1. ## Exact Equation

$\displaystyle (e^x \sin{y} -3x^2) \,dx +(e^x \cos{y} + \frac{y^{-2/3}}{3}) \, dy = 0$

$\displaystyle (e^x \sin{y} -3x^2) +(e^x \cos{y} + \frac{y^{-2/3}}{3}) \frac{dy}{dx} = 0$

$\displaystyle M = e^x \sin{y} -3x^2 \,, N = e^x \cos{y} + \frac{y^{-2/3}}{3}$

$\displaystyle \frac{\partial M}{\partial y} = e^x \cos{y}$

$\displaystyle \frac{\partial N}{\partial x} = e^x \cos{y}$

Since
$\displaystyle \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$,
there exists an F such that
$\displaystyle \frac{d}{dx}F(x, y) = 0$

$\displaystyle F = \int{M dx}$
$\displaystyle F = \int{e^x \sin{y} -3x^2 dx} = e^x \sin{y} -x^3 + h(y)$

$\displaystyle \frac{\partial F}{\partial y} = N$

$\displaystyle e^x \cos{y} + \frac{y^{-2/3}}{3} = e^x \sin{y} -x^3 + h'(y)$

I can't find any way to get past this point. Is there any possible function $\displaystyle h'(y)$ that would satisfy this equality? Or have I made an error?

2. Your working is a little off...

Since $\displaystyle \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

that means

$\displaystyle \frac{\partial F}{\partial x} = M$ and $\displaystyle \frac{\partial F}{\partial y} = N$.

So

$\displaystyle \frac{\partial F}{\partial x} = e^x\sin{y} - 3x^2$

$\displaystyle F = \int{e^{x}\sin{y} - 3x^2\,dx}$

$\displaystyle = e^x\sin{y} - x^3 + g(y)$.

$\displaystyle \frac{\partial F}{\partial y} = e^x\cos{y} + \frac{y^{-\frac{2}{3}}}{3}$

$\displaystyle F = \int{e^x\cos{y} + \frac{y^{-\frac{2}{3}}}{3}\,dy}$

$\displaystyle = e^x\sin{y} + y^{\frac{1}{3}} + h(x)$.

So we have at the same time...

$\displaystyle F = e^x\sin{y} - x^3 + g(y)$ and $\displaystyle F = e^x\sin{y} + y^{\frac{1}{3}} + h(x)$.

This means $\displaystyle g(y) = y^{\frac{1}{3}} + C_1$ and $\displaystyle h(x) = -x^3 + C_2$.

Putting everything together gives

$\displaystyle F = e^x\sin{y} - x^3 + y^{\frac{1}{3}} + C$ where $\displaystyle C = C_1 + C_2$.

3. Thanks.