= Solution
For $x\ne0$, divide by $x$ and use the <integrating factor> $\mu(x)=e^x|x|^\alpha$, separately on each side of zero. The equation becomes $(\mu y)'=|x|^\alpha/x$. Thus
$$
\boxed{y=e^{-x}\left(\frac1\alpha+C|x|^{-\alpha}\right)\quad(\alpha\ne0),\qquad y=e^{-x}(\ln|x|+C)\quad(\alpha=0).}
$$
For $x>0$ the absolute values may be omitted. The original equation is singular as a first-order evolution equation at zero, so these formulas describe regular intervals with $x\ne0$; extension through zero needs a separate regularity condition.
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