Important Solve the bernoullis

Make the given differential equation into the standard form of first order ordinary differential equation. Now determine the integrating factor.
Substitute it in the solution form. Simplify the equation in proper form.
First rewrite the equation and convert it into first order ordinary differential equation.
${\displaystyle \frac{dy}{dx}(x^2{y}^3+xy)=1}$
${\displaystyle \frac{dy}{dx}=\frac{1}{(x^2{y}^3+xy)}}$
${\displaystyle \frac{dx}{dy}=x^2{y}^3+xy}$
${\displaystyle \frac{dx}{dy}-xy=x^2{y}^3}$
${\displaystyle \frac{1}{x^2}\frac{dx}{dy}-\frac{y}{x}=y^3}$
Now let ${\displaystyle \frac{1}{x}=u}$ then on differentiating this equation on both the sides with respect to y, ${\displaystyle -\frac{1}{x^2}\frac{dx}{dy}=\frac{du}{dy}}$.
Substituting it in the equation, it will be ${\displaystyle \frac{du}{dx}+uy=-y^3}$.

Now this is an equation of first order differential equation ${\displaystyle \frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)}$ thus determine the integrating factor (IF) determine as ${\displaystyle IF=e^{\int P\left(x\right)dx}}$ where the solution is ${\displaystyle y{e}^{\int P\left(x\right)dx}=-\int Q\left(x\right)e^{\int P\left(x\right)dx}dx.}$
${\displaystyle IF=e^{\int ydx}=e^{\frac{y^2}{2}}}$
Substitute it in the solution form and simplify.
${\displaystyle u{e}^{\frac{y^2}{2}}=-\int y^2{e}^{\frac{y^2}{2}}dy}$
${\displaystyle \frac{e^{\frac{y^2}{2}}}{x}=-2(\frac{y^2}{2}-1)e^{\frac{y^2}{2}}+c}$
${\displaystyle \frac{1}{x}=2-y^2+c{e}^{-\frac{y^2}{2}}}$
Hence the solution is ${\displaystyle \frac{1}{x}=2-y^2+c{e}^{-\frac{y^2}{2}}}$