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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?

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  • $\begingroup$ 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. $\endgroup$ – avs Jan 18 '17 at 19:20
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    $\begingroup$ reduce the order by putting $u'=y$ $\endgroup$ – tired Jan 18 '17 at 19:20
  • $\begingroup$ The solution is $u = a + b/x$ $\endgroup$ – Gribouillis Jan 18 '17 at 19:30
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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.

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  • $\begingroup$ 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. $$xu' + (d-2)u = \lambda$$ $\endgroup$ – Monstrous Moonshine Jan 18 '17 at 19:26
  • $\begingroup$ 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. $\endgroup$ – Chinny84 Jan 18 '17 at 19:27
  • $\begingroup$ Nevermind, I see what you mean now. I think I got it. $\endgroup$ – Monstrous Moonshine Jan 18 '17 at 19:29
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We can also put a new unknown function $z(x)=u'(x)$. With this substitution we arrive at the 1st order homoheneous linear equation $xz'+2z=0$.

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