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Astrophysics

  • one year ago

@dan815 @ganeshie8 @IrishBoy123

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  1. sidmanfu
    • one year ago
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    @Hero

  2. Astrophysics
    • one year ago
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    Bernoulli 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 \[v=y^{1-n}\], 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.

  3. Astrophysics
    • one year ago
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    I 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

  4. Astrophysics
    • one year ago
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    ammmmgggg its zep!

  5. zepdrix
    • one year ago
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    +_+

  6. Astrophysics
    • one year ago
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    :d

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

  8. zepdrix
    • one year ago
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    Oh 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.

  9. zepdrix
    • one year ago
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    I'm not sure why they don't just call this \(\large\rm y^{-n}y'+p(x)y^{1-n}=q(x)\) the Bernoulli equation. Maybe because it's ugly

  10. Astrophysics
    • one year ago
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    Ah, 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

  11. zepdrix
    • one year ago
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    I mean, you can make the substitution right away, it won't affect anything,\[\large\rm u=y^{1-3},\qquad u=y^{-2}\]\[\large\rm u'=-2y^{-3}y'\]But you won't be able to plug in any of the pieces until you do the division.

  12. zepdrix
    • one year ago
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    Oh I'm watching the video :O He did this kind of backwards.. hmm interesting

  13. Astrophysics
    • one year ago
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    Yeah I did it that way

  14. Astrophysics
    • one year ago
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    My substitution was \[v = y^{-2} \implies y = \frac{ 1 }{ \sqrt{v} }\]

  15. zepdrix
    • one year ago
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    \(\large\rm u=y^{-2}\qquad\to\qquad y=u^{-1/2}\) Ah yes I see :) You get all these nasty fractional powers though +_+ weird stuff

  16. Astrophysics
    • one year ago
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    \[y' = - \frac{ 1 }{ 2 }v^{-3/2} \frac{ dv }{ dt }\]

  17. Astrophysics
    • one year ago
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    Yeah haha

  18. Astrophysics
    • one year ago
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    It didn't work out so well

  19. Astrophysics
    • one year ago
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    I think I'll just stick to \[\large\rm y^{-n}y'+p(x)y^{1-n}=q(x)\] haha

  20. zepdrix
    • one year ago
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    \[\large\rm -\frac{1}{2}u^{-3/2}u'+\frac{2}{t}u^{-1/2}=\frac{1}{t^2}u^{-3/2}\]Oh boy, I don't like this very much XD

  21. zepdrix
    • one year ago
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    ya, at least for this one :)

  22. Astrophysics
    • one year ago
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    Yup haha, maybe I made a fractional mistake somewhere and it was actually suppose to be the same...ehh

  23. Astrophysics
    • one year ago
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    Anyways, thanks zepdrix!

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