Solve the first-order linear ODE y' + y = e^{2x} with y(0) = 1. Which is the correct solution?

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

Answer: (A)

This first-order linear ODE uses an integrating factor to turn the left-hand side into a derivative of a product. For y' + y = e^{2x}, the integrating factor is e^{x}. Multiplying through gives (e^{x} y)' = e^{3x}. Integrating, e^{x} y = (1/3) e^{3x} + C, so y = (1/3) e^{2x} + C e^{-x}.

Apply the initial condition y(0) = 1: 1 = (1/3) + C, so C = 2/3. Therefore the solution is y = (1/3) e^{2x} + (2/3) e^{-x}.

This matches the particular solution for the nonhomogeneous term e^{2x} and the homogeneous solution C e^{-x}, with the constant fixed by the initial value.