# Solving autonomous equation

1. Jun 29, 2012

### Fredh

1. The problem statement, all variables and given/known data
y*y'' = (y')² + y²*y'

y(0) = -1/2 ,

y'(0) = 1

2. Relevant equations

3. 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

2. Jun 29, 2012

### HallsofIvy

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

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"?

3. Jun 29, 2012

### 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

4. Jun 30, 2012

### SammyS

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

5. Jun 30, 2012

### Fredh

Edit: trying to solve it again

Last edited: Jun 30, 2012
6. Jun 30, 2012

### Fredh

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

7. Jun 30, 2012

### SammyS

Staff Emeritus
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.

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

8. Jun 30, 2012

### Fredh

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