Using appropriately chosen substitutions, integrate…
Using appropriately chosen substitutions, integrate $y'+ay^2=\dfrac{b}{x^2}$.
Explanation of Method
Try a particular solution $y_1=c/x$; matching coefficients gives $c$ as a root of $ac^2-c-b=0$. Reduce via $y=y_1+1/u$; the resulting linear equation for $u$ has two cases depending on whether $2ac=1$.
Step 1: Find $c$
Substituting $y_1=c/x$: $ac^2-c=b$, so $c=\dfrac{1\pm\sqrt{1+4ab}}{2a}$.
Step 2: Reduce via $y=c/x+1/u$
With $Q=0$, $R=-a$: $Q+2Ry_1=-\dfrac{2ac}{x}$, giving $u'-\dfrac{2ac}{x}u=a$.
Step 3a: Case $2ac\ne1$
Integrating factor $x^{-2ac}$: $\dfrac{d}{dx}\!\left(\dfrac{u}{x^{2ac}}\right)=\dfrac{a}{x^{2ac}}$, giving $u=\dfrac{ax}{1-2ac}+Cx^{2ac}$.
Step 3b: Case $2ac=1$
Then $u'-\dfrac{u}{x}=a$, so $\dfrac{d}{dx}\!\left(\dfrac{u}{x}\right)=\dfrac{a}{x}$, giving $u=x(a\ln|x|+C)$.
Result
Final answer:
With $c=\dfrac{1\pm\sqrt{1+4ab}}{2a}$:
Case $2ac\ne1$: \[\boxed{\,y=\frac{c}{x}+\frac{1-2ac}{ax+C(1-2ac)x^{2ac}}}\]
Case $2ac=1$: \[\boxed{\,y=\frac{c}{x}+\frac{1}{x(a\ln|x|+C)}}\]
(both verified symbolically).