Find the general integral of the Bernoulli equation…

Find the general integral of the Bernoulli equation $xy'+y=y^2\ln x$.

Step 1: Substitute $v=1/y$

Dividing $y'+\dfrac{y}{x}=\dfrac{\ln x}{x}y^2$ by $y^2$: $-v'+\dfrac{v}{x}=\dfrac{\ln x}{x}$, i.e. $v'-\dfrac{v}{x}=-\dfrac{\ln x}{x}$.

Step 2: Solve the linear equation

With $\mu(x)=1/x$: $\dfrac{d}{dx}\!\left(\dfrac{v}{x}\right)=-\dfrac{\ln x}{x^2}$. By parts, $\displaystyle\int-\frac{\ln x}{x^2}dx=\frac{\ln x+1}{x}+C$, so $v=\ln x+1+Cx$.