Every solution of $\frac{dy}{dx} + 4y = 8$ approaches the same value as $x \to \infty$. What is that value?

Explanation

1. 01

   The factor $e^{4x}$ gives $ye^{4x} = 2e^{4x} + C$, so $y = 2 + Ce^{-4x}$.

2. 02

   As $x \to \infty$, $Ce^{-4x} \to 0$, so $y \to \frac{8}{4} = 2$.