Solve the Bernoulli Differential equations. xy′ - y = xy^5

Here the given differential equation is
${\displaystyle x{y}^{{}^{\prime }}-y=x{y}^5}$
which also can be written as
${\displaystyle y^{{}^{\prime }}-\frac{y}{x}=y^5}$.
Let ${\displaystyle v=y^{-4}thend\frac{v}{dx}=-4y^{-5}\frac{dy}{dx}}$.The above De reduces to
${\displaystyle -\frac{y^5}{4}\frac{dv}{dx}-\frac{y}{x}=y^5}$
${\displaystyle \Rightarrow -\frac{1}{49}\frac{dv}{dx}-\frac{y^{-4}}{x}=1}$
${\displaystyle \Rightarrow \frac{dv}{dx}+\frac{4v}{x}=-4}$
${\displaystyle \Rightarrow x^4\frac{dv}{dx}+4v{x}^3=-4x^4}$
${\displaystyle \Rightarrow \frac{d}{dx}\left(v{x}^4\right)=-4x^4}$
${\displaystyle \Rightarrow v{x}^4=-4\int x^{4}dx+c}$
${\displaystyle \Rightarrow v{x}^4=-\frac{4x^5}{5}+c}$
${\displaystyle \Rightarrow v=-\frac{4x}{5}+c{x}^{-4}}$
${\displaystyle \Rightarrow y^{-4}=-\frac{4x}{5}+c{x}^{-4}}$
${\displaystyle \Rightarrow y={(-\frac{4x}{5}+c{x}^{-4})}^4}$
where c is a arbitrary constant.