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Help me pleaseby judefrance
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#1
May2304, 03:38 AM

P: 4

Can you help me to solve this:
(dē e(r)/drē)+(1/r)*(d e(r)/dr)=0 There is no initials conditions, please use general form 


#2
May2304, 03:44 AM

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


#3
May2304, 03:57 AM

P: 4

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



#4
May2304, 04:10 AM

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


#5
May2304, 04:50 AM

P: 4

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 ? 


#6
May2304, 05:56 AM

P: 4

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



#7
May2304, 07:50 AM

P: 199

I would also do it this way:
Rewriting: [tex]e''+\frac{1}{r}e'=0[/tex] I would then multiply through by r^2: [tex]r^2e''+re'=0[/tex] I would recognize this as a d.e. of the EulerCauchy form: [tex]x^2y''+axy' + by=0[/tex] In the case of the given equation, a=1 and b=0. The characteristic equation for the EulerCauchy is: [tex]m^2+(a1)m+b=0[/tex] In our case: [tex] \begin{align*} m^2&=0\\ m&=0 \end{align*} [/tex] For the case of a real double root in the characteristic equation, the general solution for the EulerCaucy is given as: [tex]y=(A + B\ln x)x^m[/tex] So in our case: [tex] \begin{align*} e(r)&=(A + B\ln r)x^0\\ e(r)&=A + B\ln r \end{align*} [/tex] I guess this solution depends on having the EulerCauchy form available to you in your course, which may not be the case. 


#8
May2304, 04:32 PM

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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)+ C_{1} or u= e'= C_{1}/r. Integrating again, e= C_{1}lnr+ C_{2}. 


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