What do I do with the +- in y'= +-(2y^3)^1/2?

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Discussion Overview

The discussion revolves around the handling of the ± sign in the context of solving an ordinary differential equation (ODE) represented as yy'' - (y')² = y³. Participants explore the implications of taking the square root of (y')² = 2y³ and the subsequent steps in integrating to find y, including considerations of initial conditions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions how to proceed after obtaining y' = ±(2y³)^(1/2), asking if they should solve two different integrals or assume a positive root.
  • Another participant challenges the derivation of (y')² = 2y³, suggesting a reevaluation of the initial steps.
  • A participant explains their method involving the substitution y' = p(y) and the application of Bernoulli's method, leading to the same result.
  • There is a suggestion to keep the ± sign during integration, with a later comment noting that squaring the equation will ultimately resolve the sign issue.
  • Participants discuss the impact of initial conditions on the integration process, with one asserting that a constant of integration can be disregarded due to the specific conditions provided.
  • One participant expresses confusion about the implications of squaring the expression that includes the constant of integration, leading to a clarification about handling the ± sign in relation to the constant.

Areas of Agreement / Disagreement

Participants express differing views on how to handle the ± sign and the role of the constant of integration, indicating that the discussion remains unresolved regarding the best approach to take after obtaining the ± sign.

Contextual Notes

There are limitations related to the assumptions made about the constants of integration and the handling of the ± sign, which are not fully resolved in the discussion.

manenbu
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So there's this ODE:
yy''-(y')2 = y3
After doing some work it gets to this point:
(y')2=2y3
Taking the root:
y' = +-(2y3)1/2

My question is, what do I do with the +-?
Do I solve 2 different integrals? Assume that this is positive?
Does it matter if it has starting conditions (y(x)=a, y'(x)=b)?
 
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Hi manenbu! :smile:
manenbu said:
yy''-(y')2 = y3
After doing some work it gets to this point:
(y')2=2y3

dunno how you got that. :confused:

Start again.

(Hint: can you see the anti-derivative of yy'' plus (y')2 ? if so, adapt it for minus :wink:)
 
How I got that?

By the substitution of y' = p(y), and then y''=p'y'=p'p.

Then I solve whatever I get with bernoulli's method, and end up with this:
(y')2=2y3
After reduction of order.

If I solve it assuming that the root is positive, I get to a solution which is correct (according to the answers). It can be also solved as a homogenous problem, reaching the same result.
That's not the point (but if you really, want I can solve in both methods and scan the solution for you).

The point is what I do after I get the +- thingy after taking the root.
 
ah i see now …

ok, you have dy/y3/2 = ±dt√2 …

just keep the ± :smile:

(doesn't it get squared in the end anyway?)
 
I need to find an expression for y, that is to integrate y'.
So do I have 2 options for this?
 
(just got up :zzz: …)

Show us what you get. :smile:

(btw, haven't you missed out a constant of integration in your (y')2=2y3 ?)
 
There was a starting condition, so no constant.
Can't find the problem right now, but trust me on this. Whatever constant there was equals 0.
 
manenbu said:
Whatever constant there was equals 0.

ok, but show us what you get. :smile:
 
I get
-2/y1/2 = +-21/2x + c

I don't get your point, sorry.
 
  • #10
(have a ± :wink:)
manenbu said:
I get
-2/y1/2 = +-21/2x + c

ok, now square it to get y. :smile:
 
  • #11
I found the original problem!yy''-(y')2 = y3
y(0) = 2
y'(0) = 4

Now, if we continue from where you stopped, squaring won't help because I'm squaring ±x + c and not just ±x.
c does not equal zero.
 
  • #12
manenbu said:
Now, if we continue from where you stopped, squaring won't help because I'm squaring ±x + c and not just ±x.
c does not equal zero.

erm :redface: … (±x + c)2 = (x ± c)2 :wink:
 
  • #13
oh - and then it's a ± before the constant, so I can drop it.

Now I figured it out.
thanks a lot :)
 

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