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

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

Explanation of Method

Substitute $y=a/x$ to fix $a$, then reduce via $y=y_1+1/u$ to a linear equation for $u(x)$.

Step 1: Find $a$

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

Step 2: Reduce and solve for $u$

With $y=\dfrac2x+\dfrac1u$: $u'-\dfrac{4u}{x}=1$. With $\mu(x)=x^{-4}$: $\dfrac{d}{dx}\!\left(\dfrac{u}{x^4}\right)=\dfrac{1}{x^4}$, so $\dfrac{u}{x^4}=-\dfrac{1}{3x^3}+C$, i.e. $u=Cx^4-\dfrac{x}{3}$.

Result

Final answer:

\[\boxed{\,y=\frac{2}{x}+\frac{3}{x(3Cx^3-1)}=\frac{6Cx^3+1}{x(3Cx^3-1)},\qquad C\in\mathbb{R}}\]