Find the general integral of 4y'+y^2+4x^(-2)=0 , given that…

Find the general integral of $4y'+y^2+4x^{-2}=0$, given that a particular integral has the form $y=\dfrac{a}{x}$.

Step 1: Find $a$

Substituting: $-4a+a^2+4=0\Rightarrow(a-2)^2=0\Rightarrow a=2$, so $y_1=2/x$ (a repeated root).

Step 2: Reduce and solve for $u$

With $y=\dfrac2x+\dfrac1u$: $4u'-\dfrac{4u}{x}=1$, i.e. $u'-\dfrac{u}{x}=\dfrac14$. With $\mu(x)=1/x$: $\dfrac{d}{dx}\!\left(\dfrac{u}{x}\right)=\dfrac{1}{4x}$, so $\dfrac{u}{x}=\dfrac{\ln|x|}{4}+C$, i.e. $u=x\!\left(\dfrac{\ln|x|}{4}+C\right)$.