= Solution
Let $M=x+y$ and $N=x$. Their cross <partial derivatives> agree: $M_y=N_x=1$. More directly, construct the <scalar> <function>
$$
F(x,y)=\frac{x^2}{2}+xy,\qquad dF=(x+y)\,dx+x\,dy.
$$
This exhibits the expression as an <exact differential>. Along a solution of the <differential equation>,
$$
\frac d{dx}F(x,y(x))=x+y+xy'=0.
$$
Thus the <general solution> is
$$
\boxed{\frac{x^2}{2}+xy=C,\qquad y=\frac Cx-\frac x2\quad(x\ne0).}
$$
For $C=0$, the graph $y=-x/2$ also extends smoothly through $x=0$ and satisfies the original, undivided equation there. Nonzero $C$ cannot extend through zero as a finite-valued <differentiable> solution. The extra vertical component $x=0$ of the implicit zero-level equation is not a graph $y(x)$ on an <interval> and is not an additional solution of this <ordinary differential equation>.
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