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The general solution of $\dfrac{dy}{dx} + y = e^{-x}$ is:

A$y \cdot e^x = x + c$
B$y = e^{-x}(x + c)$
CBoth A and B are equivalent
D$y = x e^x + c$
Answer & Solution
Correct answer: C. Both A and B are equivalent
Linear DE with $P = 1, Q = e^{-x}$. IF = $e^x$. $y \cdot e^x = \int e^{-x}\cdot e^x dx + c = x + c$. So $y \cdot e^x = x + c$, equivalent to $y = e^{-x}(x + c)$.
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