Solve the differential equation y' - 2y = e^{3x} by integrating factors. Which is the general solution?

Solve the differential equation y' - 2y = e^{3x} by integrating factors. Which is the general solution?

• A. y = e^{3x} + C e^{x}
• B. y = e^{3x} + C e^{3x}
• C. y = e^{3x} + C e^{2x}
• D. y = e^{2x} + C e^{3x}

Answer: (C)

The method uses an integrating factor to turn the left side into the derivative of a product. Start with y' - 2y = e^{3x} in standard form y' + p y = q with p = -2. The integrating factor is μ = e^{∫ p dx} = e^{-2x}. Multiply through: (e^{-2x} y)' = e^{x}. Integrate both sides: e^{-2x} y = e^{x} + C. Solve for y: y = e^{2x}(e^{x} + C) = e^{3x} + C e^{2x}. Therefore, the general solution is y = e^{3x} + C e^{2x}.

The method uses an integrating factor to turn the left side into the derivative of a product. Start with y' - 2y = e^{3x} in standard form y' + p y = q with p = -2. The integrating factor is μ = e^{∫ p dx} = e^{-2x}. Multiply through: (e^{-2x} y)' = e^{x}. Integrate both sides: e^{-2x} y = e^{x} + C. Solve for y: y = e^{2x}(e^{x} + C) = e^{3x} + C e^{2x}. Therefore, the general solution is y = e^{3x} + C e^{2x}.