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    Solve (x+y)^2(x\frac{dy}{dx}+y)=xy(1+\  frac{dy}{dx})

    rearranging the formula would give,

    x[(x+y)^2-y]\frac{dy}{dx}+y[(x+y)^2-x]=0

 x[(x+y)^2-y]dy+y[(x+y)^2-x]dx=0

    I have tried multiplying an integrating factor but I can't find one depending only on x or y. Is there a way of finding an integrating factor involving both x and y? or is there a substitution or a different method for these type of equations?

    Thanks in advance for any help
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    (Original post by r2enigma)
    ...
    Differential equations is not my best subject area, but if you divide both sides of the original equation by the terms that don't involve the derivative, then you may recognize the form of each side and be able to integrate directly.
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    (Original post by r2enigma)
    Solve (x+y)^2(x\frac{dy}{dx}+y)=xy(1+\  frac{dy}{dx})

    rearranging the formula would give,

    x[(x+y)^2-y]\frac{dy}{dx}+y[(x+y)^2-x]=0

 x[(x+y)^2-y]dy+y[(x+y)^2-x]dx=0

    I have tried multiplying an integrating factor but I can't find one depending only on x or y. Is there a way of finding an integrating factor involving both x and y? or is there a substitution or a different method for these type of equations?

    Thanks in advance for any help
    Use separating the variables.
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    (Original post by jonathan3909)
    Use separating the variables.
    thanks for your reply. How do I separate x and y variables into two sides? Any hints?
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    (Original post by r2enigma)
    thanks for your reply. How do I separate x and y variables into two sides? Any hints?
    The hint is in your above post:rolleyes:
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    (Original post by r2enigma)
    thanks for your reply. How do I separate x and y variables into two sides? Any hints?
    Ok I'll help.Expand the above brackets and multiply throughout first.
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    x^{3} y' + yx^{2} + 2yx^{2}y' + 2xy^{2} + xy^{2}y' + y^{3} = xy + xyy' where  y' = \frac{dy}{dx}.

    Now write the whole thing as : \frac {dy}{dx} = .....
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    x^{3} y' + 2yx^{2}y' + xy^{2}y' - xyy' = xy - 2xy^{2} - yx^{2} - y^{3}

    \frac {dy}{dx} (x^{3} + 2yx^{2} + xy^{2} -xy) = xy - 2xy^{2} - yx^{2} - y^{3}
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    And put the y variables on dy side and x variables on dx side and then integrate to get y=....
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    (Original post by jonathan3909)
    And put the y variables on dy side and x variables on dx side
    How?
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    (Original post by ghostwalker)
    Differential equations is not my best subject area, but if you divide both sides of the original equation by the terms that don't involve the derivative, then you may recognize the form of each side and be able to integrate directly.
    Agreed.

    For the OP:

    Spoiler:
    Show
    Ignoring the terms directly involving the derivative, you have two other expressions in the original equation. You might want to find the derivatives of those expressions as a starting point.
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    (Original post by DFranklin)
    Agreed.

    For the OP:

    Spoiler:
    Show
    Ignoring the terms directly involving the derivative, you have two other expressions in the original equation. You might want to find the derivatives of those expressions as a starting point.
    I'll try that thanks
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    (Original post by DFranklin)
    How?
    I agree-It's too complex but not impossible to do
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    (Original post by jonathan3909)
    I agree-It's too complex but not impossible to do
    I don't see any way of doing it. If you have a method, I suggest you post it.
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    (Original post by DFranklin)
    I don't see any way of doing it. If you have a method, I suggest you post it.
    Haven't you done separating the variables?
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    (Original post by DFranklin)
    Agreed.

    For the OP:

    Spoiler:
    Show
    Ignoring the terms directly involving the derivative, you have two other expressions in the original equation. You might want to find the derivatives of those expressions as a starting point.
    Found the solution. Thanks
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    (Original post by jonathan3909)
    I agree-It's too complex but not impossible to do
    It is impossible because no amount of algebraic manipulation will allow us to express the original DE in the form:
    p(x)q(y)\frac{dy}{dx}+r(x)s(y)=0.
    What the others have outlined is the best approach.
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    You're right-
 
 
 
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