# Question #9651f

Apr 17, 2017

(a)

$y ' = {e}^{x} / {e}^{y}$

This is separable :

${e}^{y} y ' = {e}^{x}$

Integrate both sides wrt x:

$\int {e}^{y} y ' \mathrm{dx} = \int {e}^{x} \mathrm{dx}$

$= \int \frac{d}{\mathrm{dx}} \left({e}^{y}\right) \mathrm{dx} = {e}^{x} + C$

$\implies {e}^{y} = {e}^{x} + C$

$y \left(0\right) = 1 \implies e = 1 + C \implies C = e - 1$

So:

${e}^{y} = {e}^{x} + e - 1$

Or:

$y = \ln \left({e}^{x} + e - 1\right)$

(b)

$x ' = {e}^{x} / {e}^{y}$

Again separable :

${e}^{- x} x ' = {e}^{- y}$

This time, integrate both wides wrt y:

$\int {e}^{- x} x ' \mathrm{dy} = \int {e}^{- y} \mathrm{dy}$

$\int \frac{d}{\mathrm{dy}} \left(- {e}^{- x}\right) \mathrm{dy} = - {e}^{- y} + C$

$- {e}^{- x} = - {e}^{- y} + C$

$y \left(0\right) = 1 \implies - 1 = - \frac{1}{e} + C \implies C = \frac{1}{e} - 1$

$- {e}^{- x} = - {e}^{- y} + \frac{1}{e} - 1$

${e}^{- y} = {e}^{- x} + \frac{1}{e} - 1$

$- y = \ln \left({e}^{- x} + \frac{1}{e} - 1\right)$

$y = \ln \left(\frac{1}{{e}^{- x} + \frac{1}{e} - 1}\right)$