Integrate second order DE once Given the vorticity equation $$\frac{D \omega}{Dt}=(\omega \cdot \nabla)\textbf{u}+ν\nabla^2ω$$ and $\textbf{u} =  (−αr/2,v(r),αz) $ in cylindrical polars where alpha is positive constant. 
Find $\omega = \xi (r) \hat{\textbf z}$. I got it as $$\xi =\frac{\partial v}{\partial r}+\frac vr$$
Use the vorticity equation to show that $\xi(r)$ satisﬁes a second-order ordinary differential equation.
I got it as $$ \frac{\partial ^2 \xi}{\partial r^2}+\frac 1r \frac{\partial \xi}{\partial r}+ \frac{\alpha}{ν} \xi=0$$
Integrate the equation once to derive a ﬁrst-order ordinary differential equation for $\xi (r)$. You may assume that $\xi '(r)$ is ﬁnite as $r→0$. 

How can you integrate it "once"...
 A: First, you are missing the contribution from the $z-$component of the vortex convection term $\mathbb{u} \cdot \nabla \mathbb{\omega},$ which is $-\frac{ \alpha}{2}r \frac{d \xi}{d r}.$
The correct steady-state differential equation is 
$$\frac{d^2 \xi}{dr^2}+\frac 1r \frac{d \xi}{dr}+ \frac{\alpha}{ν} \xi= -\frac{\alpha}{2 \nu}r \frac{d \xi}{d r}.$$
This can be written as
$$\frac{1}{r}\frac{d}{dr}\left(r \frac{d \xi}{dr} \right) +\frac{\alpha}{2 \nu}r \frac{d \xi}{d r} + \frac{\alpha}{ν} \xi= 0.$$
The key step is to observe that this equation can be recast as 
$$\frac{1}{r}\frac{d}{dr}\left(r \frac{d \xi}{dr} + \frac{\alpha}{2 \nu}r^2 \xi \right) = 0.$$
Integrate "once" to obtain
$$r \frac{d \xi}{dr} + \frac{\alpha}{2 \nu}r^2 \xi = A,$$
and,
$$\frac{d \xi}{dr} + \frac{\alpha}{2 \nu}r \xi = \frac{A}{r}$$
where  $A$ is an integration constant.
To solve this first-order differential equation, multiply both sides by the integrating factor $\exp\left(\frac{\alpha r^2}{4 \nu} \right)$ to obtain
$$\frac{d}{dr}\left( \exp\left(\frac{\alpha r^2}{4 \nu} \right)\xi \right) = \frac{A}{r}\exp\left(\frac{\alpha r^2}{4 \nu}\right).$$
If we integrate again, the term on the RHS will introduce an exponential integral that is unbounded in the far-field. Hence, we enforce the boundary condition $\xi(r) \to 0$ as $r \to \infty$ by setting $A= 0,$ to obtain the solution
$$\xi(r) = B\exp\left(-\frac{\alpha r^2}{4 \nu}\right),$$
where $B$ is a second integration constant.
