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

Question

Accepted 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 $y = e^{-x}(x + c)$.**

Answer

$y \cdot e^x = x + c$

Answer

$y = e^{-x}(x + c)$

Answer

$y = x e^x + c$

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

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