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

[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)?

 

[tiny-tim]

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 )

 

[manenbu]

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.

 

[tiny-tim]

ah i see now …

ok, you have dy/y^3/2 = ±dt√2 …

just keep the ±

(doesn't it get squared in the end anyway?)

 

[manenbu]

I need to find an expression for y, that is to integrate y'.
So do I have 2 options for this?

 

[tiny-tim]

(just got up :zzz: …)

Show us what you get.

(btw, haven't you missed out a constant of integration in your (y')²=2y³ ?)

 

[manenbu]

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.

 

[tiny-tim]

Whatever constant there was equals 0.

ok, but show us what you get.

 

[manenbu]

I get
-2/y^1/2 = +-2^1/2x + c

I don't get your point, sorry.

 

[tiny-tim]

(have a ± )

I get
-2/y^1/2 = +-2^1/2x + c

ok, now square it to get y.

 

[manenbu]

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.

 

[tiny-tim]

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

 

[manenbu]

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

Now I figured it out.
thanks a lot :)