Bernoulli First Order ODE
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.
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.