How to solve : #dy/dx +1 = e^(x+y)# ?

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#dy/dx +1 = e^(x+y)#

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Answer 1 · 2018-01-22T09:55:59

#y = -(ln(C_1-x)+x)#

#d/(dx)(y+x)=e^(y+x)#

Making #z = x+y->z'=1+y'# then

#y'+1=e^(x+y) ->z'=e^z# or

#dz/e^z = dx# or #e^-z dz = dx# and integrating both sides

#-e^-z = x + C_0# or

#y = -(ln(C_1-x)+x)#