At the end you wrote the reciprocal of y instead of y. It should be: y=(x²+c)/2W((x²+c)/2e)
@gadxxxx
8 ай бұрын
I agree.
@archerrusksjejandjajw
9 ай бұрын
Can I ask what I did wrong in the following work? We got different answers so I'm confused: yln(y) - y = (x^2+C)/2 y(ln(y)-1)=(x^2+C)/2 (e^ln(y))(ln(y)-1)=(x^2+C)/2 (e^(ln(y)-1))(ln(y)-1)=(x^2+C)/2e After taking lambert of both sides: ln(y)-1 = W((x^2+C)/2e) ln(y)=W((x^2+C)/2e)+1 y=e^(W((x^2+C)/2e)+1)
@Anmol_Sinha
9 ай бұрын
It's correct and both give the same answer. Let's simplify the video's answer furthur(after correcting a minor mistake) Let the lambert term be m (I am tired lol) So, their expression is y = (x²+C)/2m But (x²+C)/2 was y(lny-1) ! Substitute: y = y(lny-1)/m 1 = (lny-1)/m y = e^(m+1) which is the result you got! Nice job. I liked your approach
@verytiredanimal
9 ай бұрын
sick
@alexandercarroll9707
9 ай бұрын
Is multiplying the dx like that really a valid step?
@DavidPumpernickel
9 ай бұрын
It is, but it's not necessarily obvious. Off the top of my head, I think you could show it using some results relating to the Radon-Nikodym derivative, defined in measure theory.
@DavidPumpernickel
9 ай бұрын
Actually it's even simpler than this. You can justify it to yourself with chain rule lol
@alexandercarroll9707
9 ай бұрын
Ok I trust you lol Im only in calc 1 haven't even gotten close to touching measure theory@@DavidPumpernickel
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