= Solution
An <integrating factor for a differential one-form> is a nonzero function $\mu(x,y)$ for which $\mu(dy+f\,dx)$ is an <exact differential>, say $dF$. Then solutions lie on <level sets> $F=\text{constant}$. Locally, the compatibility condition is $\partial_x\mu=\partial_y(\mu f)$.
Here multiplication by $2ye^x$, which is nonzero for $y>0$, produces
$$
2ye^x\,dy+e^x(2x+x^2+y^2)\,dx=d\left[e^x(x^2+y^2)\right].
$$
The <initial condition> makes this <first integral> equal to $a^2$. Taking the positive branch yields \b[the solution]
$$
\boxed{y(x)=\sqrt{a^2e^{-x}-x^2},}
$$
on the connected interval containing zero where the radicand is positive. This domain restriction matters because the original <differential equation> is singular at $y=0$.
Completing the square gives $2x+x^2=(x+1)^2-1\ge-1$. As $y>0$, the <differential equation> consequently implies
$$
y'=-\frac{2x+x^2+y^2}{2y}\le\frac{1-y^2}{2y}.
$$
At $x=0$, $y'=-a/2<0$. Thus, moving a little to the left from $y(0)=a\ge1$, the curve rises above $a$. Throughout $y>1$ every segment of the <direction field> points downwards as $x$ increases, by the inequality above. Tracing backwards therefore makes $y$ rise further, so it cannot return to $y=1$. This graphical barrier argument proves $y>a\ge1$ and $y'<0$ for every $x<0$ on the solution. The explicit <first integral> also shows this branch extends throughout $x<0$: writing $t=-x>0$, the minimum of $e^t/t^2$ is $e^2/4>1$, so $a^2e^t-t^2>0$ for $a\ge1$. Including the initial point, \b[$y'<0$ for all $x\le0$.]
For $a=1$ the curve is $y=\sqrt{e^{-x}-x^2}$ on $(-\infty,b)$, where the strictly increasing function $x^2e^x$ has the unique positive root $x^2e^x=1$ at $b\simeq0.703467$. The curve decreases through $(0,1)$ with slope $-1/2$ and reaches $(b,0)$ only as a limiting endpoint. Its derivative is
$$
y'=\frac{-e^{-x}-2x}{2\sqrt{e^{-x}-x^2}}.
$$
As $x\to-\infty$, $y\sim e^{-x/2}$ and $y'\sim-\tfrac12e^{-x/2}\to-\infty$. As $x\uparrow b$, the numerator tends to $-b^2-2b<0$ and the denominator to zero from above, so again $y'\to-\infty$. More precisely, $y\sim\sqrt{b(b+2)(b-x)}$ at the endpoint. \b[Both ends have the requested unbounded negative slope.]
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ia/paper-2-integrating-factor.png]
{title=The positive solution for initial value one, with the horizontal barrier and the steep limiting endpoint}
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