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  • $\begingroup$ I think there is a problem with the change of variables here. if $z=mygo^{l_0}$, then $mygo=z^{1/l_0}$. If you insert this into the equation, you wont get the one that you found. $\endgroup$ Commented Apr 19, 2017 at 16:14
  • $\begingroup$ 'mygo' is definitely a function but in your expression $mygo(a,b,c...)^{1/l_0}$, it is taken as function composition.Actually, it is just multiplication by this function, namely $mygo*(a,b,c...)^{1/l_0}$ . Therefore, as I said, the simplification is unfortunately wrong. $\endgroup$ Commented Apr 19, 2017 at 16:21
  • $\begingroup$ @SeyhmusGüngören Sorry, I do not understand your concerns. $\endgroup$ Commented Apr 19, 2017 at 18:32
  • $\begingroup$ the answer is simply wrong $\endgroup$ Commented Apr 19, 2017 at 18:33
  • $\begingroup$ @SeyhmusGüngören I disagree. Please insert some random values for the constants to verify that g0 solves the third equation in your question. $\endgroup$ Commented Apr 19, 2017 at 21:48