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Math Help - Exact equation. Solve?

  1. #1
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    Exact equation. Solve?

     <br />
(y^2-xe^{x^2}) dx + (2xy+cos(y)) dy = 0<br />
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  2. #2
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    First you need to check if \frac{\partial}{\partial y}(y^2 - x\,e^{x^2}) = \frac{\partial}{\partial x}(2xy + \cos{y}).

    \frac{\partial}{\partial y}(y^2 - x\,e^{x^2}) = 2y.

    \frac{\partial}{\partial x}(2xy + \cos{y}) = 2y.


    So the condition is satisfied. Since this is true, that means \frac{\partial f}{\partial x} = y^2 - x\,e^{x^2} and \frac{\partial f}{\partial y} = 2xy + \cos{y}.

    Since \frac{\partial f}{\partial x} = y^2 - x\,e^{x^2}

    f(x, y) = \int{y^2 - x\,e^{x^2}\,dx}

     = xy^2 - \frac{1}{2}e^{x^2} + g(y).


    Since \frac{\partial f}{\partial y} = 2xy + \cos{y}

    f(x, y) = \int{2xy + \cos{y}\,dy}

     = xy^2 + \sin{y} + h(x).


    Putting these together, we can see that

    f(x, y) = xy^2 - \frac{1}{2}e^{x^2} + \sin{y} + C where C is a constant.
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  3. #3
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    Weird.

    The solution says:

    sinx + xy- cos y = C.

    Is it possible to get it to look like that?
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  4. #4
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    No I don't believe so. The answer you have been given does not take into account the exponential function...
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