# Help me please

by judefrance
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 P: 4 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
 Sci Advisor HW Helper PF Gold P: 12,016 Hint: Convince yourself of the following equality: $$\frac{d^{2}e}{dr^{2}}+\frac{1}{r}\frac{de}{dr}=\frac{1}{r}\frac{d}{dr}( r\frac{de}{dr})$$
 P: 4 And, if i want to find the general form of e(r)?
 Sci Advisor HW Helper PF Gold P: 12,016 Help me please You get: $$\frac{d}{dr}(r\frac{de}{dr})=0$$ This differential equation can be directly integrated to find the general solution.
 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 ?
 P: 4 THANKS!!!!! you save me!!!
 P: 199 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.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,310 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.