= Solution
For \b[every $a\ne0$ and every real $b$, there is exactly one solution on the maximal connected interval containing $a$]. The formula in part (i) proves existence on the entire appropriate half-line, and the initial condition fixes its sole constant uniquely. This agrees with the local <Picard-Lindelöf theorem>.
For \b[$a=0$, there is no classical solution on an interval containing the initial point], whatever $b$ is: the printed equation is not defined there. Even if one only imposes the equation for $x\ne0$ and asks for a finite continuous extension across zero, the negative side rules it out. For $x\to0^-$:
* If $C\ne0$, then $C+\arctan x\to C$, while $x^{-2}e^{-1/x}$ diverges in magnitude.
* If $C=0$, then $\arctan x\sim x$, and $y(x)\sim e^{-1/x}/x$, which also diverges in magnitude.
Thus no choice of the negative-side integration constant admits a finite left limit.
There is a useful distinction with a separately extended one-sided problem. On $x>0$, every member of the solution family tends to zero with all its derivatives as $x\downarrow0$, since the exponential dominates every inverse power of $x$. If the forcing is separately defined to be zero at the right endpoint, that modified one-sided problem has infinitely many smooth solutions with $y(0)=0$, and none with $y(0)\ne0$. \b[Those one-sided extensions do not give solutions on the two-sided intervals in the question.]
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