How do you find the general solution to #dy/dx+e^(x+y)=0#?

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Answer 1 · 2016-07-24T17:34:53

#y = ln ( 1/(e^x + C) )#

#dy/dx+e^(x+y)=0#

this is separable

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

#dy/dx= - e^x e^y#

#e^(-y) dy/dx= - e^x#

#int\ e^(-y) dy/dx \ dx=int - e^x \ dx#

#int\ e^(-y) \ dy=- int e^x \ dx#

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

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

#e^(y) = 1/(e^x + C)#

#y = ln ( 1/(e^x + C) )#