$\displaystyle

(y^2-xe^{x^2}) dx + (2xy+cos(y)) dy = 0

$

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- Sep 10th 2010, 06:25 PMA Beautiful MindExact equation. Solve?
$\displaystyle

(y^2-xe^{x^2}) dx + (2xy+cos(y)) dy = 0

$ - Sep 10th 2010, 06:38 PMProve It
First you need to check if $\displaystyle \frac{\partial}{\partial y}(y^2 - x\,e^{x^2}) = \frac{\partial}{\partial x}(2xy + \cos{y})$.

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

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

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

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

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

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

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

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

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

Putting these together, we can see that

$\displaystyle f(x, y) = xy^2 - \frac{1}{2}e^{x^2} + \sin{y} + C$ where $\displaystyle C$ is a constant. - Sep 10th 2010, 07:14 PMA Beautiful Mind
Weird.

The solution says:

$\displaystyle sinx + xy- cos y = C$.

Is it possible to get it to look like that? - Sep 10th 2010, 07:54 PMProve It
No I don't believe so. The answer you have been given does not take into account the exponential function...