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    I need to find the general solution in implicit form, of this equation:

     dy/dx = - [(x + y + 2)/(2x+2y-1)]

    Ive then got:

     -2x -2y +1 dy = -x-y-2 dx

    But dont know where to go next?
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    It's not separable, so there's no use multiplying through by the denominator (at least not yet).

    Notice that you can write it as \dfrac{dy}{dx} = -\dfrac{(x+y)+2}{2(x+y)-1}, which indicates that the substitution u=x+y may prove to be useful.
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    From the offset, try making a substituting that involves both x and y (motivation: "implicit". And it looks like it'll make things nicer).
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    (Original post by bikinmad)
    I need to find the general solution in implicit form, of this equation:

     dy/dx = - [(x + y + 2)/(2x+2y-1)]

    Ive then got:

     -2x -2y +1 dy = -x-y-2 dx

    But dont know where to go next?
    Possibly the substitution z = x + y
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    so i let u = x + y

    And i got up to the point:

    



du/dx = 1 + dy/dx



du/dx - 1 = -[u+2/2u-1]



1= u+2/2u-1 + 1(du/dx)

    Is this correct so far?
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    (Original post by bikinmad)
    so i let u = x + y

    And i got up to the point:

    



du/dx = 1 + dy/dx



du/dx - 1 = -[u+2/2u-1]



1= u+2/2u-1 + 1(du/dx)

    Is this correct so far?
    It's correct, but there's no need for the last step. Your aim is to separate the variables - so put du/dx on one side, everything else on the other and use separation of variables (you'll want to simplify a bit before dividing across)
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    (Original post by bikinmad)
    so i let u = x + y

    And i got up to the point:

    



du/dx = 1 + dy/dx



du/dx - 1 = -[u+2/2u-1]



1= u+2/2u-1 + 1(du/dx)

    Is this correct so far?
    Yup, although what I'd recommend doing is combining the fraction and the 1 into a single fraction. Then you can separate things.
 
 
 
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