The Burger's vortex in 2 Dimension - solving Differential equation

After simplifying the vortex equation, I get to this equation:

$$-\alpha y \partial_y \omega = \alpha \omega + \nu \partial_{yy} \omega$$

where the $\alpha$ and $\nu$ are constant values and $\omega(y)$. I want to solve the equation for $\omega(y)$. How can I do it? I appreciate in advance for kind helps.

Regards,

Ehsan

$$-\alpha y \dfrac{d\omega}{dy} = \alpha\omega + \nu \dfrac{d^2\omega}{dy^2}$$

this is the same as

$$\nu \dfrac{d^2\omega}{dy^2} + \alpha \dfrac{d}{dy}(y\omega) = 0$$ thus

$$\nu \omega ' + \alpha y \omega + C = 0$$

now if $C=0$ then the solution is separable but if not then we have to solve $$\frac{d\omega}{dy} + \frac{\alpha}{\nu} y \omega= -C$$ using integrating factor $$\omega \mathrm{e}^{\frac{\alpha}{\nu}\frac{y^2}{2}} = -\int_0^y C \mathrm{e}^{\frac{\alpha}{\nu}\frac{y'^2}{2}}dy'$$

or

$$\omega(y) = -\left(\int_0^y C \mathrm{e}^{\frac{\alpha}{\nu}\frac{y'^2}{2}}dy'\right)\mathrm{e}^{-\frac{\alpha}{\nu}\frac{y^2}{2}}$$

• Very well done indeed, plus one!!! – Robert Lewis Nov 27 '14 at 5:25
• @robertlewis Thanks! Too kind. – Chinny84 Nov 27 '14 at 7:18
• Well, I was wondering if there wasn't some clever way of solving this, and you found it! – Robert Lewis Nov 27 '14 at 7:32

Maple finds solutions $$\omega(y) = \left(c_1 + c_2\; \text{erfi}\left(\sqrt{-\alpha/(2\nu)} y\right)\right) e^{-\alpha y^2/(2 \nu)}$$