Exact Differential Equation? I tried to solve this equation so far, since the partial derivative respect to $x$ and $y$ are not exact, I have to find the $u(x)$ to make them exact    
$e^x \cos(y)dx+\sin(y)dy=0$
Partial derivative of $y = e^x\sin(y)$
partial derivative of $x = -\sin(y)$
Now i have to find the $u(x)$ to make them exact
$u(x)(e^x\sin(y))=u'(x)(-\sin(y))+u(x)(-\sin(y))$
$u(x)\sin(y)(e^x+1)=u'(x)-\sin(y)$
$u(x)(e^x+1) = \frac{du}{dx} - 1$
$(e^x + 2)dx = \frac 1 u du $
$e^x + 2x = \ln(u)$
$u = e^{e^x + 2x}$
this $u(x)$ is too strange to make them exact...
If anyone can help to understand why it doesn't work.....

You don't need to make this equation exact, it's seperable. First subtract $sin(y)dy$ to get 
$e^x\cos ydx=-\sin y dy$
Then divide both sides by $\cos y$ to get 
$e^xdx=-tanydy$.
Proceed to integrate both sides.
