Roots of a derivative

  • Thread starter manenbu
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  • #1
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Main Question or Discussion Point

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)?
 

Answers and Replies

  • #2
tiny-tim
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Hi manenbu! :smile:
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:)
 
  • #3
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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.
 
  • #4
tiny-tim
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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?)
 
  • #5
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I need to find an expression for y, that is to integrate y'.
So do I have 2 options for this?
 
  • #6
tiny-tim
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(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 ?)
 
  • #7
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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.
 
  • #8
tiny-tim
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Whatever constant there was equals 0.
ok, but show us what you get. :smile:
 
  • #9
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I get
-2/y1/2 = +-21/2x + c

I don't get your point, sorry.
 
  • #10
tiny-tim
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(have a ± :wink:)
I get
-2/y1/2 = +-21/2x + c
ok, now square it to get y. :smile:
 
  • #11
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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
tiny-tim
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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
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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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