Solution: This is a differential equation in the form of \[ M\mathrm{d}x + N \mathrm{d}y = 0. \] Note that \[ \frac{\partial M}{\partial y} = -1 = \frac{\partial N}{\partial x}. \] Therefore, the equation is exact.

Let $F(x,y) = 0$ is a solution then \[ \frac{\partial F}{\partial x} = M \text{ and } \frac{\partial F}{\partial y} = N. \] We have \begin{align*} \frac{\partial F}{\partial x} = M & \implies \frac{\partial F}{\partial x} = 2x - y + 1 \\ & \implies F(x,y) = x^2 - xy + x + f(y), \end{align*} where $f$ is any function of $y$ only. Now we also have \begin{align*} \frac{\partial F}{\partial y} = N & \implies -x + f^\prime (y) = 2y - x - 1 \\ & \implies f^\prime (y) = 2y - 1 \\ & \implies f(y) = y^2 - y + c, \end{align*} where $c$ is an arbitrary constant. Therefore, \[ F(x,y) = x^2 -xy + x + y^2 - y + c. \] Thus, a solution of the given differential equation will be \[ x^2 + y^2 - xy + x - y = c . \]