Question

@dan815 @ganeshie8 @IrishBoy123

Answer 1

@Hero

Answer 2

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.

Answer 3

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

Answer 4

ammmmgggg its zep!

Answer 5

+_+

Answer 6

:d

Answer 7

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

Answer 8

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.

Answer 9

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

Answer 10

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

Answer 11

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.

Answer 12

Oh I'm watching the video :O
He did this kind of backwards.. hmm interesting

Answer 13

Yeah I did it that way

Answer 14

My substitution was \[v = y^{-2} \implies y = \frac{ 1 }{ \sqrt{v} }\]

Answer 15

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

Answer 16

\[y' = - \frac{ 1 }{ 2 }v^{-3/2} \frac{ dv }{ dt }\]

Answer 17

Yeah haha

Answer 18

It didn't work out so well

Answer 19

I think I'll just stick to \[\large\rm y^{-n}y'+p(x)y^{1-n}=q(x)\] haha

Answer 20

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

Answer 21

ya, at least for this one :)

Answer 22

Yup haha, maybe I made a fractional mistake somewhere and it was actually suppose to be the same...ehh

Answer 23

Anyways, thanks zepdrix!