Problem: Solve the initial value problem
\[
\frac{dy}{dx} + 2y = e^{-x}, \quad y(0)=1.
\]
Solution: We start with the differential equation
\[
\frac{dy}{dx} + 2y = e^{-x}.
\]
This is a first-order linear equation, so we first compute the integrating factor:
\[
\mu(x) = e^{\int 2\,dx} = e^{2x}.
\]
Multiplying through by \(\mu(x)\) gives
\[
e^{2x}\frac{dy}{dx} + 2e^{2x}y = e^{2x}e^{-x} = e^{x}.
\]
Notice that the left-hand side is the derivative of \(e^{2x}y\):
\[
\frac{d}{dx}\Bigl(e^{2x}y\Bigr) = e^{x}.
\]
Integrate both sides with respect to \(x\):
\[
e^{2x}y = \int e^{x}\,dx = e^{x} + C.
\]
Thus,
\[
y(x) = e^{-2x}\Bigl(e^{x} + C\Bigr) = e^{-x} + C e^{-2x}.
\]
Applying the initial condition \(y(0)=1\) we obtain:
\[
1 = e^0 + C e^0 = 1 + C,
\]
so that \(C = 0\). Hence, the solution is:
\[
\textcolor{blue}{\boxed{y(x) = e^{-x}.}}
\]