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 ))#