Sci Advisor HW Helper PF Gold P: 12,016 Hint: Convince yourself of the following equality: $$\frac{d^{2}e}{dr^{2}}+\frac{1}{r}\frac{de}{dr}=\frac{1}{r}\frac{d}{dr}( r\frac{de}{dr})$$
 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: 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.