Find the general integral of the Bernoulli equation…
Find the general integral of the Bernoulli equation $y^{n-1}(ay'+y)=x$.
Explanation of Method
Expand and substitute $v=y^n$ (so $v'=ny^{n-1}y'$) to reduce to a linear equation for $v(x)$.
Step 1: Expand and substitute $v=y^n$
Expanding: $ay^{n-1}y'+y^n=x$. With $v=y^n$, $v'=ny^{n-1}y'$, so $y^{n-1}y'=v'/n$:
\[\frac{a}{n}v'+v=x\quad\Longrightarrow\quad v'+\frac{n}{a}v=\frac{n}{a}x.\]
Step 2: Solve the linear equation
With $\mu(x)=e^{nx/a}$: $\dfrac{d}{dx}\!\left(e^{nx/a}v\right)=\dfrac{n}{a}xe^{nx/a}$. Integrating by parts, $\displaystyle\int\frac{n}{a}xe^{nx/a}dx=e^{nx/a}\!\left(x-\frac{a}{n}\right)+C$, so $v=x-\dfrac{a}{n}+Ce^{-nx/a}$.
Result
Final answer:
\[\boxed{\,y^n=x-\frac{a}{n}+Ce^{-nx/a},\qquad C\in\mathbb{R}}\]
Note: a commonly circulated form of this answer, $y^n=n(x-a)+Ce^{-x/a}$, does not satisfy the original equation — direct substitution shows a nonzero residual. The boxed formula above has been verified symbolically.