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Math Help - Easy Equation, stuck on the constant.

  1. #1
    Super Member craig's Avatar
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    Easy Equation, stuck on the constant.

    Let \omega = (3y^2 + 2y)dx + (6xy + 2x + y^2)dy.

    Find potential function f for \omega.

    We know  \frac{\partial f}{\partial x} = 3y^2 + 2y, so f = 3xy^2 + 2xy + g(y)

    Differentiating wrt y, \frac{\partial f}{\partial y} = 6xy + 2x + g'(y) = 6xy + 2x + y^2.

    So we have g(x) = \frac{1}{3}y^3 + constant.

    The answers then simply state that the constant is just zero, not quite sure how they get to this?

    Thanks in advance
    Last edited by craig; January 4th 2011 at 03:56 PM.
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  2. #2
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    Quote Originally Posted by craig View Post
    Let \omega = (3y^2 + 2y)dx + (6xy + 2x + y^2)dy.

    Find potential function f for \omega.

    We know  \frac{\partial f}{\partial x} = 3y^2 + 2y, so f = 3xy^2 + 2xy + g(x)

    Differentiating wrt y, \frac{\partial f}{\partial y} = 6xy + 2x + g'(x) = 6xy + 2x + y^2.

    So we have g(x) = \frac{2}{3}y^3 + constant.

    The answers then simply state that the constant is just zero, not quite sure how they get to this?

    Thanks in advance
    Dear craig,

    You have a arbitary constant. Therefore you can take any value you like for the constant. For simplicity they may have taken it to by equal to zero.
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  3. #3
    Super Member craig's Avatar
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    Ahh thankyou. I was debating whether you could basically choose your value of the constant.

    Cheers for the quick reply.
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  4. #4
    Super Member General's Avatar
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    That g(x) should be g(y).
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    Quote Originally Posted by craig View Post
    Let \omega = (3y^2 + 2y)dx + (6xy + 2x + y^2)dy.

    Find potential function f for \omega.

    We know  \frac{\partial f}{\partial x} = 3y^2 + 2y, so f = 3xy^2 + 2xy + g(x)

    Differentiating wrt y, \frac{\partial f}{\partial y} = 6xy + 2x + g'(x) = 6xy + 2x + y^2.

    So we have g(x) = \frac{2}{3}y^3 + constant.

    The answers then simply state that the constant is just zero, not quite sure how they get to this?

    Thanks in advance
    You set your \displaystyle\frac{df}{dy}=N(x,y)

    Then you have

    \displaystyle 6xy+2x+g'(y)=6xy+2x+y^2\Rightarrow g'(y)=y^2

    \displaystyle \int g'(y)dy=\int y^2dy\Rightarrow g(y)=\frac{y^3}{3}+c

    I have a different solution for your g(y) solution then what you obtained. I don't think you should have a 2/3 just a 1/3.
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  6. #6
    Super Member craig's Avatar
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    Thanks for the last two posts. You're both right, typos due to typing far too quickly lol.

    Cheers for spotting those
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