How do you find the general solution to #dy/dx=xe^y#?

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Answer 1 · 2016-07-24T15:34:07

#y = ln (2/ (C - x^2))#

#y' = x e^y#

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

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

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

#- e^(-y) =x^2/2 + C#

#e^(-y) =C - x^2/2#

#e^(y) = 1/ (C - x^2/2)#
#= 2/ (C - x^2)#

#y = ln (2/ (C - x^2))#