Solve y' + 2y = e^{−x}, with y(0)=0. Which is y(x)?

• A. y = (1/2) e^{2x} − (1/2) e^{−x}
• B. y = (1/3) e^{2x} − (1/3) e^{−x}
• C. y = −(1/3) e^{2x} + (1/3) e^{−x}
• D. y = (1/3) e^{2x} + (1/3) e^{−x}

Answer: (B)

This is a linear first‑order differential equation. Use an integrating factor to combine the derivative and the y-term.

Take the integrating factor μ(x) = e^{2x}. Multiply through: (e^{2x} y)' = e^{2x} e^{-x} = e^{x}. Integrate both sides to get e^{2x} y = e^{x} + C, so y = e^{-x} + C e^{-2x}.

Apply y(0) = 0: y(0) = 1 + C = 0, hence C = -1. The solution is y(x) = e^{-x} - e^{-2x}, which can also be written as e^{-x}(1 - e^{-x}).

Note that this form involves a term with e^{-2x}; the given options are linear combinations of e^{2x} and e^{-x}, so they do not match the correct solution.