How do you find the general solution to $dy/dx=xe^y$ ?

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Answer 1 · 2016-07-24T15:34:07

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

$y' = x e^y$

$e^(-y)y' = x$

$int \ e^(-y)y' \ dx =int \ x \ dx$

$int \ e^(-y)\ dy =int \ x \ dx$

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

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

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

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

How do you find the general solution to dy/dx=xe^ydy/dx=xe^y?