= Solution
With $a=2$, integrate the $dx$ coefficient with respect to $x$:
$$
\Phi(x,y)=\frac{x^3}{3}+\frac{x^2y}{2}-xy^2+g(y).
$$
Its other <partial derivative> is $\Phi_y=x^2/2-2xy+g'(y)$. Matching the $dy$ coefficient gives $g'=0$. Therefore the <differential equation> is $d\Phi=0$, and its implicit <general solution> is
$$
\boxed{\frac{x^3}{3}+\frac{x^2y}{2}-xy^2=C}.
$$
These are <level curves> of the <potential of a conservative vector field>. Where $N\ne0$, they give the usual graph solution $dy/dx=-M/N$. At $N=0$, the implicit description retains possible vertical tangents instead of dividing by zero.
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