Answer to Question #229243 in Chemical Engineering for Lokika

Question

The solution of ye^{x}dx + (2y + e^{x})dy=0, y(0) = -1 is, choose the correct answer A) y = e^{x}. B) y=e^{-x} C) y = 1 D) y = -1

Answer 1

Given DE:

$y e^xdx+(2y+e^x)dy=0$

This equation is equivalent to the next

$d(y^2+ye^x)=0$

So,

$y^2+ye^x=C$

Initial condition gives

$(-1)^2+(-1)e^0=C$

$C=0$

Hence, we obtain

$y^2+ye^x=0$

or

$y=0,\;{\rm and}\; y=-e^{x}$