Using appropriately chosen substitutions, integrate…

Using appropriately chosen substitutions, integrate $y'+y^2+\dfrac{1}{x}y-\dfrac{4}{x^2}=0$.

Explanation of Method

Try a particular solution $y_1=c/x$; matching coefficients gives $c=\pm2$. Reduce via $y=y_1+1/u$ to a linear equation for $u(x)$.

Step 1: Find $c$

Substituting $y_1=c/x$: $c^2-4=0\Rightarrow c=\pm2$. Take $y_1=2/x$.

Step 2: Reduce and solve for $u$

With $Q=-1/x$, $R=-1$: $Q+2Ry_1=-\dfrac{5}{x}$, giving $u'-\dfrac{5u}{x}=1$. With $\mu(x)=x^{-5}$: $\dfrac{d}{dx}\!\left(\dfrac{u}{x^5}\right)=\dfrac{1}{x^5}$, so $\dfrac{u}{x^5}=-\dfrac{1}{4x^4}+C$, i.e. $u=Cx^5-\dfrac{x}{4}$.

Result

Final answer:

\[\boxed{\,y=\frac{2}{x}+\frac{4}{x(4Cx^4-1)}=\frac{8Cx^4+2}{x(4Cx^4-1)},\qquad C\in\mathbb{R}}\]