- #1

Benny

- 584

- 0

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 Benny
- Start date

- #1

Benny

- 584

- 0

- #2

marlon

- 3,777

- 11

Is x function of y ? is y function of x ? Or both x and y function of some other argument ?

Your are going to need to be a bit more specific here

marlon

- #3

heman

- 361

- 0

- #4

Benny

- 584

- 0

- #5

saltydog

Science Advisor

Homework Helper

- 1,591

- 3

Benny said:

Hey Benny, you got this? It's a particular form of the Riccati Equation right?

[tex]y^{'}+Q(x)y+R(x)y^2=P(x)[/tex]

so yours is:

[tex]y^{'}-y^2=x^2[/tex]

Thus make the substitution:

[tex]y=\frac{u^{'}}{-u}[/tex]

Turn the crank and get:

[tex]u^{''}-x^2u=0[/tex]

Then solve via power series, then take the derivative, form the quotient, then a plot. Put it all into Mathematica and back-substitute to make sure it's correct, then compare with numerical results. Or just do what you want.

- #6

Benny

- 584

- 0

Share:

- Last Post

- Replies
- 15

- Views
- 343

- Replies
- 1

- Views
- 367

- Replies
- 2

- Views
- 278

- Last Post

- Replies
- 4

- Views
- 380

- Replies
- 4

- Views
- 689

- Replies
- 14

- Views
- 459

- Replies
- 36

- Views
- 1K

- Last Post

- Replies
- 6

- Views
- 155

- Last Post

- Replies
- 5

- Views
- 382

- Replies
- 23

- Views
- 217