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

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The discussion revolves around solving the ordinary differential equation (ODE) yy'' - (y')² = y³, which simplifies to (y')² = 2y³. The participants explore the implications of the ± sign after taking the square root, leading to two potential integrals. They conclude that both positive and negative roots must be considered, especially when initial conditions are provided, such as y(0) = 2 and y'(0) = 4. The conversation emphasizes the importance of correctly handling constants of integration and the implications of squaring expressions in the solution process.

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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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