What is a solution to the differential equation #dy/dx=xe^y#?

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Answer 1 · 2016-08-29T20:36:36

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

#y' = xe^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 #

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

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

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