Why Does the Cylindrical Wall Heat Equation Solution Include Logarithms?

clava345
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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
 
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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
 
Thanks
 
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