How Do You Solve This Differential Equation Without Initial Conditions?

[judefrance]

Help me please!

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

 

[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})

 

[judefrance]

Yes, but...

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

 

[arildno]

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

 

[judefrance]

Maybe this?

If I understand, you've made:
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 ?

 

[judefrance]

Thanks

THANKS! you save me!

 

[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:

<br /> \begin{align*}<br /> m^2&amp;=0\\<br /> m&amp;=0<br /> \end{align*}<br />

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:

<br /> \begin{align*}<br /> e(r)&amp;=(A + B\ln r)x^0\\<br /> e(r)&amp;=A + B\ln r<br /> \end{align*}<br />

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

 

[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.