Roots of a derivative

• manenbu
In summary, after doing some work, you get to this point: (y')2=2y3; taking the root yields y'=+-(2y3)1/2. This can be solved as a homogenous problem or assuming that the root is positive, yielding the same solution.f

manenbu

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)?

Hi manenbu!
yy''-(y')2 = y3
After doing some work it gets to this point:
(y')2=2y3

dunno how you got that.

Start again.

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

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 ±

(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.

(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.

Whatever constant there was equals 0.

ok, but show us what you get.

I get
-2/y1/2 = +-21/2x + c

I don't get your point, sorry.

(have a ± )
I get
-2/y1/2 = +-21/2x + c

ok, now square it to get y.

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.

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 … (±x + c)2 = (x ± c)2

oh - and then it's a ± before the constant, so I can drop it.

Now I figured it out.
thanks a lot :)