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.

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