Solving autonomous equation

  • Thread starter Fredh
  • Start date
  • #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
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,847
965

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"?
 
  • #3
10
0
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
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,377
1,038
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 .
 
  • #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
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 [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!
 

Related Threads on Solving autonomous equation

  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
3
Views
1K
Replies
1
Views
987
Replies
2
Views
4K
Replies
0
Views
3K
Replies
3
Views
1K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
915
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
4
Views
5K
Top