Find the general solution of dy/dx+3y=e^3x.

Use the integrating factor method.

μ dy/dx + 3μy = μ e^{3x}

Let μ satisfy dμ/dx = 3μ.

Substitution yields μ dy/dx + y dμ/dx = μe^{3x}.

The left side is a product rule for (d/dx) μy so you can integrate both sides.

μy = ∫ μe^{3x} dx

y = (1/μ) ∫ μe^{3x} dx

Now find μ from it's differential equation which is separable.

dμ/dx = 3μ

dμ/μ = 3 dx.

∫ dμ/μ = ∫ 3 dx

ln |μ| = 3x + C

μ = e^{3x+C} = e^{c}e^{3x} = Ae^{3x}

Plug in any specific μ, say μ = e^{3x} into the original solution of y.

y = e^{-3x} ∫ e^{3x}e^{3x} dx = e^{-3x} ∫ e^{6}^{x}dx = e^{-3x} (e^{6}^{x}/6 + C) =
**e ^{3x}/6 + Ce^{-3x}**

Check:

(d/dx) (**e ^{3x}/6 + Ce^{-3x}**) + 3(

**e**) = e

^{3x}/6 + Ce^{-3x}^{3x}/2 - 3Ce

^{-3x}+ e

^{3x}/2 + 3Ce

^{-}

^{3x}= e

^{3x}.

## Comments

Do you have another way to do this?

Never mind, I got it.