Exact equation. Solve?

**A Beautiful Mind** · September 10th 2010, 06:25 PM

$<br /> (y^2-xe^{x^2}) dx + (2xy+cos(y)) dy = 0<br />$

**Prove It** · September 10th 2010, 06:38 PM

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.

**A Beautiful Mind** · September 10th 2010, 07:14 PM

Weird.

The solution says:

$sinx + xy- cos y = C$ .

Is it possible to get it to look like that?

**Prove It** · September 10th 2010, 07:54 PM

No I don't believe so. The answer you have been given does not take into account the exponential function...

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