# Solving the ODE $xu'' + 2u' = 0$

As stated in the title, I'm trying to solve this ODE. I know how to solve the ODE $xu'' + u' = (xu')' = 0$, but this trick doesn't work when we replace $u'$ with $2u'$. How can we find the general solution for this equation?

• The function $u$, without derivatives, does not occur in your ODE. Can we take $v = u'$? In that case, we would get the ODE $v' + v / x = 0$, which considerably simplifies looking for the integrating factor. – avs Jan 18 '17 at 19:20
• reduce the order by putting $u'=y$ – tired Jan 18 '17 at 19:20
• The solution is $u = a + b/x$ – Gribouillis Jan 18 '17 at 19:30

Same trick applies - but slightly modified. $$xu'' + u' + u' = \frac{d}{dx}\left[xu' + u\right] = 0$$ so we have $$xu' + u = \lambda$$ which you can solve by integration factor.
• What do you mean when you say, solve it by integration factor? Let's say we make the equation a bit more general, e.g. $$xu' + (d-2)u = \lambda$$ – Monstrous Moonshine Jan 18 '17 at 19:26
• Sorry I meant integrating factor have you come across solving first order linear differential equations? I assume you have, if not I can update the post. – Chinny84 Jan 18 '17 at 19:27
We can also put a new unknown function $z(x)=u'(x)$. With this substitution we arrive at the 1st order homoheneous linear equation $xz'+2z=0$.