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Please check my work (exact equation, then i solved)

  1. Apr 15, 2010 #1
    1. The problem statement, all variables and given/known data
    (y-x^3) + (x+y^3)y' = 0

    equation is convenienty already in the required form, that is M_(x,y) + N_x(x,y)dy/dx = 0

    so..

    M_y = 1 = N_x therefore equation is exact. Therefore I now solve...

    I am solving for a function, f(x,y) whose partial derivative with respect to y = M
    and whose partial with respect to x = N.
    Is this correct?

    assuming correct so far, my next step is to integrate the M term, with respect to x

    integ M_x (y-x^3) = -(1/4)x^4 + h(y)

    i have a questions about this step though:

    if y's are treated as constants when integrating, wouldn't the y in this equation become yx?

    so that integ_x (y-x^3) = yx -(1/4)x^4 + h(y)
    ??

    assuming yx is not in the integrated result, my next step is to take derivative of the just integrated M term, now with respect to y:

    derivative of just-integ M_y [-(1/4)x^4 + h(y) ] = h'(y)

    so i know that h'(y) = N term which is:

    h'(y) = (x+y^3)

    But i think im going to stop here because it looks sort of weird and i want to make sure its right so far.
    thanks for any help

    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
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 15, 2010 #2

    Char. Limit

    User Avatar
    Gold Member

    First, yes, the y in your equation becomes yx. This is true.

    Second, now that you know this, keep going! You almost have it, I can tell.
     
  4. Apr 15, 2010 #3
    ok so slight rework to fix the integration error...

    integ M_x (y-x^3) = yx -(1/4)x^4 + h(y)

    derivative of just-integ M_y [yx -(1/4)x^4 + h(y) ] = x + h'(y)

    x + h'(y) = (x+y^3) (the N term)

    h'(y) = y^3

    h(y) = (1/4)y^4

    correct so far? thank you
     
  5. Apr 15, 2010 #4
    so, final answer: f(x,y) = x + (1/4)y^4

    ??
     
  6. Apr 15, 2010 #5

    Char. Limit

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    Gold Member

    Well, I've cleaned up your answers for latex format, and it looks like your final answer is missing a term. Just look at the first two of the quotes by you and the answer should be right there.
     
  7. Apr 15, 2010 #6
    for some reason im confused as to where final answer comes from:
    Is it simply N term + whatever I worked out h(y) to be?

    If so, answer should be f(x,y) = x + y^3 + (1/4)y^4.

    Is this correct and also is answer: N-term + h(y)?
    thanks
     
  8. Apr 16, 2010 #7

    Mark44

    Staff: Mentor

    The answer should be f(x, y) = C, where f(x, y) is the function you found by integrating M and N. I haven't checked this work, but the idea is exactly the same as in the other thread you posted.
     
  9. Apr 16, 2010 #8

    Char. Limit

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    Gold Member

    Not quite.

    The answer is [itex]\int \psi_x dx = \psi + h(y)[/itex]

    where psi_x is the partial derivative of psi, which means f(x,y).

    You found psi, and you found h(y). Those were the first two items I quoted in my last post.
     
  10. Apr 16, 2010 #9

    Mark44

    Staff: Mentor

    Since psi means f(x,y), then why not use f(x,y) instead of introducing another symbol that muddies the waters and cannot be typed on standard keyboards?
     
  11. Apr 16, 2010 #10

    Char. Limit

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    Because I saw psi used as the symbol for f(x,y), and assumed it was common.

    Also, how do you write the partial derivative of f(x,y) with respect to x?

    Is it [itex]{f(x,y)}_x[/itex]?
     
  12. Apr 16, 2010 #11

    Mark44

    Staff: Mentor

    The simplest ways use subscripts - fx or fx(x, y).

    Or you can use this notation
    [tex]\frac{\partial f}{\partial x}[/tex]
     
  13. Apr 16, 2010 #12

    Char. Limit

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    Ah.

    Just to clear things up, let me resay what I said before, but in more standard notation:

    [itex]\int f_x dx = f(x,y) + h(y)[/itex]

    Thanks for clearing that up, Mark. I'll try to be less confusing from now on.
     
  14. Apr 16, 2010 #13

    Mark44

    Staff: Mentor

    Not a problem.
     
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