't understand them well, I figured learning how to do the originally posted problem would help. Now, I'm not sure if this problem can be solved as written, since both derivates are with respect to y.
     
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