What is a solution to the differential equation #y'=-2xe^y#?

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Answer 1 · 2016-07-11T20:41:24

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

#y'=-2xe^y#

this is separable

# e^-y \ y' =-2x#

#int e^-y \ y' \ dx =-2 int x \ dx#

#int \ e^-y \ dy =-2 int x \ dx#

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

# e^-y \ = x^2 + C#

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

# ln (e^y) \ = ln(1/(x^2 + C))#

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