What is a solution to the differential equation #y'-xy=0#?

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Answer 1 · 2016-07-10T09:02:01

#y = Ce^(x^2)#

this is separable

#y' = xy#

#1/y \ y'= x#

#int \ 1/y \ y' dx = int x \ dx#

#int \ 1/y \ dy = int x \ dx#

#lny = x^2+ C#

#y = e^(x^2+ C) = Ce^(x^2)#

NB using C as a generic pointer for constant term.