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

1. sidmanfu

@Hero

2. 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.

3. 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

4. Astrophysics

ammmmgggg its zep!

5. zepdrix

+_+

6. Astrophysics

:d

7. 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

8. 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.

9. 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

10. 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

11. 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.

12. zepdrix

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

13. Astrophysics

Yeah I did it that way

14. Astrophysics

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

15. 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

16. Astrophysics

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

17. Astrophysics

Yeah haha

18. Astrophysics

It didn't work out so well

19. Astrophysics

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

20. 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

21. zepdrix

ya, at least for this one :)

22. Astrophysics

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

23. Astrophysics

Anyways, thanks zepdrix!