- #1

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## 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

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- Thread starter Fredh
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- #1

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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

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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

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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

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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

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

Edit: trying to solve it again

Last edited:

- #6

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

- #7

SammyS

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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

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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!

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