Solve a homogeneous first-order differential equation

$y^\prime=xu^\prime+u$, while the right-hand side is $u^2+u+1$. Hence $xu^\prime=u^2+1$, so $(u^2+1)^{-1}u^\prime=1/x$.

$\int du/(1+u^2)=\int dx/x$ gives $\tan^{-1}u=\ln x+c$ on a suitable branch. Thus $u=\tan(\ln x+c)$ and $y=x\tan(\ln x+c)$. A solution is taken on an interval with $x>0$ avoiding the poles where $\ln x+c=\pi/2+n\pi$, $n\in\mathbb Z$.