Solve $\dfrac{dy}{dx} = e^{x+y}$.
A$e^{-y} = -e^x + c$
B$e^y = e^x + c$
C$y = e^x + c$
D$\log y = x + c$
Answer & Solution
Correct answer: A. $e^{-y} = -e^x + c$
$dy/e^y = e^x\,dx$ ⇒ $\int e^{-y}\,dy = \int e^x\,dx$ ⇒ $-e^{-y} = e^x + c'$ ⇒ $e^{-y} = -e^x + c$.
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