Finding the Integrating Factor

  1. Hi ok so te question is asking for me to fin the integrating factor of (y2x+y)dy + (x2y+2x)dy = 0 the only thing i know is that the integrating factor should be a function of xy

    Can some one pleas explain how to do this? Thank you.
     
  2. jcsd
  3. Defennder

    Defennder 2,616
    Homework Helper

    You appear to have mistyped your problem. There 2 "dy"s here but no "dx".
     
  4. HallsofIvy

    HallsofIvy 40,504
    Staff Emeritus
    Science Advisor

    Assuming that you mean [itex](y^2x+ y)dx+ (x^2y+ 2x)dy= 0[/itex], if you know that the integrating factor is a function of xy, the obvious thing to do is to put multiply that equation by u(xy) and see what happens. In order that [itex]u(xy)(y^2x+ y)dx+ u(xy)(x^2y+ 2x)dy= 0[/itex] be exact, we must have [itex](u(y^2x+ y))_y[/itex] and [itex](u(x^2y+ 2x)_x)[/itex] equal. That is, [itex]u_x(y^2x+ y)+ u(2xy+1)= u_x(x^2y+ 2x)+ u(2xy+ 2)[/itex]. Since u is a function of xy specifically, [itex]u_x= yu'[/itex] and [itex]u_y= xu'[/itex]. That is, u must satisfy [itex]xu(y^2x+ y)+ u(2xy+1)= yu'(x^2y+ 2x)+ u(2xy+ 2)[/itex]. That gives you a differential equation for u which may or may not be solvable, depending upon whether there really is an integrating factor that is a function of xy only.
     
  5. Yes the original equation is (y2x+y)dx + (x2y+2x)dy = 0, i have taken your sugestion but it did not work, i tryed a more general form of xnym and multiplied it through to see if i could solve for the n and m and equate the two of them but the two are not equal i end up getting (m+2)=(n+2) and (m+1)=2(n+1).

    so can any one also give me some advice on how to solve this problem i am having. Thank you.
     
  6. Defennder

    Defennder 2,616
    Homework Helper

    I think HallsofIvy's advice is spot on. Equate the two derivatives after differentiating the product of u and the expression preceding the dx and dy. Use the chain rule when differentiating u(xy) with respect to x and y only by denoting product xy as v. Simplify the resulting equation and you'll notice it reduces to a simple separable differential equation. Solve that and express u in terms of x,y only and you're done. I tried it out and it works like a charm.
     
  7. HallsofIvy

    HallsofIvy 40,504
    Staff Emeritus
    Science Advisor

    You told us, originally, that the integrating factor was a function of xy. Why are you now trying xmyn? How do you know the integrating factor is either a function of xy or of the form xmyn?
     
  8. Yeah it did work sorry about that, and as for havening it as xmyn other times you can use it to find out what power the x and y are to making it easer. mind you you have to know xy are the integrating factor.
     
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