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

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

Step 1: Find $a$

With $xy=a$: $-a+(a-2)^2=0\Rightarrow a^2-5a+4=0\Rightarrow(a-4)(a-1)=0$. Take $a=1$, so $y_1=1/x$.

Step 2: Reduce and solve for $u$

Expanding gives $y'=-y^2+\dfrac{4y}{x}-\dfrac{4}{x^2}$. With $y=\dfrac1x+\dfrac1u$: $u'+\dfrac{2u}{x}=1$. With $\mu(x)=x^2$: $\dfrac{d}{dx}(x^2u)=x^2$, so $x^2u=\dfrac{x^3}{3}+C$, i.e. $u=\dfrac{x}{3}+\dfrac{C}{x^2}$.