Question #9651f

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Answer 1 · 2017-04-17T17:56:08

(a)

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

This is separable :

$e^y y' = e^x$

Integrate both sides wrt x:

$int e^y y' dx =int e^x dx$

$= int d/dx (e^y ) dx = e^x + C$

$implies e^y = e^x + C$

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

So:

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

Or:

$y = ln (e^x + e - 1)$

(b)

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

Again separable :

$e^(-x) x' = e^(-y)$

This time, integrate both wides wrt y:

$int e^(-x) x' dy=int e^(-y) dy$

$int d/dy ( - e^(-x)) dy=- e^(-y) + C$

$- e^(-x) = - e^(-y) + C$

$y(0) = 1 implies -1 = - 1/e + C implies C = 1/e -1$

$- e^(-x) = - e^(-y) + 1/e -1$

$e^(-y) = e^(-x) + 1/e -1$

$-y = ln ( e^(-x) + 1/e -1 )$

$y = ln ( 1/( e^(-x) + 1/e -1 ))$