Answer
$y = \pm (\frac{2}{5x}+Cx^4)^{-1/2}$
Work Step by Step
$y'+\frac{2}{x}y = \frac{y^3}{x^2}$
Note that $n=3$
Let $u = y^{1-n} = y^{-2}$
A first-order linear differential equation is one that can be put into the form:
$\frac{dy}{dx} + P(x)~y = Q(x)$
We can consider the given equation with $u$:
$\frac{du}{dx} + (-2)(\frac{2}{x})u = (-2)\frac{1}{x^2}$
$\frac{du}{dx} + (\frac{-4}{x})u = \frac{-2}{x^2}$
Let $P(x) = \frac{-4}{x}$
Let $Q(x) = \frac{-2}{x^2}$
We can find the integrating factor:
$I(x) = e^{\int \frac{-4}{x}~dx} = e^{-4~ln~x} = e^{ln~x^{-4}} = \frac{1}{x^4}$
We can multiply by the integrating factor:
$\frac{1}{x^4}~\frac{du}{dx}+(\frac{-4}{x^5})~u = \frac{-2}{x^6}$
$\frac{d}{dx}(\frac{1}{x^4}~u) = \frac{-2}{x^6}$
We can integrate both sides of the equation:
$\frac{1}{x^4}~u = \int \frac{-2}{x^6}~dx$
$\frac{1}{x^4}~u = \frac{2}{5x^5}+C$
$u = \frac{2}{5x}+Cx^4$
We can use the substitution $u = y^{-2}$:
$y^{-2} = \frac{2}{5x}+Cx^4$
$y = \pm (\frac{2}{5x}+Cx^4)^{-1/2}$
