To determine the potential function $\phi(x,y)$ of the given vector field:

$\mathbf{F} = (6xy - e^{2y}\cos x)\mathbf{i} + 2(x^3 + e^{2y}\sin x)\mathbf{j},$ we note that a potential function $\phi(x,y)$ satisfies:

$\frac{\partial \phi}{\partial x} = F_1 \quad \text{and} \quad \frac{\partial \phi}{\partial y} = F_2,$ where $F_1$ and $F_2$ are the components of the vector field.

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Step 1: Integrate $F_1$ with respect to $x$

Given $F_1 = 6xy - e^{2y}\cos x$, we integrate with respect to $x$ (treating $y$ as constant):

$\phi(x,y) = \int (6xy - e^{2y}\cos x) \, dx.$

• For $6xy$, the integral is $3x^2y$ (since $\int x \, dx = \frac{x^2}{2}$).
• For $-e^{2y}\cos x$, the integral is $-e^{2y}\sin x$ (since $\int \cos x \, dx = \sin x$).

Thus:

$\phi(x,y) = 3x^2y - e^{2y}\sin x + h(y),$ where $h(y)$ is an arbitrary function of $y$ (since the partial derivative with respect to $x$ eliminates any purely $y$-dependent terms).

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Step 2: Differentiate $\phi$ with respect to $y$

To determine $h(y)$, differentiate $\phi(x,y)$ with respect to $y$:

$\frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y} \left( 3x^2y - e^{2y}\sin x + h(y) \right).$

This gives:

$\frac{\partial \phi}{\partial y} = 3x^2 - 2e^{2y}\sin x + h'(y).$

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Step 3: Equate $\frac{\partial \phi}{\partial y}$ to $F_2$

From the given vector field, $F_2 = 2(x^3 + e^{2y}\sin x)$. Equating this to $\frac{\partial \phi}{\partial y}$:

$3x^2 - 2e^{2y}\sin x + h'(y) = 2x^3 + 2e^{2y}\sin x.$

Simplify the equation:

[ h'(y) = 2x^3 + x

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