- #1

- 187

- 0

"Thus e^y + e^-y =2x or

e^2y - 2xe^y + 1 = 0"

how did they go from the first to the second part?

- Thread starter thenewbosco
- Start date

- #1

- 187

- 0

"Thus e^y + e^-y =2x or

e^2y - 2xe^y + 1 = 0"

how did they go from the first to the second part?

- #2

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,916

- 19

For example, their second equation has an e^(2y) in it

1: I assume you meant e^(2y) and not e^2y (which is the same as (e^2)y)

- #3

- 18

- 0

You should really try to figure this out yourself. What's the difference between the two equations?

- #4

- 187

- 0

[tex]e^y + exp(-y)=2x[/tex]

to

[tex]e^2y - 2xe^y + 1 = 0[/tex]

help please

- #5

- 81

- 0

I'm not sure, but that y should be raised too....e^(2y) not (e^2)y

- #6

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,916

- 19

Have you tried any of our hints?

- Replies
- 6

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 1K

- Replies
- 6

- Views
- 2K

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 11

- Views
- 18K

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 17

- Views
- 4K

- Replies
- 1

- Views
- 10K

- Replies
- 6

- Views
- 1K