Astrophysics
  • Astrophysics
@dan815 @ganeshie8 @IrishBoy123
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
sidmanfu
  • sidmanfu
@Hero
Astrophysics
  • Astrophysics
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.
Astrophysics
  • Astrophysics
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

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Astrophysics
  • Astrophysics
ammmmgggg its zep!
zepdrix
  • zepdrix
+_+
Astrophysics
  • Astrophysics
:d
zepdrix
  • zepdrix
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
zepdrix
  • zepdrix
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.
zepdrix
  • zepdrix
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
Astrophysics
  • Astrophysics
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
zepdrix
  • zepdrix
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.
zepdrix
  • zepdrix
Oh I'm watching the video :O He did this kind of backwards.. hmm interesting
Astrophysics
  • Astrophysics
Yeah I did it that way
Astrophysics
  • Astrophysics
My substitution was \[v = y^{-2} \implies y = \frac{ 1 }{ \sqrt{v} }\]
zepdrix
  • zepdrix
\(\large\rm u=y^{-2}\qquad\to\qquad y=u^{-1/2}\) Ah yes I see :) You get all these nasty fractional powers though +_+ weird stuff
Astrophysics
  • Astrophysics
\[y' = - \frac{ 1 }{ 2 }v^{-3/2} \frac{ dv }{ dt }\]
Astrophysics
  • Astrophysics
Yeah haha
Astrophysics
  • Astrophysics
It didn't work out so well
Astrophysics
  • Astrophysics
I think I'll just stick to \[\large\rm y^{-n}y'+p(x)y^{1-n}=q(x)\] haha
zepdrix
  • zepdrix
\[\large\rm -\frac{1}{2}u^{-3/2}u'+\frac{2}{t}u^{-1/2}=\frac{1}{t^2}u^{-3/2}\]Oh boy, I don't like this very much XD
zepdrix
  • zepdrix
ya, at least for this one :)
Astrophysics
  • Astrophysics
Yup haha, maybe I made a fractional mistake somewhere and it was actually suppose to be the same...ehh
Astrophysics
  • Astrophysics
Anyways, thanks zepdrix!

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