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Homework Help: Verifying an integrating factor -please check my work , thanks.

  1. May 31, 2010 #1
    Verifying an integrating factor --please check my work , thanks.

    1. The problem statement, all variables and given/known data
    Verify that 1/y^4 is an integrating factor for (3x^2-y^2)dy/dx - 2xy = 0, and then use it to solve the equation

    (1/y^4)* (3x^2-y^2)dy/dx = 2xy* (1/y^4)

    =3x^2 / y^4 dy = 2x/y^3 dx

    =3x^2 / y dy = 2x dx

    =integ (1/y)dy = (2/3)*integ (1/x)dx

    =ln|y| = (2/3)ln|x| + c

    =y = xe^(2/3 +c)


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 31, 2010 #2

    Mark44

    Staff: Mentor

    Re: Verifying an integrating factor --please check my work , thanks.

    What happened to the -y^2 term on the left side in the next line?

    Also, don't connect equations with =. Expressions are equal to one another, but equations are not equal to equations.
     
  4. May 31, 2010 #3
    Re: Verifying an integrating factor --please check my work , thanks.

    (1/y^4)* (3x^2-y^2)dy/dx = 2xy* (1/y^4)

    3x^2 / y^4 - 1/y^2 dy = 2x / y^3 dx

    ((3x^2/y) - y)dy = 2x dx

    so, at this point i am finding out that it is not separable afterall, right? thanks
     
  5. May 31, 2010 #4
    Re: Verifying an integrating factor --please check my work , thanks.

    so i should test for exactness?

    that is, M_y(x,y) ?=? (N_x(x,y)

    ?
    thanks
     
  6. May 31, 2010 #5

    Mark44

    Staff: Mentor

    Re: Verifying an integrating factor --please check my work , thanks.

    Yes. After you multiply by the integrating factor, test for exactness.
     
  7. May 31, 2010 #6
    Re: Verifying an integrating factor --please check my work , thanks.

    [3x^2 / y^4 - 1/y^2] dy - [2x / y^3] dx = 0
    ....M......................................N..........= 0

    M_x = 6x/y^4 = N_y

    ok so now i know it is exaxct and the integrating factor makes it exact, but i have question before i proceed.

    If i were to partially differentiate M wrt y instead of x, (and then N wrt y) i get a totally different result.. The equation does not pass the exatness test.

    How do i know which variable to differentiate wrt? thanks
     
  8. May 31, 2010 #7

    Mark44

    Staff: Mentor

    Re: Verifying an integrating factor --please check my work , thanks.

    You have your M and N switched. M is associated with dx and N is associated with dy. My = 6x/y^4 = Nx.
     
  9. May 31, 2010 #8
    Re: Verifying an integrating factor --please check my work , thanks.

    sorry i overlooked that..

    but i still dont understand why M_x is not equal to N_y/?
     
  10. May 31, 2010 #9

    Mark44

    Staff: Mentor

    Re: Verifying an integrating factor --please check my work , thanks.

    There's no reason for Mx to be equal to Ny, but there is a good reason for My to be equal to Nx.

    If the equation M dx + N dy = 0 is exact, then the left side represents the total differential of some unknown function F(x, y), so that we have dF(x, y) = 0 ==> F(x, y) = C.

    In the differential equation M = Fx(x, y) and N = Fy(x, y). If F and its first and second partials are continuous, the mixed partials will be equal. When you check for exactness, you are checking to see whether My = Nx. IOW, you are verifying that
    [tex]\frac{\partial ^2 F}{\partial x \partial y} = \frac{\partial ^2 F}{\partial y \partial x}[/tex]
     
  11. May 31, 2010 #10
    Re: Verifying an integrating factor --please check my work , thanks.

    thanks for the explanation. My final answer is now matching answer key, so im moving on to my other prob..
    thanks.
     
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