Find the general integral of the equation y'=(y)/(2yln y+y-x)

Find the general integral of the equation $y'=\dfrac{y}{2y\ln y+y-x}$.

Step 1: Invert and rearrange

$\dfrac{dx}{dy}=2\ln y+1-\dfrac{x}{y}$, i.e. $\dfrac{dx}{dy}+\dfrac{x}{y}=2\ln y+1$.

Step 2: Apply the integrating factor

With $\mu(y)=e^{\int dy/y}=y$: $\dfrac{d}{dy}(xy)=2y\ln y+y$.

Step 3: Integrate

By parts, $\displaystyle\int 2y\ln y\,dy=y^2\ln y-\frac{y^2}{2}$, so $xy=y^2\ln y-\dfrac{y^2}{2}+\dfrac{y^2}{2}+C=y^2\ln y+C$.