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

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

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

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

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

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

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

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

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

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