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21 hours ago comment added Harold It's a good idea to see it as a Bernoulli equation there. Actually your method and mine are similar in the beginning. But after I got $p'-\frac p{2y}=\frac y{2p}$ I went straight to another substitution $t=\frac p y$ without identifying that it is actually a Bernoulli equation. Thanks for your observation!
yesterday history edited Nasser CC BY-SA 4.0
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yesterday history answered Nasser CC BY-SA 4.0