I'm reading on wikipedia (http://en.wikipedia.org/wiki/Vibrations_of_a_circular_drum#The_radially_symmetric_case) that the solution to the ODE [itex]R''+(1/r)R'-KR=0[/itex], by letting [itex]K=-\lambda ^2[/itex] is a linear combination of the Bessel functions of order 0, namely [itex]R(r)=c_1 J_0 (\lambda r ) + c_2 Y_0 (\lambda r)[/itex]. Then they say that c_2 must equal 0 for the solution to make physical sense (because Y_0 diverges at r=0) and thus the solution to the ODE is of the form [itex]R(r)=c_1J_0 (\lambda r )[/itex]. So far so good.(adsbygoogle = window.adsbygoogle || []).push({});

However in the problem of my thread https://www.physicsforums.com/showthread.php?t=652277, I had the equation [itex]R''+\frac{2R'}{r}+CR=0[/itex] which is very similar to the ODE in wikipedia and I've determined that the physically acceptable solution is of the form [itex]R(r)=c_2\frac{\sin (\sqrt C r )}{r}[/itex] (I've checked that it indeed satisfies the ODE). I find this unbelievable that the solution to the ODE of my problem differs so much from the one in wikipedia. What's going on here? I'm totally clueless.

Edit: Here's a graph between J_0(x) and sin (x) /x : http://www.wolframalpha.com/input/?i=graph+J_0+(x)+and+sin+(x)+/x+from+0+to+100. They are very similar! Is it possible to write the Bessel function of the first kind of order 0 as a constant times sin (kx) /x?

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# Solutions to 2 similar ODE's. Why does they differ that much?

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