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?
Solving the ODE $xu'' + 2u' = 0$
0
$\begingroup$
real-analysis
ordinary-differential-equations
-
0The 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. – 2017-01-18
-
1reduce the order by putting $u'=y$ – 2017-01-18
-
0The solution is $u = a + b/x$ – 2017-01-18
2 Answers
5
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.
-
0What do you mean when you say, solve it by integration factor? Let's say we make the equation a bit more general, e.g. $$xu' + (d-2)u = \lambda$$ – 2017-01-18
-
0Sorry 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. – 2017-01-18
-
0Nevermind, I see what you mean now. I think I got it. – 2017-01-18
0
We can also put a new unknown function $z(x)=u'(x)$. With this substitution we arrive at the 1st order homoheneous linear equation $xz'+2z=0$.