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Homework Help: Finding the integrating factor

  1. Jun 28, 2011 #1
    I am having trouble with the below:

    [ 4* (x^3/y^2) + (3/y)] dx + [3*(x/y^2) +4y]dy=0

    I found My= -8x^3y^-3 - 3y^-2 and Nx= 3y^-2
    i then subtracted Nx from My and divided by [3*(x/y^2) +4y]


    [-8x^3y^-3 - 6y^-2] / [3*(x/y^2) +4y]. can you guys give me a hint as to where my error is?
     
  2. jcsd
  3. Jun 29, 2011 #2
    So I multiplied the original equation by y^2 on both sides and got:

    (4x^3+3y) dx + (3x+4y^3) dy = 0

    and this equation is exact. You have to remember that I(x) = exp^(Integral[(M_y-N_x)/N,x]). In this case, since M_y=N_x, you get exp(Integral[0,x]) = 1.
     
  4. Jun 29, 2011 #3
    Thanks. I figured that (Nx-My)/M did the trick
     
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