Answer
$f(x,y)=x^2-3xy+2y^2-8y+k$
Work Step by Step
Given: $F(x,y)=(2x-3y)i+(-3x+4y-8)j$
when $F(x,y)=ai+bj$ is a conservative field, then throughout the domain $D$, we get
$\dfrac{\partial a}{\partial y}=\dfrac{\partial b}{\partial x}$
Here, $a$ and $b$ are first-order partial derivatives on the domain $D$.
Then, we have $a_x=-3; b_y=-3$
Here, $\dfrac{\partial a}{\partial y} = \dfrac{\partial b}{\partial x}$
Thus, the vector field $F$ is conservative.
Need to determine $f$ with $F=\nabla f$
$f_x=(2x-3y) \\ f_y=(-3x+4y-8) $
But $f(x,y)=xy^2-x^2+g(y)$
where, $g(y)$ is a constant of integration with respect to $x$ with a function of $y$.
Take derivative of $f(x,y)=xy^2-x^2+g(y)$ with respect to $y$.
$f_y(x,y)=-3x+g'(y)$
Thus, we find that $g'(y)=4y-8$
W can now integrate it with respect to $y$.
we get $g(y)=2y^2-8y+k$
Here, $k$ is a constant.
Hence, our answer is:
$f(x,y)=x^2-3xy+2y^2-8y+k$
