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Special Integrating Factor?

  1. Sep 20, 2007 #1
    1. Solve the equation
    [sin(xy) + xycos(xy)]dx + [1 +x[tex]^{2}[/tex]cos(xy)dy = 0

    2. Relevant equations



    3. The attempt at a solution
    [sin(xy) + xycos(xy)]dx = M(x,y)
    [1 +x[tex]^{2}[/tex]cos(xy)dy = N(x,y)

    dM/dy
    sin(xy) = xcos(xy)

    xycos(xy) = uv' + vu' = (xy)(-xsin(xy)) + (x)(cos(xy))
    u = xy du = x v = cos(xy) dv=-xsin(xy)

    dM/dy= xcos(xy) + (xy)(-xsin(xy)) + (x)(cos(xy)) = 2xcos(xy) - x[tex]^{2}[/tex]ysin(xy)


    dN/dy
    u = x[tex]^{2}[/tex] du = 2x v= cos(xy) dv = -ysin(xy)
    x[tex]^{2}[/tex](-xsin(xy))+ (2x)cos(xy) = (2x)cos(xy) - x[tex]^{2}[/tex]y(sin(xy))

    Since dM/dy = dN/dx, the equation is exact.

    F(x,y) = [tex]\int[/tex][1 + x[tex]^{2}[/tex]cos(xy)dy + h(x)
    dy = y
    x[tex]^{2}[/tex]cos(xy)dy = x[tex]^{2}x[/tex]sin(xy) = x[tex]^{3}[/tex]sin(xy)

    F(x,y) = y + x[tex]^{3}[/tex]sin(xy) + h(x)

    dF/dx(x,y) = N(x,y) = 3x[tex]^{2}[/tex]cos(xy) + h'(x)

    N(x,y) = M (x,y)

    3x[tex]^{2}[/tex]cos(xy) + h'(x) = sin(xy) + xycos(xy)

    h'(x) = sin(xy) + xycos(xy) - 3x[tex]^{2}[/tex]cos(xy)

    h(x) = [tex]\int[/tex][sin(xy) + xycos(xy) - 3x[tex]^{2}[/tex]cos(xy)]

    sin(xy) = -ycos(xy)

    xycos(xy) = uv - [tex]\int[/tex]vdv = xy[tex]^{2}[/tex]sin(xy) - [tex]\int[/tex]ysin(u)(du) = xy[tex]^{2}[/tex]sin(xy) + ycos(xy)
    u = xy du = y dv = cos(xy) v=ysin(xy)

    3x[tex]^{2}[/tex]cos(xy) = 3x[tex]^{2}[/tex]ysin(xy) - [tex]\int[/tex]ysin(xy)(6x)
    u = 3x[tex]^{2}[/tex] du = 6x dv=cos(xy) v = ysin(xy)

    I got until there. How do I integrate [tex]\int[/tex]ysin(xy)(6x)? Or did I missed something?
     
  2. jcsd
  3. Sep 20, 2007 #2

    Dick

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    You are doing some pretty odd differentiating. For the common derivative of M and N, I get 2xcos(xy)-x^3*sin(xy). And when I integrate x^2*cos(xy)dy I get x*sin(xy). I sort of looks like you are mixing up pieces of differentiation and integration.
     
  4. Sep 20, 2007 #3
    :(

    How did you get (x^3)(sinxy)?

    xycos(xy)

    u = xy du = x v = cos(xy) dv=-xsin(xy)

    = uv' + vu' = (xy)(-xsin(xy)) + (x)(cos(xy))

    I still have this 2xcos(xy) - x^2ysin(xy) :(
     
  5. Sep 20, 2007 #4

    Dick

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    OOOOPS. Apologies, I sent the wrong thing. You are right on the differentiation. But I still don't like the integral of x^2*cos(xy)dy. Can you check that one again. I still get x*sin(xy).
     
  6. Sep 20, 2007 #5
    Unfortunately, the integral is correct.

    I guess I should switch to M(x,y).

    May I ask how to ingrate [tex]\int[/tex][sin(xy)dx?

    I want to set xy = u, which means du = y, but I don't have du.
     
  7. Sep 20, 2007 #6

    Dick

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    d/dy(x*sin(xy))=x^2*cos(xy). d/dy(x^3*sin(xy))=x^4*cos(xy). x^3*sin(xy) is not the correct antiderivative! To do the integral, sure u=xy, du=y*dx, dx=du/y. So integrate sin(u)*du/y (y is constant this time). So I get -cos(u)/y=-cos(xy)/y.
     
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