Find the general integral of the Bernoulli equation…

Find the general integral of the Bernoulli equation $xy'-4y-x^2\sqrt{y}=0$.

Step 1: Substitute $v=\sqrt{y}$

With $y=v^2$, $y'=2vv'$: $2vv'-\dfrac{4v^2}{x}=xv$. Dividing by $2v$: $v'-\dfrac{2v}{x}=\dfrac{x}{2}$.

Step 2: Solve the linear equation

With $\mu(x)=1/x^2$: $\dfrac{d}{dx}\!\left(\dfrac{v}{x^2}\right)=\dfrac{1}{2x}$, so $\dfrac{v}{x^2}=\dfrac12\ln|x|+C$, i.e. $v=\dfrac{x^2}{2}(\ln|x|+2C)$.