Question

Question #332a8

Answer 1

I got: `y=-ln(e^-x+1/e)`

Explanation:

We can separate and write:
`dx/e^x=dy/e^y`
or:
`e^-xdx=e^-ydy`
integrate:
`inte^-xdx=inte^-ydy`
`-e^-x+c=-e^-y`
`e^-y=e^-x+c`
and taking the natural log of both sides:
`ln(e^-y)=ln(e^-x+c)`
`-y=ln(e^-x+c)`
`y=-ln(e^-x+c)`
when
`x=0` then `y=1` so:
`1=-ln(c)`
`c=1/e`
and finally:
`y=-ln(e^-x+1/e)`