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Is there a mistake here so far?

  1. Aug 17, 2008 #1
    question attached

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  2. jcsd
  3. Aug 17, 2008 #2
    Mail it to my on wing-tales@gmail.com (without the - between wing and tales). Let me have a look.

    You might want to include the topic of your problem in the title.
  4. Aug 17, 2008 #3
    ok emailed
  5. Aug 17, 2008 #4
    Could you write the starting problem here? The first sentence is a little hard to read.
  6. Aug 17, 2008 #5


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    The problem is to solve
    [tex]\frac{dy}{dx}= \frac{3x- y+ 1}{x+ y+ 1}[/tex]

    What franky2727 has done is to set u= 3x- y+1 and v= x+ y+ 1 so that, from the equation, y'= u/v. Then from the first equation u'= 3- y'= 3- u/v. I frankly don't see how adding more variables is going to help, especially since he then defines z= u/v so he winds up with a total of 5 variables!

    franky2727, rewrite your equation as (x+y+ 1)dy = (3x-y+1)dx so that (x+y+1)dy- (3x-y+1)dx= (x+y+1)dy+ (-3x+ y- 1)dx= 0. Since (x+y+1)x= 1= (-3x+y-1)y, that is an exact equation. That is, there exist F(x,y) so that
    [tex]dF= \frac{\partial F}{\partial y}dy+ \frac{\partial F}{\partial x}dx= (x+y+1)dy+ (-3x+ y- 1)dx[/tex]
    Since dF= 0, F(x,y)= constant. Do you know how to find F from
    [tex]\frac{\partial F}{\partial y}= x+ y+ 1[/tex]
    [tex]\frac{\partial F}{\partial x}= -3x+ y- 1[/tex]?
  7. Aug 17, 2008 #6
    erm, no not a clue the way my lecturer did it was to add u,v, and z and then back substitute when he got to the end but it seemed long winded and hard. i see what your doing adding DF to emiminate but i dont have a clue where you would go after this? isnt it easier to back substitute cos i'm quite confused here
  8. Aug 17, 2008 #7
    oh and also just out of curiosity have i logged out that last line correctly?
  9. Aug 17, 2008 #8


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    [tex]\frac{\partial F}{\partial y}= x+ y+ 1[/tex]
    the, integrating with respect to y (treating x as if it were a constant)
    Notice that the "constant of integration" here may be a function of x because we are treating x as a constant. Differentiating that with respect to x,
    [tex]\frac{\partial F}{\partial x}= y+ g'(x)[/tex]
    and since we know the partial derivative of with respect to x must be -3x+ y+ 1,
    [tex]\frac{\partial F}{\partial x}= y+ g'(x)= -3x+ y+ 1[/tex]
    we have
    [tex]g'(x)= -3x+ 1[/tex]
    [tex]g(x)= -\frac{3}{2}x^2+ x[/tex]
    (I've ignored the constant of integration- which really is a constant since g is a function of x only)
    Putting those together,
    [tex]F= xy+ \frac{1}{2}y^2+ y-\frac{3}{2}x^2+ x[/tex]
    and since dF= 0 means F= C,
    [tex]xy+ \frac{1}{2}y^2+ y-\frac{3}{2}x^2+ x= C[/tex]
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