Question

#JEE




solve using Variable separable


$\huge {\red{ \tt{ \frac {dy}{dx} ={ e }^{x - y} + {x}^{2} . {e}^{ - y} }}}$
​

Answer 1

Step-by-step explanation:

ex +2x.ex is this one correct

Answer 2

Step-by-step explanation:

We have,

$\rm\frac {dy}{dx} ={ e }^{x - y} + {x}^{2} . {e}^{ - y}$

$\rm \implies\frac {dy}{dx} ={ e }^{x}. {e}^{ - y} + {x}^{2} . {e}^{ - y}$

$\rm \implies\frac {dy}{dx} = ({x}^{2} + {e}^{x} ). {e}^{ - y}$

$\rm \implies\frac {dy}{ {e}^{ - y}} = ({x}^{2} + {e}^{x} )dx$

$\rm \implies{e}^{ y}dy= ({x}^{2} + {e}^{x} )dx$

$\rm \implies \int{e}^{ y}dy= \int({x}^{2} + {e}^{x} )dx$

$\rm \implies {e}^{ y}= \frac{{x}^{3} }{3}+ {e}^{x} + c$