How's the ODE is turned to this auxiliary equation?

Question

I have the following ODE

$$ r\frac{d^{2}u}{dr^{2}} + 2\frac{du}{dr} = 0$$

i.e. $ru''+2u'=0$.

How is the ODE converted to this form of auxiliary equation: $m(m-1)+2m=0$

I have no idea on how to proceed. Thanks for any help!

Answer 1

HINT

Here it is another way to solve it. Multiply both sides by $r$ in order to obtain \begin{align*} ru^{\prime\prime} + 2u^{\prime} = 0 \Longleftrightarrow r^{2}u^{\prime\prime} + 2ru^{\prime} = 0 \Longleftrightarrow (r^{2}u^{\prime})^{\prime} = 0 \end{align*}

Answer 2

To solve $ru'' + 2u'= 0$ apply the change of function $$u(r)=r^m $$ so that $$u'(r)=mr^{m-1}, ~~u''(r)=m(m-1)r^{m-2}$$ in order to obtain $$rm(m-1)r^{m-2}+2mr^{m-1}=0$$ which simplifies to $$r^{m-1}\Big(m(m-1)+2m\Big)=0$$ where the auxillary equation is $$m(m-1)+2m=0$$

Answer 3

This is an Euler-Cauchy DE, where the standard solution method is to try to find solutions of the power type. In many cases this gives a full set of basis solutions. In your case where $m=0,-1$ these basis solutions are $u_1(r)=1$ and $u_2(r)=r^{-1}$. Any other solution is a linear combination.