- #1

- 10

- 0

## 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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Fredh
- Start date

- #1

- 10

- 0

y*y'' = (y')² + y²*y'

y(0) = -1/2 ,

y'(0) = 1

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

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 965

Perhaps this is a typo but it should be y(p'p)= p## 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

You should have yp'= p+ yDividing both sides by p

p'p = p + y²

And then I'm stuck

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

- #3

- 10

- 0

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

SammyS

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,377

- 1,038

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

- #5

- 10

- 0

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

Edit: trying to solve it again

Last edited:

- #6

- 10

- 0

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

- #7

SammyS

Staff Emeritus

Science Advisor

Homework Helper

Gold Member

- 11,377

- 1,038

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

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

In specific that means that [itex]\displaystyle y'(0)=y(0)\left(y(0)+k\right)\ .[/itex]

Plug the initial values into that & solve for k.

[itex]\displaystyle (1)=\left(-\frac{1}{2}\right)\left(\left(-\frac{1}{2}\right)+k\right)[/itex]

- #8

- 10

- 0

Oh!

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

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

Share: