The solution $y^{n-1}\left(a{y}^{\mathrm{\prime }}+y\right)=x$ is 

$a{y}^{n-1}{y}^{\mathrm{\prime }}+y^n=x$, Put $y^n=u$

$\Rightarrow n{y}^{n-1}\frac{dy}{dx}=\frac{du}{dx}$. The given equation reduces to 

$\frac{a}{n}\frac{du}{dx}+u=x\Rightarrow \frac{du}{dx}+\frac{n}{a}u=\frac{n}{a}x$ 

I.F. $=e^{\frac{n}{a}x}$ Multiplying with $e^{\frac{n}{a}x}$ we have 

$\frac{d}{dx}\left(u{e}^{\frac{n}{a}x}\right)=\frac{n}{a}\left[\begin{array}{c}\frac{n}{x}x\\ x{e}^a\end{array}\right]$

$u{e}^{\frac{n}{a}x}=\frac{n}{a}\left[\frac{x{e}^{\frac{n}{a}}x}{n/a}-\frac{1}{(n/a{)}^2}{e}^{\frac{n}{a}x}\right]+C$ 

So $u=x-\frac{a}{n}+C{e}^{-\frac{n}{a}x}$ 

$\Rightarrow n{y}^n=nx-a+C{e}^{-\frac{n}{a}x}$