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Homework Help: Finding an exact solution

  1. Apr 15, 2010 #1
    1. The problem statement, all variables and given/known data

    Solve (x+2)sin(y)dx + xcos(y)dy = 0 by finding an integrating factor u(x).


    2. Relevant equations

    I'll bring them in as I solve.

    3. The attempt at a solution

    I assume there's some parent function ƒ(x,y)=c.

    ---> ∂ƒ/∂x, ∂ƒ/∂y=0

    Now I'll just put it into the form M(x,y) + N(x,y)dy/dx = 0.

    (x+2)sin(y) + xcos(y)dy/dx = 0.

    An integrating factor is just some multiplied by the equation that helps us solve it. In this case I want ∂M/∂y=∂N/∂x, so I'll just modify M and N by multiplying them by a function of x (because the problem says to use u(x)).

    ∂/∂y [u(x)(x+2)sin(y)] = ∂/∂x [u(x)xcos(y)]

    -----> u(x)(x+2)cos(y) = [u(x)*1+x*u'(x)]cos(y) (product rule used on x terms)

    ------> u(x)(x+2)=u(x)+x*u'(x)

    ------>u(x)(x+1)=x*u'(x)

    ------> u'(x)/u(x) = 1 + 1/x

    ------> d/dx ln|u(x)| = 1 + 1/x

    -------> ln|u(x)| = x + ln|x| + K

    --------> u(x) = exeln|x| (Don't need constant ek)

    ----------> u(x) = xex

    ----------> ∂/∂y [xex(x+2)sin(y)] = ∂/∂x [xexxcos(y)]

    -----------> ∂M/∂y=∂N/∂x. Mission Accomplished. Almost.

    -----------> (x2ex + 2xex)sin(y) + x2excos(y) dy/dx = 0.

    Integrate ∂ƒ/∂y with respect to y.

    -------------> ƒ(x,y) = -x2exsin(y) + g(x)

    -------------> ∂f/∂x = -(2x + x2)exsin(y) + g'(x) = (x2ex + 2xex)sin(y).

    But wait! This is supposed to cancel out the y terms; it doesn't.

    Help, please!
     
  2. jcsd
  3. Apr 15, 2010 #2

    vela

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    You have the wrong sign on the first term. The integral of cosine is +sine.
     
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