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# Differential equations question watch

1. Given that y=1 and x=0, express y in terms of x if dx/dy=e^x/e^y
The answer I get is y=-ln(e^-x+e^1,-1) however the answer in my book says y=x+1 -ln(e^-x+e^1.-1).
My question is where does the x+1 come from?
2. I get the same answer as you.
3. (Original post by DFranklin)
I get the same answer as you.
Surely it must be a mistake in the book?
4. (Original post by gooner1010)
Surely it must be a mistake in the book?
I reckon. (I'm also not 100% sure I do get the same answer as you - what exactly y=-ln(e^-x+e^1,-1) is supposed to mean (specifically e^1,-1) is unclear).

FWIW, I'll post my working:

. When x = 0, y = 1, so 1 + C = 1/e, so C = 1/e -1

take logs: , or

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Updated: February 12, 2017
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