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clava345
Dec4-11, 11:25 AM
I'm a little bit rusty with my differential equations, and can't seem to see how solving for 1/r d/dr (r dT/dr)=0 has the solution T(r)=C_1*ln⁡(r)+C_2

JJacquelin
Dec4-11, 02:44 PM
integration of d/dr (r dT/dr)=0 leads to (r dT/dr) = C1
dT = C1 dr/r
integration of dT=C1 dr/r leads to t = C1 ln(r) + C2

clava345
Dec4-11, 04:27 PM
Thanks