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So we have the two ODE solutions are the cosine/sine and $r^n$ since it was a Cauchy Euler type.

For the steady state, the solution is just a constant since it has to have period 2pi. But with $r^n$, how with lambda equal to zero does $\ln r$ come into play?

If my question is hard to follow, go here http://web.mit.edu/6.013_book/www/chapter5/5.7.html

Look at (9) and you will see ln r. I don't see we get that.

For the steady state, the solution is just a constant since it has to have period 2pi. But with $r^n$, how with lambda equal to zero does $\ln r$ come into play?

If my question is hard to follow, go here http://web.mit.edu/6.013_book/www/chapter5/5.7.html

Look at (9) and you will see ln r. I don't see we get that.

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