Conquering Autonomous Equations: Solving for y(0) and y'(0)

[Fredh]

Homework Statement

y*y'' = (y')² + y²*y'

y(0) = -1/2 ,

y'(0) = 1

Homework Equations

The Attempt at a Solution

y' = p
y'' = p'p

p(p'p) = p² + y²p

Dividing both sides by p

p'p = p + y²

And then I'm stuck

 

[HallsofIvy]

Homework Statement

y*y'' = (y')² + y²*y'

y(0) = -1/2 ,

y'(0) = 1

Homework Equations

The Attempt at a Solution

y' = p
y'' = p'p

p(p'p) = p² + y²p

Perhaps this is a typo but it should be y(p'p)= p²+ y²p

Dividing both sides by p

p'p = p + y²

And then I'm stuck

You should have yp'= p+ y². Also, the "p' " indicates differentiation with respect to y rather than the original independent variable.

Now that is certainly non-trivial but at least is linear. Can you find the "integrating factor"?

 

[Fredh]

Thanks!

It was not a typo, anyway, I solved yp' = p + y²:

dp/dy - p/y = y

Integrating factor = 1/y

p'/y - p/y² = 1

And that gives me y' = y(y+k)

But that does not fit the initial conditon y'(0) = 1

 

[SammyS]

Thanks!

It was not a typo, anyway, I solved yp' = p + y²:

dp/dy - p/y = y

Integrating factor = 1/y

p'/y - p/y² = 1

And that gives me y' = y(y+k)

But that does not fit the initial condition y'(0) = 1

It fits just fine with k = -3/2 .

 

[Fredh]

It fits just fine with k = -3/2 .

Edit: trying to solve it again

 

[Fredh]

I don't know what to do, can you tell me how does it fit?

 

[SammyS]

Thanks!

It was not a typo, anyway, I solved yp' = p + y²:

dp/dy - p/y = y

Integrating factor = 1/y

p'/y - p/y² = 1

And that gives me y' = y(y+k)

But that does not fit the initial conditon y'(0) = 1

So, you have y' = y(y+k).

In specific that means that $\displaystyle y'(0)=y(0)\left(y(0)+k\right)\ .$

Plug the initial values into that & solve for k.

$\displaystyle (1)=\left(-\frac{1}{2}\right)\left(\left(-\frac{1}{2}\right)+k\right)$

 

[Fredh]

Oh!
Of course, it didn't occur to me it was y(0) I had to use, thank you very much!