Determine if the differential equation (3x2y + e^y)dx + (x3 + xe^y − 2y)dy = 0 is exact and solve it.

The differential equation is in the form $M(x,y)dx + N(x,y)dy = 0$.

$$M(x,y) = 3x^2y + e^y$$ $$N(x,y) = x^3 + xe^y - 2y$$

Calculate the partial derivatives $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$.

$$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(3x^2y + e^y) = 3x^2 + e^y$$ $$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x^3 + xe^y - 2y) = 3x^2 + e^y$$

Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the differential equation is exact.

Integrate $M(x,y)$ with respect to $x$, treating $y$ as a constant:

$$F(x,y) = \int (3x^2y + e^y) dx = x^3y + xe^y + g(y)$$

Now, differentiate $F(x,y)$ with respect to $y$ and set it equal to $N(x,y)$:

$$\frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(x^3y + xe^y + g(y)) = x^3 + xe^y + g'(y)$$

Equating this to $N(x,y)$:

$$x^3 + xe^y + g'(y) = x^3 + xe^y - 2y$$ $$g'(y) = -2y$$

Integrate $g'(y)$ with respect to $y$ to find $g(y)$:

$$g(y) = \int -2y dy = -y^2 + C_0$$

Substitute $g(y)$ back into $F(x,y)$:

$$F(x,y) = x^3y + xe^y - y^2 + C_0$$

The general solution is $F(x,y) = C$, where $C$ is an arbitrary constant.

$$\boxed{x^3y + xe^y - y^2 = C}$$

The differential equation is in the form $M(x,y)dx + N(x,y)dy = 0$.

$$M(x,y) = \sin y - y \sin x$$ $$N(x,y) = x \cos y + \cos x$$

Calculate the partial derivatives $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$.

$$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(\sin y - y \sin x) = \cos y - \sin x$$ $$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x \cos y + \cos x) = \cos y - \sin x$$

Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the differential equation is exact.

Integrate $M(x,y)$ with respect to $x$, treating $y$ as a constant:

$$F(x,y) = \int (\sin y - y \sin x) dx = x \sin y - y(-\cos x) + g(y)$$ $$F(x,y) = x \sin y + y \cos x + g(y)$$

Now, differentiate $F(x,y)$ with respect to $y$ and set it equal to $N(x,y)$:

$$\frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(x \sin y + y \cos x + g(y)) = x \cos y + \cos x + g'(y)$$

Equating this to $N(x,y)$:

$$x \cos y + \cos x + g'(y) = x \cos y + \cos x$$ $$g'(y) = 0$$

Integrate $g'(y)$ with respect to $y$ to find $g(y)$:

$$g(y) = \int 0 dy = C_0$$

Substitute $g(y)$ back into $F(x,y)$:

$$F(x,y) = x \sin y + y \cos x + C_0$$

The general solution is $F(x,y) = C$, where $C$ is an arbitrary constant.

$$\boxed{x \sin y + y \cos x = C}$$