# Solve ODE for Y: yy'' = (y')2 - (y')3

• manenbu
In summary, the conversation discusses the equation yy'' = (y')2 - (y')3 and the steps taken to solve it. It includes the use of the substitution y' = p(y) and the application of the chain rule. The conversation also addresses a mistake in the solution and concludes with a thank you for pointing it out.

## Homework Statement

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

## 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.

manenbu said:

## Homework Statement

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

## 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?
manenbu said:
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.

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. :)

manenbu said:
dp/p + dp/(1-p) = dy/y

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

Errm... $\frac{d}{dp}\ln|1-p|=-\frac{1}{1-p}\neq\frac{1}{1-p}$

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! :)