In solving y' - 2y = e^{3x} by integrating factor, which integrating factor μ(x) is used?

In solving y' - 2y = e^{3x} by integrating factor, which integrating factor μ(x) is used?

• A. e^{2x}
• B. e^{x}
• C. e^{-2x}
• D. e^{-x}

Answer: (C)

When solving a linear first-order ODE of the form y' + P(x) y = Q(x), use the integrating factor μ(x) = e^{∫ P(x) dx}. Here the equation y' - 2y = e^{3x} can be written as y' + (-2) y = e^{3x}, so P(x) = -2. The integrating factor is μ(x) = e^{-2x}. Multiply both sides by μ(x): e^{-2x} y' - 2 e^{-2x} y = e^{3x} e^{-2x} = e^{x}. The left-hand side becomes the derivative of μ(x) y, since (e^{-2x} y)' = e^{-2x} y' + (-2) e^{-2x} y. Thus (e^{-2x} y)' = e^{x}. Integrate: e^{-2x} y = ∫ e^{x} dx = e^{x} + C. Solve for y: y = e^{2x} (e^{x} + C) = e^{3x} + C e^{2x}. So the integrating factor is e^{-2x}, which is the correct choice.

When solving a linear first-order ODE of the form y' + P(x) y = Q(x), use the integrating factor μ(x) = e^{∫ P(x) dx}. Here the equation y' - 2y = e^{3x} can be written as y' + (-2) y = e^{3x}, so P(x) = -2. The integrating factor is μ(x) = e^{-2x}.

Multiply both sides by μ(x): e^{-2x} y' - 2 e^{-2x} y = e^{3x} e^{-2x} = e^{x}. The left-hand side becomes the derivative of μ(x) y, since (e^{-2x} y)' = e^{-2x} y' + (-2) e^{-2x} y.

Thus (e^{-2x} y)' = e^{x}. Integrate: e^{-2x} y = ∫ e^{x} dx = e^{x} + C. Solve for y: y = e^{2x} (e^{x} + C) = e^{3x} + C e^{2x}.

So the integrating factor is e^{-2x}, which is the correct choice.