Question

(x ) dx + ye^y dy = 0 ,donde y(0) = 1 (ED Separables)

Answer 1

La función es:

${e}^{y} (y - 1) = \frac{ {x}^{2} }{2}$

con valor inicial y(0) = 1.

Explicación paso a paso:

$xdx = - y {e}^{y} dy \\ integrando \\ \frac{ {x}^{2} }{2} + c = - (y {e}^{y} - {e}^{y} ) \\ {e}^{y} (y - 1) = - \frac{ {x}^{2} }{2} + c \\ y(0) = 1 \: ... \: {e}^{1} (1 - 1) = - \frac{0}{2} + c \: ... \: c = 0 \\ {e}^{y} (y - 1) = - \frac{ {x}^{2} }{2}$