1. May 23, 2004

### judefrance

Can you help me to solve this:

(d² e(r)/dr²)+(1/r)*(d e(r)/dr)=0

There is no initials conditions, please use general form

2. May 23, 2004

### arildno

Hint:
Convince yourself of the following equality:
$$\frac{d^{2}e}{dr^{2}}+\frac{1}{r}\frac{de}{dr}=\frac{1}{r}\frac{d}{dr}(r\frac{de}{dr})$$

3. May 23, 2004

### judefrance

Yes, but....

And, if i want to find the general form of e(r)?

4. May 23, 2004

### arildno

You get:
$$\frac{d}{dr}(r\frac{de}{dr})=0$$
This differential equation can be directly integrated to find the general solution.

Last edited: May 23, 2004
5. May 23, 2004

### judefrance

Maybe this?

1/r*d/dr*(r*de/dr)=0
so:
d/dr*(r*de/dr)=0

If i integrate, i find:
e(r)=A*ln(r)+B

Wrong or not ?

6. May 23, 2004

### judefrance

Thanks

THANKS!!!!! you save me!!!

7. May 23, 2004

### TALewis

I would also do it this way:

Rewriting:

$$e''+\frac{1}{r}e'=0$$

I would then multiply through by r^2:

$$r^2e''+re'=0$$

I would recognize this as a d.e. of the Euler-Cauchy form:

$$x^2y''+axy' + by=0$$

In the case of the given equation, a=1 and b=0. The characteristic equation for the Euler-Cauchy is:

$$m^2+(a-1)m+b=0$$

In our case:

\begin{align*} m^2&=0\\ m&=0 \end{align*}

For the case of a real double root in the characteristic equation, the general solution for the Euler-Caucy is given as:

$$y=(A + B\ln x)x^m$$

So in our case:

\begin{align*} e(r)&=(A + B\ln r)x^0\\ e(r)&=A + B\ln r \end{align*}

I guess this solution depends on having the Euler-Cauchy form available to you in your course, which may not be the case.

8. May 23, 2004

### HallsofIvy

Another way to do this problem is let u= e' so that u'= e" and the equation reduces to the separable first order equation u'+ (1/r)u= 0. Then du/u= -dr/r and so
ln(u)= -ln(r)+ C1 or u= e'= C1/r. Integrating again, e= C1ln|r|+ C2.