Strange way of solving a linear 2nd order DE

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SUMMARY

The discussion focuses on solving the linear second-order differential equation (DE) of the form \(\Phi^{''} + \frac{6}{\eta}\Phi^{'} = 0\). The solution involves recognizing that \(\Phi^{'} \propto \eta^{-6}\), leading to the conclusion that \(\Phi \propto \eta^{-5} + C\), where \(C\) is a constant. A participant initially attempted to solve the DE by substituting \(\Phi^{'} = x\) and separating variables, but encountered difficulties. Ultimately, the correct approach was confirmed by another participant, emphasizing the straightforward nature of the solution.

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Alexrey
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Homework Statement


I was given a DE of the form: \Phi^{''}+(6/\eta)\Phi^{'}=0 where the next step was given as \Phi^{'} \propto \eta^{-6} where the answer came out to be \Phi \propto \eta^{-5} + constant

The Attempt at a Solution


My attempt was to set \Phi^{'}=x where I would then get x^{'}=-(6/\eta)x and then solve for a seperable DE, but my answer was incorrect. Any help would be appreciated, thanks guys.

Any help would be appreciated, thanks guys.
 
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Alexrey said:
x^{'}=-(6/\eta)x
To clarify, η is the independent variable, and differentiation is wrt that? If so dx/x=-6d\eta/\eta yes? Isn't it straightforward from there?
 
About 5 minutes after posting this I figured it out. :/ Thanks for replying though, I appreciate it, that's exactly what I got. Cheers!
 

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