Bernoulli First Order ODE I want to know if my answer is equivalent to the one in the back of the book. if so what was the algebra? if not then what happened?
$$x^2y'+ 2xy = 5y^3$$
$$y' = -\frac{2y}{x} + \frac{5y^3}{x^2}$$
$n = 3$
$v = y^{-2}$
$-\frac{1}{2}v'=y^{-3}$
$$\frac{-1}{2}v'-\frac{2}{x}v = \frac{5}{x^2}$$
$$v'+\frac{4}{x}v=\frac{-10}{x^2}$$
this is now a first order linear ODE where:
$$y(x)=\frac{1}{u(x)}\int u(x)q(x)$$
$u(x)=e^{4\int\frac{1}{x}}=x^4$
$q(x) = \frac{-10}{x^2}$
$$\frac{1}{x^4}\int x^4 \frac{-10}{x^2}=\frac{1}{x^4}\frac{-10x^{2+1}}{2+1}+C$$
which leaves us with :
$$\frac{1}{y^2} = \frac{-10}{3x}+x^{-4}C$$
naturally 
$$y^2= \frac{1}{\frac{-10}{3x}+x^{-4}C}$$
The book states the answer as being:
$$y^2= \frac{x}{2+Cx^5}$$ 

You switched one sign too many in
$$
-\frac12v'+\frac2xv=\frac5{x^2}
$$
Then
$$
\left(\frac{v}{x^4}\right)'=\frac{v'}{x^4}-\frac{4v}{x^5}=-\frac{10}{x^6}
\implies
\frac{v}{x^4}=\frac2{x^5}+C
$$
etc.
