Answer to Question #229243 in Chemical Engineering for Lokika

Question #229243

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


1
Expert's answer
2021-09-06T00:54:01-0400

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}"

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