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An ODE solving for Y

  • Thread starter manenbu
  • Start date
  • #1
103
0

Homework Statement



yy'' = (y')2 - (y')3

Homework Equations





The Attempt at a Solution



y' = p(y)
y'' - p'p

yp' = p - p2

dp/p + dp/(1-p) = dy/y

ln|p|+ln|1-p| = ln|y|+c

p-p2 = cy

y' - y'2 = cy

now what?
how do I solve for y?
I think I'm missing some stupid algebra thingy here, but can't figure it out.
 

Answers and Replies

  • #2
33,164
4,848

Homework Statement



yy'' = (y')2 - (y')3

Homework Equations





The Attempt at a Solution



y' = p(y)
y'' - p'p
What does p(y) mean? I would normally take this to mean "p of y". Do you mean y' = py; i.e. p times y?
What is y'' - p'p? That's not an equation. How does it relate to the equation above it?
yp' = p - p2

dp/p + dp/(1-p) = dy/y

ln|p|+ln|1-p| = ln|y|+c

p-p2 = cy

y' - y'2 = cy

now what?
how do I solve for y?
I think I'm missing some stupid algebra thingy here, but can't figure it out.
 
  • #3
103
0
y' = p(y)
p is a function of y
should be y = p'p, not y - p'p.
y'' = p'(y)y' (chain rule)
y'' = p'(y)p(y) or p'p. :)
 
  • #4
gabbagabbahey
Homework Helper
Gold Member
5,002
6
dp/p + dp/(1-p) = dy/y

ln|p|+ln|1-p| = ln|y|+c
Errm... [itex]\frac{d}{dp}\ln|1-p|=-\frac{1}{1-p}\neq\frac{1}{1-p}[/itex] :wink:
 
  • #5
103
0
oh. of course. stupid me. Now it all comes together.
Thanks for pointing this out.

It's always the little stuff that makes it problematic.
Have a nice day! :)
 

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