- #1

thenewbosco

- 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?

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- Thread starter thenewbosco
- Start date

- #1

thenewbosco

- 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

- 14,967

- 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

Dr Avalanchez

- 18

- 0

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

- #4

thenewbosco

- 187

- 0

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

to

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

help please

- #5

asdf60

- 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

- 14,967

- 19

Have you tried any of our hints?

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