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Astrophysics
 one year ago
@dan815 @ganeshie8 @IrishBoy123
Astrophysics
 one year ago
@dan815 @ganeshie8 @IrishBoy123

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Bernoulli I have this question \[y' +\frac{ 2y }{ t }=\frac{ y^3 }{ t^2 }\] and bernoulli's form is \[y'+p(t)y=q(t)y^n\] but how come I can't make the substitution right off the bat here as it's already in Benoulli's form and the substitution is \[v=y^{1n}\], but instead I have to divide by y^3? I see if I do divide it would make it sort of linear right, and then it works that way...I'm just a bit confused on why it works one way and not the other.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I saw a video recently and it did not divide by y^n and it made a substitution right away but pauls online notes requires you to divide hmm...unless this isn't in the right form

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0ammmmgggg its zep!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Hmm I'm not sure I understand the question. You want to make this substitution \(\large\rm u=y^{1n}\) right from the start? Or some other substitution and avoid the division all together? :o hmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Oh oh, so you're asking, "Isn't it already in the form that we need?" The answer is, no. The first step for these Bernoulli's is to do the division before substituting.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5I'm not sure why they don't just call this \(\large\rm y^{n}y'+p(x)y^{1n}=q(x)\) the Bernoulli equation. Maybe because it's ugly

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Ah, yeah that's what pauls online notes told me to haha, but I did it without and got something else, because I watched this https://www.youtube.com/watch?v=7MmhoqvM9_Q and went this way

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5I mean, you can make the substitution right away, it won't affect anything,\[\large\rm u=y^{13},\qquad u=y^{2}\]\[\large\rm u'=2y^{3}y'\]But you won't be able to plug in any of the pieces until you do the division.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5Oh I'm watching the video :O He did this kind of backwards.. hmm interesting

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I did it that way

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0My substitution was \[v = y^{2} \implies y = \frac{ 1 }{ \sqrt{v} }\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5\(\large\rm u=y^{2}\qquad\to\qquad y=u^{1/2}\) Ah yes I see :) You get all these nasty fractional powers though +_+ weird stuff

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0\[y' =  \frac{ 1 }{ 2 }v^{3/2} \frac{ dv }{ dt }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0It didn't work out so well

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0I think I'll just stick to \[\large\rm y^{n}y'+p(x)y^{1n}=q(x)\] haha

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5\[\large\rm \frac{1}{2}u^{3/2}u'+\frac{2}{t}u^{1/2}=\frac{1}{t^2}u^{3/2}\]Oh boy, I don't like this very much XD

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.5ya, at least for this one :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Yup haha, maybe I made a fractional mistake somewhere and it was actually suppose to be the same...ehh

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Anyways, thanks zepdrix!
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